From 56abf2fe8cc5704178c4be85823214346915ab40 Mon Sep 17 00:00:00 2001 From: Markus Maximilian Rudolf Tripp Date: Wed, 12 Aug 2026 16:02:02 +0200 Subject: [PATCH 1/8] first commit --- .gitignore | 259 +++++++++++++++++++++++++++++++++++++ ncr-of-monomial-curves.tex | 249 +++++++++++++++++++++++++++++++++++ references.bib | 59 +++++++++ 3 files changed, 567 insertions(+) create mode 100644 .gitignore create mode 100644 ncr-of-monomial-curves.tex create mode 100644 references.bib diff --git a/.gitignore b/.gitignore new file mode 100644 index 0000000..8f5414b --- /dev/null +++ b/.gitignore @@ -0,0 +1,259 @@ +## Core LaTeX auxiliary files +*.aux +*.lof +*.log +*.lot +*.fls +*.out +*.toc +*.fmt +*.fot +*.cb +*.cb2 +.*.lb + +## Bibliography +*.bbl +*.bcf +*.blg +*-blx.aux +*-blx.bib +*.run.xml + +## Build tool directories/files +*.fdb_latexmk +*.synctex +*.synctex(busy) +*.synctex.gz +*.synctex.gz(busy) +*.pdfsync + +## Build tool auxiliary files +*.alg +*.loa + +## Auxiliary and intermediate files from other packages +# algorithms +*.alg +*.loa + +# achemso +acs-*.bib + +# amsthm +*.thm + +# beamer +*.nav +*.pre +*.snm +*.vrb + +# changes +*.soc + +# cprotect +*.cpt + +# elsarticle (documentclass of Elsevier journals) +*.spl + +# endnotes +*.ent + +# fixme +*.lox + +# feynmf/feynmp +*.mf +*.mp +*.t[1-9] +*.t[1-9][0-9] +*.tfm + +#(r)(e)ledmac/(r)(e)ledpar +*.end +*.?end +*.[1-9] +*.[1-9][0-9] +*.[1-9][0-9][0-9] +*.[1-9]R +*.[1-9][0-9]R +*.[1-9][0-9][0-9]R +*.eledsec[1-9] +*.eledsec[1-9]R +*.eledsec[1-9][0-9] +*.eledsec[1-9][0-9]R +*.eledsec[1-9][0-9][0-9] +*.eledsec[1-9][0-9][0-9]R + +# glossaries +*.acn +*.acr +*.glg +*.glo +*.gls +*.glsdefs +*.lzo +*.lzs +*.slg +*.slo +*.sls + +# uncomment this for glossaries-extra (will ignore makeindex's style files!) +# *.ist + +# gnuplot +*.gnuplot +*.table + +# gnuplottex +*-gnuplottex-* + +# gregoriotex +*.gaux +*.gtex + +# htlatex +*.4ct +*.4tc +*.idv +*.lg +*.trc +*.xref + +# hyperref +*.brf + +# knitr +*-concordance.tex +*.tikz +*-tikzDictionary + +# listings +*.lol + +# luatexja-ruby +*.ltjruby + +# makeidx +*.idx +*.ilg +*.ind + +# minitoc +*.maf +*.mlf +*.mlt +*.mtc[0-9]* +*.slf[0-9]* +*.slt[0-9]* +*.stc[0-9]* + +# minted +_minted* +*.pyg + +# morewrites +*.mw + +# newpax +*.newpax + +# nomencl +*.nlg +*.nlo +*.nls + +# pax +*.pax + +# pdfpcnotes +*.pdfpc + +# sagetex +*.sagetex.sage +*.sagetex.py +*.sagetex.scmd + +# scrwfile +*.wrt + +# svg +svg-inkscape/ + +# sympy +*.sout +*.sympy +sympy-plots-for-*.tex/ + +# pdfcomment +*.upa +*.upb + +# pythontex +*.pytxcode +pythontex-files-*/ + +# tcolorbox +*.listing + +# thmtools +*.loe + +# TikZ & PGF +*.dpth +*.md5 +*.auxlock + +# titletoc +*.ptc + +# todonotes +*.tdo + +# vhistory +*.hst +*.ver + +# easy-todo +*.lod + +# xcolor +*.xcp + +# xmpincl +*.xmpi + +# xindy +*.xdy + +# xypic precompiled matrices and outlines +*.xyc +*.xyd + +# endfloat +*.ttt +*.fff + +# Latexian +TSWLatexianTemp* + +## Editors / OS junk +*~ +*.swp +*.bak +.DS_Store +Thumbs.db + +## Generated PDF/output — comment these two lines out if you WANT to track the compiled PDF +*.pdf +!final/*.pdf + +## latexmk +*.fdb_latexmk +*.fls + +## Other common temp files +*.tmp +*.log~ diff --git a/ncr-of-monomial-curves.tex b/ncr-of-monomial-curves.tex new file mode 100644 index 0000000..f60c7c2 --- /dev/null +++ b/ncr-of-monomial-curves.tex @@ -0,0 +1,249 @@ +\documentclass[10pt]{amsart} + +\usepackage[T1]{fontenc} +\usepackage{mathtools} +\usepackage{todonotes} +\usepackage{blindtext} +%\usepackage{parskip} + +\usepackage[ + backend=biber, + style=alphabetic, % common in math; change if journal requires + sorting=nyt, + giveninits=true, + eprint=false, url=false, doi=false, isbn=false, + maxbibnames=9 +]{biblatex} +\addbibresource{references.bib} +\AtEveryBibitem{% + \clearfield{month}% + \clearfield{day}% + \clearlist{language}% +} + +\newtheorem{theorem}{Theorem}[section] +\newtheorem{lemma}[theorem]{Lemma} +\newtheorem{proposition}[theorem]{Proposition} +\newtheorem{corollary}[theorem]{Corollary} +\theoremstyle{definition} +\newtheorem{definition}[theorem]{Definition} +\newtheoremstyle{remstyle} + {\medskipamount} % space above (half of standard topsep) + {\medskipamount} % space below (half of standard topsep) + {} % body font (normal/upright) + {} % indent + {\itshape} % head font (italic) + {.} % punctuation after label + {.5em} % space after label + {} % head spec +\theoremstyle{remstyle} +\newtheorem{remark}[theorem]{Remark} +\DeclareMathOperator{\supp}{supp} + +\DeclareMathOperator{\Frac}{Frac} +\DeclareMathOperator{\spn}{span} +\DeclareMathOperator{\Min}{Min} +\DeclareMathOperator{\hgt}{ht} +\DeclareMathOperator{\depth}{depth} +\DeclareMathOperator{\gldim}{gl.dim} +\DeclareMathOperator{\End}{End} +\DeclareMathOperator{\id}{id} +\newcommand{\qpower}[2][q]{{}^{#1}#2} +\newcommand{\gapset}{\NN_0\setminus\SGR} +\newcommand{\kk}{\mathsf{k}} +\newcommand{\SGR}{\mathsf{S}} +\newcommand{\NN}{\mathbb{N}} +\newcommand{\ZZ}{\mathbb{Z}} +\newcommand{\QQ}{\mathbb{Q}} +\newcommand{\A}{\mathbb{A}} + +\title{Noncommutative Resolutions of Monomial Curves} +\author{Markus M. R. Tripp} +\date{} + +\begin{document} +\maketitle + +\begin{abstract} +\end{abstract} + +\section{Monomial curves} + +\section{Frobenius maps on affine semigroups} +Let $\SGR$ be an affine semigroup, i.e., a finitely generated submonoid of $\ZZ^d$ for some $d \in \NN$, and let $R \coloneqq \kk[\SGR]$ denote the associated semigroup algebra. Fix a positive integer $q$, and write $q^{-1}\SGR \coloneqq \{ a/q \mid a \in \SGR \} \subseteq \QQ^d$ for the affine semigroup $\SGR$ refined by $q$. Consider the semigroup algebra +\[ R^{1/q} \coloneqq \kk[q^{-1}\SGR] = \spn_\kk\{ x^{a/q} \mid a \in \SGR \} \] +regarded as a subring of $\kk[\QQ^d] = \spn_\kk \{ x^\alpha \mid \alpha \in \QQ^d \}$, which carries a natural $\QQ^d$-grading. While the semigroup ring $R^{1/q}$ is defined for arbitrary $q \in \NN$ in any characteristic, the notation is motivated by the case $q = p^e$ with $\kk$ perfect of characteristic $p$, where $R^{1/q}$ is the ring of $q$-th roots of elements in $R = \kk[\SGR]$. Further, for $r = \sum_{a \in \QQ^d} k_a x^{a}$, we set $\qpower{r} \coloneqq \sum_{a \in \QQ^d} k_a x^{aq}$, and write $\qpower{F}$ for the resulting \emph{$q$-th pseudo-Frobenius endomorphism} on $R$, which maps $r$ to $\qpower{r}$. + +\begin{remark} +% PREVIOUS VERSION % Note that even in positive characteristic $p$ with $q = p^e$, the $q$-th pseudo-Frobenius twist need not agree with the usual $q$-th Frobenius twist, where $R$-scalar multiplication is pulled back via the $e$-times iterated Frobenius endomorphism $F^e \colon r \mapsto r^{p^e}$ on $R$. The two agree precisely when the underlying field $\kk$ is finite, in which case $\kk \subseteq R$ is fixed pointwise by $F^e$. +Note that even in positive characteristic $p$ with $q = p^e$, the $q$-th pseudo-Frobenius endomorphism $\qpower{F}$ need not agree with the usual $e$-times iterated Frobenius endomorphism $F^e \colon R \to R$, mapping $r$ to $r^{q}$ . The two agree precisely when the underlying field $\kk$ is finite, in which case $\kk \subseteq R$ is fixed pointwise by $F^e$. +\end{remark} + +\begin{definition} +Let $M$ be an $R$-module. The \textbf{$q$-th pseudo-Frobenius twist} of $M$, denoted $\qpower{M}$, is the $R$-module obtained by restricting scalars along the \mbox{$q$-th} pseudo-Frobenius endomorphism $\qpower{F}$. That is, $\qpower{M}$ is the $R$-module with the same underlying abelian group as $M$, but with $R$-action given by~$r\cdot m \coloneqq \qpower{r}m$ +% EXPLICIT DECLARATION % \[ \Bigl( \sum_{a \in S} k_a x^a \Bigr) \cdot_q \; m \coloneqq \bigl( \sum_{a \in S} k_a x^{aq} \bigr) m \] +for all $r \in R$ and $m \in M$. +\end{definition} + +\begin{lemma} +There is a cannonical $R$-module isomorphism +\[ R^{1/q} \xrightarrow{\;\sim\;} \qpower{R}, \] +defined by $r = \sum_{a \in \SGR} k_a x^{a/q} \mapsto \qpower{r} = \sum_{a \in \SGR} k_a x^a$. +\end{lemma} + +\begin{proof} +By abuse of notation, we also write $\qpower{F}$ for the map $R^{1/q} \to \qpower{R}$ above; it is clearly additive and bijective. To verify compatibility with the respective $R$-module structures, write $r = \sum_{a \in \SGR} k_a x^a \in R$ and $f = \sum_{a \in \SGR} \ell_a x^{a/q} \in R^{1/q}$. Then +\[ \qpower{F}(rf) = \qpower{F}\Bigl(\sum_{a, b \in \SGR} k_a\ell_b x^{a + b/q}\Bigr) = \sum_{a, b \in \SGR} k_a\ell_b x^{aq + b}, \] +which is precisely $r\cdot\qpower{F}(f)$ in $\qpower{R}$. +\end{proof} + +\begin{definition} +We say that $R$ is \textbf{$q$-split} if the Frobenius map $\qpower{F} \colon R \to \qpower{R}$ splits as a map of $R$-modules. Moreover, we call $R$ \textbf{$q$-weakly normal} if, for every $r \in \kk[G(\SGR)]$\todo{$\Frac(R)$?}, $\qpower{r} \in R$ implies $r \in R$. +\end{definition} + +\begin{remark} +In semigroup-thereotic terms, $R = \kk[\SGR]$ is $q$-weakly normal if and only if, for every $a \in G(\SGR)$, $qa \in S$ implies $a \in S$. In particular, $\SGR$ (or equivalently, $R$) is normal if and only if $R$ is $q$-weakly normal for all $q \in \NN$. +\end{remark} + +In positive characteristic $p$, the Frobenius endomorphism yields $\qpower[p]{r} \in R$ if and only if $r^p \in R$, from which we obtain the following: + +\begin{corollary} +If $\kk$ has characteristic $p >0$, then $R$ is $p$-weakly normal +% MORE INFORMATION % (or, equivalently, $p^e$-weakly normal for all $e \in \NN$) +if and only if $R$ is weakly normal (i.e., for any $r \in \Frac(R)$, $r^p$ implies $r \in R$). In particular, $R$ is seminormal in this case. +\end{corollary} + +\begin{proposition}\textnormal{\cite{Hochster1976}, \cite[Prop. 5.3]{Yasuda2009}} +$R$ is $q$-split if and only if it is $q$-weakly normal. +\end{proposition} + +\begin{proof} +% FAKE PROOF % Suppose $R$ is $q$-split, with $\sigma \colon \qpower{R} \to R$ an $R$-module homomorphism satisfying $\sigma \circ \qpower{F} = \id_R$; in particular, $\sigma(1) = 1$. Now let $r \in \Frac(R)$ be such that $\qpower{r} \in R$. Then $r = r\sigma(1) = \sigma(\qpower{r}) \in R$. + +Suppose $R$ is $q$-split, with $\sigma \colon \qpower{R} \to R$ an $R$-module homomorphism satisfying $\sigma \circ \qpower{F} = \id_R$. Let $r \in \kk[G(\SGR)]$ be such that $\qpower{r} \in R$, and choose $s \in R^\bullet$ with $sr \in R$. Then +\[ sr = \sigma\bigl( \qpower{F}(sr) \bigr) = \sigma(\qpower{s}\qpower{r}) = s\sigma(\qpower{r}), \] +and cancelling $s$ yields $r = \sigma(\qpower{r}) \in R$. + +Conversely, suppose that $R$ is $q$-weakly normal. Let $\sigma \colon \qpower{R} \to R$ be the map that sends $f \in \qpower{R}$ to the unique $g \in R$ with $f = \qpower{g}$, if such a $g$ exists, and to $0$ otherwise. Clearly $\sigma \circ \qpower{F} = \id_R$, so it remains to show that $\sigma$ is a homomorphism of $R$-modules. As $\sigma$ is additive, it suffices to check that $\sigma(x^{aq+b}) = x^a \sigma(x^b)$ for $a$, $b \in \SGR$, which is straightforward using the assumption that $R$ is $q$-weakly normal. % MORE EXPLANATION % Suppose first that $b = qc$ for some $c \in \SGR$. Then $\sigma(x^{aq+b}) = \sigma(x^{q(a+c)}) = x^{a+c} = x^a \sigma(x^b)$. Now suppose there exists no $c \in \SGR$ with $b = qc$, so $\sigma(x^b) = 0$. If $aq+b = qc$ for some $c \in \SGR$, then $\qpower{x^{c-a}} = x^b \in R$, so $x^{c-a} \in R$ by assumption, giving $c-a \in \SGR$ and $b = q(c-a)$, a contradiction. Hence $\sigma(x^{aq+b}) = 0 = x^a\sigma(x^b)$. +\end{proof} + +\begin{corollary}\textnormal{\cite[Cor. 6.3]{Bruns2006}} +$R$ is normal if and only if $R$ is $q$-split for all $q \in \NN$. +\end{corollary} + +\section{$R$-module decomposition of $R^{1/q}$} + +We now specialize to the case where $\SGR = \langle a_1, \dots, a_n \rangle \subseteq \NN_0$ is a numerical semigroup. Following Shibuta~\cite{Shibuta2017}, we sketch how $R^{1/q} = \kk[x^{a_1/q}, \dots, x^{a_n/q}]$ decomposes as an $R$-module. Observe that +\[ +\Gamma_q \coloneqq \left\{ \bigl( x^{a_1/q} \bigr)^{e_1} \cdots \bigl( x^{a_n/q} \bigr)^{e_n} \mid e_i \in [0,q-1] \right\} +\] +is a homogeneous generating set for $R^{1/q}$. We partition $\Gamma_q$ by degree modulo $\ZZ$: writing each degree as +$a/q$ with $a \in \NN_0$, this amounts to grouping elements by the residue of $a$ modulo $q$. This yields +\[ +\Gamma_q = \Gamma_{q,0} \sqcup \Gamma_{q,1} \sqcup \cdots \sqcup \Gamma_{q,q-1}, +\] +where $\Gamma_{q,r} \coloneqq \{ x^{a/q} \in \Gamma_q \mid a \equiv r \pmod{q}\}$. Let $I_{q,r}$ be the $R$-submodule of $\kk[x^q \mid q \in \QQ_{\geq 0}]$ +generated by $\Gamma_{q,r}$. As a $\kk$-vector space, +\begin{equation}\label{eq:ind-summand-as-vector-space} +I_{q,r} = \spn_\kk \{ x^{a/q} \mid a \in \SGR, \, a \equiv r \pmod{q} \}, +\end{equation} +and, by construction, summing over $r$ gives the $R$-module decomposition +\[ +R^{1/q} = \bigoplus_{r \in [0,q-1]} I_{q,r}, +\] +where each summand is indecomposable (see Lemma~\ref{lemma:rank-1}). + +% REDUNDANT % \begin{lemma} +%Each $R$-module $I_{q,r}$ has rank $1$ and is therefore indecomposable. +%\end{lemma} + +%\begin{proof} +%For simplicity, let $Q \coloneqq \Frac(R)$. The localization $V_{q,r} \coloneqq (R\setminus \{ 0 \})^{-1} I_{q,r}$ is isomorphic to $Q \otimes_R I_{q,r}$ as a $Q$-vector space. For a fixed element $x^{a/q} \in I_{q,r} \subseteq V_{q,r}$, we claim that $V_{q,r}$ is $1$-dimensional over $Q$, generated by $x^{a/q}$. + +%To see this, take any $x^{b/q} \in I_{q,r}$. Since $b \equiv a \pmod{q}$, we have $b-a = cq$ for some $c \in \ZZ$. Thus, $x^{b/q} = x^c x^{a/q} \in \spn_Q\{ x^{a/q} \}$. The claim follows because any element of $V_{q,r}$ is a $Q$-linear combination of homogeneous elements of $I_{q,r}$. +%\todo{note $\Frac{R} = \kk(x)$} +%\end{proof} + +Moreover, writing $m_{q,r}/q$ for the minimal degree occurring among +elements of $\Gamma_{q,r}$, we have +\[ +I_{q,r} = x^{m_{q,r}/q} \cdot \langle x^\alpha \mid \alpha \in \NN_0, \, m_{q,r}/q + \alpha \in \Gamma_{q,r} \rangle_R, +\] +from which it follows that every indecomposable summand of $R^{1/q}$ is isomorphic to $I(A) \coloneqq \langle x^\alpha \mid \alpha \in A \cup \{ 0 \} \rangle_R$ for some $A \subseteq \gapset$\todo{name?}. We may always choose $A$ to be minimal with respect to the order $\preceq_\SGR$ induced by $\SGR$, where $\alpha \preceq_\SGR \beta$ means $\beta - \alpha \in \SGR$. Write +\[ \Min(\SGR) \coloneqq \{ A \in \gapset \mid A \text{ is minimal with respect to } \preceq_\SGR \}. \] +In particular, as $q$ ranges over $\NN$, only finitely many indecomposable summands of $R^{1/q}$ occur, up to isomorphism. + +\begin{lemma}\label{lemma:rank-1} +For each $A \in \Min(\SGR)$, the $R$-module $I(A)$ has rank $1$ and is therefore indecomposable. Moreover, the assignment +\[ A \mapsto [I(A)] \] +defines a bijection between $\Min(\SGR)$ and the set of isomorphism classes of homogeneous $R$-submodules of $\bar{R} = \kk[x]$\todo{this also holds for $\kk[x,x^{-1}]$, interesting?}. +\end{lemma} + +\begin{proof} +For simplicity, set $Q \coloneqq \Frac(R) = \kk(x)$\todo{note $\Frac{R} = \kk(x)$} and fix $A \in \Min(\SGR)$. Observe that $Q \otimes_R I(A)$ is naturally isomorphic, as a $Q$-vector space, to $I(A)_{(0)} \subseteq Q$. Since $I(A)$ contains $1$, it follows that $I(A)_{(0)} = Q$ and $I(A)$ has rank $1$. + +For the second part, it is clear that $I(A)$, $A \in \Min(\SGR)$, are homogeneous fractional ideals of $R$, so the map is well-defined. To show injectivity, suppose we are given an $R$-module isomorphism $\varphi \colon I(A) \to I(B)$ for some $A$, $B \in \Min(\SGR)$. This induces an isomorphism of $Q$-vector spaces $\varphi_{(0)} \colon I(A)_{(0)} \to I(B)_{(0)}$, where both sides equal~$Q$. +%and $\varphi_{(0)}(f/1) = \varphi(f)$ for all $f \in I(A)$. +In particular, $\varphi_{(0)}$ (and therefore $\varphi$) is given by multiplication with some $g_0 \in Q$, so $\varphi(f) = g_0 f$ for all $f \in I(A)$. Taking $f=1$ yields $g_0 = \varphi(1) \in I(B)$. Since $\varphi$ is surjective and $1 \in I(B)$, a degree argument yields $g_0 \in \kk\setminus\{ 0 \}$. Consequently, +\[ I(B) = \varphi(I(A)) = g_0 \cdot I(A) = I(A), \] +and because $A$ and $B$ are both minimal with respect to $\preceq_\SGR$, it follows that~$A = B$. Now, let $I \subseteq \bar{R}$ be an arbitrary homogeneous $R$-submodule. Because $R$ is Noetherian and $\bar{R}$ is finitely generated as an $R$-module, $I$ is finitely generated, i.e, $I = \langle x^\beta \mid \beta \in B\rangle_R$ for some finite subset $B \subseteq \NN_0$. Define $m \coloneqq \min(B)$ and $A^\prime \coloneqq \{ \beta - m \mid \beta \in B \setminus \{ 0 \} \}$. Consequently, $I = x^m \cdot I(A^\prime)$, which implies that $I \cong I(A^\prime)$. The claim then follows from the fact that $I(A^\prime)=I(A)$, where $A$ denotes the set of minimal elements of $A^\prime \cap (\gapset)$ with respect to $\preceq_\SGR$. +\end{proof} + +\begin{remark} +We now present another perspective on the $R$-modules $I(A)$. For $A \in \Min(\SGR)$, the $R$-module $I(A)$ admits the $\kk$-vector space description +\[ I(A)=\spn_\kk \{x^\alpha \mid \alpha\in \SGR_A \}, \] +where $\SGR_A\coloneqq (A \cup \{ 0 \}) +\SGR$. The set $\SGR_A$ is a normalized relative ideal of $\SGR$, that is, a subset of $\NN_0$ containing $0$ that is closed under addition by elements of $\SGR$. One readily checks that $A \mapsto \SGR_A$ defines a one-to-one correspondence between the elements of $\Min(\SGR)$ and the normalized relative ideals of $\SGR$. +\todo{reference for this terminology?} +\end{remark} + +\begin{corollary} +For each $A \in \Min(\SGR)$, the $R$-module $I(A)$ is maximal Cohen--Macaulay. +\end{corollary} + +\begin{proof} +By the Cohen--Seidenberg theorems, $\dim(R) = \dim(\bar{R}) = \dim(\kk[x]) = 1$. Fix $A \in \Min(\SGR)$ and a maximal ideal $\mathfrak{m}$ of $R$. Since $R$ is a domain of Krull dimension $1$, every maximal ideal has height $1$; in particular $\dim(R_\mathfrak{m}) = \hgt(\mathfrak{m}) = 1$. The localization +$I(A)_\mathfrak{m}$ is a non-zero $R_\mathfrak{m}$-submodule of $\kk(x)$, hence torsion-free, so $1 \leq \depth(I(A)_\mathfrak{m}) \leq \dim(R_\mathfrak{m}) = 1$. Hence, $I(A)$ is maximal Cohen--Macaulay. +\end{proof} + +\todo{add direct sums are Cohen--Macaulay (Bruns/Herzog, 1993, Prop. 1.2.9)} + +\begin{lemma} +It holds that $\qpower{\bar{R}} = \bar{R}^{\otimes q}$ as $R$-module. +\end{lemma} + +\begin{proof} +Note that $\bar{R}^{1/q} = \kk[x^{1/q}] = \bigoplus_{r \in [0,q]} x^{r/q} \cdot \kk[x]$ +\end{proof} + +%\begin{example} +%\end{example} + +\begin{proposition} +For sufficiently large $q$, all indecomposable summands of $R^{1/q}$ are isomorphic to the normalization $\bar{R} = \kk[x]$ of $R$. That is, $R^{1/q} \cong \bar{R}^{\oplus q}$. +\end{proposition} + +\begin{proof} +Let $q$ be such that $\ZZ_{\geq q} \subseteq S$, and fix $r \in [0,q-1]$. For simplicity, set +\[ \supp(I_{q,r}) \coloneqq \{ a/q \mid a \in S, \, a \equiv r \pmod{q} \} \] +and observe that $\supp(I_{q,r}) \subseteq \{ m_{q,r}/q + n \mid n \in \NN_0\}$, where $m_{q,r}/q$ is the minimum of $\supp(I_{q,r})$. Conversely, for each $n \in \NN_0$, the choice of $q$ ensures $nq \in S$, and hence $m_{q,r} + nq \in S$. Since $m_{q,r}+nq \equiv r \pmod q$, we +obtain $m_{q,r}/q + n \in \supp(I_{q,r})$, and altogether +\[ \supp(I_{q,r}) = \{ m_{q,r}/q + n \mid n \in \NN_0\}. \] +It now follows from~\eqref{eq:ind-summand-as-vector-space} that +\[ I_{q,r} = \spn_\kk \{ x^{m_{q,r}/q + n} \mid n \in \NN_0 \} = x^{m_{q,r}/q} \cdot \spn_\kk\{ x^n \mid n \in \NN_0 \} \cong k[x] . \] +\end{proof} + +%\begin{proposition} +% R is not q-split +%\end{proposition} + +%\begin{proposition} +% if two consecutive powers have same indecomposable summands, then so do all larger powers +%\end{proposition} + +\section{Global dimension of endomorphism algebras} + +\todo[inline]{For $\A \coloneqq \oplus_{A \in \Min(S)} I(A)$, think about $\gldim(\End_R(\A))$ and/or $\gldim(\End_R(R^{1/q}))$ for small $q$.} + +\printbibliography + +\end{document} \ No newline at end of file diff --git a/references.bib b/references.bib new file mode 100644 index 0000000..85d4ac8 --- /dev/null +++ b/references.bib @@ -0,0 +1,59 @@ +@article{Shibuta2017, + title = {Affine semigroup rings are of finite F-representation type}, + volume = {45}, + issn = {0092-7872, 1532-4125}, + url = {https://www.tandfonline.com/doi/full/10.1080/00927872.2017.1313425}, + doi = {10.1080/00927872.2017.1313425}, + pages = {5465--5470}, + number = {12}, + journaltitle = {Communications in Algebra}, + shortjournal = {Communications in Algebra}, + author = {Shibuta, Takafumi}, + urldate = {2026-08-03}, + date = {2017-12-02}, + langid = {english}, +} + +@article{Hochster1976, +title = {The purity of the Frobenius and local cohomology}, +journal = {Advances in Mathematics}, +volume = {21}, +number = {2}, +pages = {117-172}, +year = {1976}, +issn = {0001-8708}, +doi = {https://doi.org/10.1016/0001-8708(76)90073-6}, +url = {https://www.sciencedirect.com/science/article/pii/0001870876900736}, +author = {Melvin Hochster and Joel L Roberts} +} + +@article{Yasuda2009, +title = {On monotonicity of F-blowup sequences}, +volume = {53}, +issn = {0019-2082}, +url = {https://projecteuclid.org/journals/illinois-journal-of-mathematics/volume-53/issue-1/On-monotonicity-of-F-blowup-sequences/10.1215/ijm/1264170841.full}, +doi = {10.1215/ijm/1264170841}, +number = {1}, +journaltitle = {Illinois Journal of Mathematics}, +shortjournal = {Illinois J. Math.}, +author = {Yasuda, Takehiko}, +urldate = {2026-08-11}, +date = {2009-01-01}, +} + +@article{Bruns2006, +title = {On seminormal monoid rings}, +volume = {302}, +rights = {https://www.elsevier.com/tdm/userlicense/1.0/}, +issn = {00218693}, +url = {https://linkinghub.elsevier.com/retrieve/pii/S0021869305006605}, +doi = {10.1016/j.jalgebra.2005.11.012}, +pages = {361--386}, +number = {1}, +journaltitle = {Journal of Algebra}, +shortjournal = {Journal of Algebra}, +author = {Bruns, Winfried and Li, Ping and Römer, Tim}, +urldate = {2026-08-12}, +date = {2006-08}, +langid = {english}, +} \ No newline at end of file From 7e4b5388e772e93f01876b33f6be4fc55d2de76f Mon Sep 17 00:00:00 2001 From: Markus Maximilian Rudolf Tripp Date: Wed, 12 Aug 2026 16:02:03 +0200 Subject: [PATCH 2/8] Initial commit --- README.md | 93 +++++++++++++++++++++++++++++++++++++++++++++++++++++++ 1 file changed, 93 insertions(+) create mode 100644 README.md diff --git a/README.md b/README.md new file mode 100644 index 0000000..b7e25ca --- /dev/null +++ b/README.md @@ -0,0 +1,93 @@ +# ncr-of-monomial-curves + + + +## Getting started + +To make it easy for you to get started with GitLab, here's a list of recommended next steps. + +Already a pro? Just edit this README.md and make it your own. Want to make it easy? [Use the template at the bottom](#editing-this-readme)! + +## Add your files + +* [Create](https://docs.gitlab.com/user/project/repository/web_editor/#create-a-file) or [upload](https://docs.gitlab.com/user/project/repository/web_editor/#upload-a-file) files +* [Add files using the command line](https://docs.gitlab.com/topics/git/add_files/#add-files-to-a-git-repository) or push an existing Git repository with the following command: + +``` +cd existing_repo +git remote add origin https://git.uni-graz.at/markus.tripp/ncr-of-monomial-curves.git +git branch -M main +git push -uf origin main +``` + +## Integrate with your tools + +* [Set up project integrations](https://git.uni-graz.at/markus.tripp/ncr-of-monomial-curves/-/settings/integrations) + +## Collaborate with your team + +* [Invite team members and collaborators](https://docs.gitlab.com/user/project/members/) +* [Create a new merge request](https://docs.gitlab.com/user/project/merge_requests/creating_merge_requests/) +* [Automatically close issues from merge requests](https://docs.gitlab.com/user/project/issues/managing_issues/#closing-issues-automatically) +* [Enable merge request approvals](https://docs.gitlab.com/user/project/merge_requests/approvals/) +* [Set auto-merge](https://docs.gitlab.com/user/project/merge_requests/auto_merge/) + +## Test and Deploy + +Use the built-in continuous integration in GitLab. + +* [Get started with GitLab CI/CD](https://docs.gitlab.com/ci/quick_start/) +* [Analyze your code for known vulnerabilities with Static Application Security Testing (SAST)](https://docs.gitlab.com/user/application_security/sast/) +* [Deploy to Kubernetes, Amazon EC2, or Amazon ECS using Auto Deploy](https://docs.gitlab.com/topics/autodevops/requirements/) +* [Use pull-based deployments for improved Kubernetes management](https://docs.gitlab.com/user/clusters/agent/) +* [Set up protected environments](https://docs.gitlab.com/ci/environments/protected_environments/) + +*** + +# Editing this README + +When you're ready to make this README your own, just edit this file and use the handy template below (or feel free to structure it however you want - this is just a starting point!). Thanks to [makeareadme.com](https://www.makeareadme.com/) for this template. + +## Suggestions for a good README + +Every project is different, so consider which of these sections apply to yours. The sections used in the template are suggestions for most open source projects. Also keep in mind that while a README can be too long and detailed, too long is better than too short. If you think your README is too long, consider utilizing another form of documentation rather than cutting out information. + +## Name +Choose a self-explaining name for your project. + +## Description +Let people know what your project can do specifically. Provide context and add a link to any reference visitors might be unfamiliar with. A list of Features or a Background subsection can also be added here. If there are alternatives to your project, this is a good place to list differentiating factors. + +## Badges +On some READMEs, you may see small images that convey metadata, such as whether or not all the tests are passing for the project. You can use Shields to add some to your README. Many services also have instructions for adding a badge. + +## Visuals +Depending on what you are making, it can be a good idea to include screenshots or even a video (you'll frequently see GIFs rather than actual videos). Tools like ttygif can help, but check out Asciinema for a more sophisticated method. + +## Installation +Within a particular ecosystem, there may be a common way of installing things, such as using Yarn, NuGet, or Homebrew. However, consider the possibility that whoever is reading your README is a novice and would like more guidance. Listing specific steps helps remove ambiguity and gets people to using your project as quickly as possible. If it only runs in a specific context like a particular programming language version or operating system or has dependencies that have to be installed manually, also add a Requirements subsection. + +## Usage +Use examples liberally, and show the expected output if you can. It's helpful to have inline the smallest example of usage that you can demonstrate, while providing links to more sophisticated examples if they are too long to reasonably include in the README. + +## Support +Tell people where they can go to for help. It can be any combination of an issue tracker, a chat room, an email address, etc. + +## Roadmap +If you have ideas for releases in the future, it is a good idea to list them in the README. + +## Contributing +State if you are open to contributions and what your requirements are for accepting them. + +For people who want to make changes to your project, it's helpful to have some documentation on how to get started. Perhaps there is a script that they should run or some environment variables that they need to set. Make these steps explicit. These instructions could also be useful to your future self. + +You can also document commands to lint the code or run tests. These steps help to ensure high code quality and reduce the likelihood that the changes inadvertently break something. Having instructions for running tests is especially helpful if it requires external setup, such as starting a Selenium server for testing in a browser. + +## Authors and acknowledgment +Show your appreciation to those who have contributed to the project. + +## License +For open source projects, say how it is licensed. + +## Project status +If you have run out of energy or time for your project, put a note at the top of the README saying that development has slowed down or stopped completely. Someone may choose to fork your project or volunteer to step in as a maintainer or owner, allowing your project to keep going. You can also make an explicit request for maintainers. From d231eabdf7d1ade952efd2158eebbf50c7e1ff91 Mon Sep 17 00:00:00 2001 From: Markus Maximilian Rudolf Tripp Date: Wed, 2 Sep 2026 12:36:46 +0200 Subject: [PATCH 3/8] first draft --- ncr-of-monomial-curves.tex | 152 ++++++++++++++++++++++++++++++++----- references.bib | 59 ++++++++++++++ 2 files changed, 190 insertions(+), 21 deletions(-) diff --git a/ncr-of-monomial-curves.tex b/ncr-of-monomial-curves.tex index f60c7c2..1077c10 100644 --- a/ncr-of-monomial-curves.tex +++ b/ncr-of-monomial-curves.tex @@ -2,8 +2,11 @@ \usepackage[T1]{fontenc} \usepackage{mathtools} +\usepackage{amssymb} \usepackage{todonotes} \usepackage{blindtext} +\usepackage{tikz} +\usetikzlibrary{arrows.meta,positioning,decorations.pathreplacing} %\usepackage{parskip} \usepackage[ @@ -38,8 +41,17 @@ {} % head spec \theoremstyle{remstyle} \newtheorem{remark}[theorem]{Remark} -\DeclareMathOperator{\supp}{supp} +\newtheorem{exampled}[theorem]{Example} +\newenvironment{example}{ + \renewcommand{\qedsymbol}{$\blacksquare$} + \pushQED{\qed} + \begin{exampled} + }{ + \popQED + \end{exampled} +} +\DeclareMathOperator{\supp}{supp} \DeclareMathOperator{\Frac}{Frac} \DeclareMathOperator{\spn}{span} \DeclareMathOperator{\Min}{Min} @@ -85,7 +97,7 @@ Let $M$ be an $R$-module. The \textbf{$q$-th pseudo-Frobenius twist} of $M$, den for all $r \in R$ and $m \in M$. \end{definition} -\begin{lemma} +\begin{lemma}\label{lemma:cannonc-isom} There is a cannonical $R$-module isomorphism \[ R^{1/q} \xrightarrow{\;\sim\;} \qpower{R}, \] defined by $r = \sum_{a \in \SGR} k_a x^{a/q} \mapsto \qpower{r} = \sum_{a \in \SGR} k_a x^a$. @@ -131,9 +143,22 @@ Conversely, suppose that $R$ is $q$-weakly normal. Let $\sigma \colon \qpower{R} $R$ is normal if and only if $R$ is $q$-split for all $q \in \NN$. \end{corollary} +\begin{lemma}\textnormal{\cite[Lemma 5.0.3]{Blickle2013}}\label{lemma:q-split} +Let $q = p_1^{e_1} \cdots p_k^{e_k}$, where the $p_i$ are distinct primes. Then $R$ is $q$-split if and only if $R$ is $(p_1^{\ell_1}\cdots p_k^{\ell_k})$-split for all $(\ell_1, \dots, \ell_k) \in \NN_0^k$. +\end{lemma} + +\begin{proof} +Assume that $R$ is $q$-split, i.e., there exists a $R$-module homomorphism $\sigma \colon \qpower{R} \to R$ such that $\sigma \circ \qpower{F} = \id_R$. We first claim that, for each $i \in [1,k]$, $R$ is $p_i^{\ell_i}$ split for every $\ell_i \in \NN_0$. To this end, fix an $i \in [1,k]$ and consider the $R$-module homomorphism $f_i \colon \qpower[p_i]{R} \to \qpower{R}$, defined by $r \mapsto \qpower[q/p_i]{r}$. Setting $\sigma_i \coloneqq \sigma \circ f_i$, we obtain that +\[ \sigma_i(\qpower[p_i]{F}(r)) = \sigma(f_i(\qpower[p_i]{r})) = \sigma(\qpower{r}) = r \] +for all $r \in R$, so $R$ is $p_i$-split. Now, an induction on $\ell_i$ shows that $\sigma_i^{\ell_i} \circ \qpower[p_i^{\ell_i}]{F} = \id_R$ for all $\ell_i \in \NN_0$, where $\sigma_i^{\ell_i}$ denotes the $\ell_i$-fold composition of $\sigma_i$. Moreover, $\sigma_i^{\ell_i}$, regarded as a map $\qpower[p_i^{\ell_i}]{R} \to R$, is readily verified to be $R$-linear, providing the claim. + +Finally, fix $(\ell_1, \dots, \ell_k) \in \NN_0^k$, and set $q^\prime \coloneqq p_1^{\ell_1} \cdots p_k^{\ell_k}$. By the first part, $R$ is $p_i^{\ell_i}$-split for each $i \in [0,k]$. Hence, there exist $R$-module homomorphisms $\sigma_i^\prime \colon \qpower[p_i^{\ell_i}]{R} \to R$ such that $\sigma_i^\prime \circ \qpower[p_i^{\ell_i}]{F} = \id_R$. Setting $\sigma^\prime \coloneqq \sigma_1^\prime \circ \cdots \circ \sigma_k^\prime$, a direct verification shows that $\sigma^\prime$ gives the required splitting of $\qpower[q^\prime]{F}$. +\end{proof} + \section{$R$-module decomposition of $R^{1/q}$} -We now specialize to the case where $\SGR = \langle a_1, \dots, a_n \rangle \subseteq \NN_0$ is a numerical semigroup. Following Shibuta~\cite{Shibuta2017}, we sketch how $R^{1/q} = \kk[x^{a_1/q}, \dots, x^{a_n/q}]$ decomposes as an $R$-module. Observe that +We now specialize to the case where $\SGR = \langle a_1, \dots, a_n \rangle$ %= \{ c_1a_1 + \cdots + c_na_n \mid c_i \in \NN_0\} \subseteq \NN_0 \} +is a numerical semigroup. Following Shibuta~\cite{Shibuta2017}, we sketch how $R^{1/q} = \kk[x^{a_1/q}, \dots, x^{a_n/q}]$ decomposes as an $R$-module. Observe that \[ \Gamma_q \coloneqq \left\{ \bigl( x^{a_1/q} \bigr)^{e_1} \cdots \bigl( x^{a_n/q} \bigr)^{e_n} \mid e_i \in [0,q-1] \right\} \] @@ -176,11 +201,11 @@ In particular, as $q$ ranges over $\NN$, only finitely many indecomposable summa \begin{lemma}\label{lemma:rank-1} For each $A \in \Min(\SGR)$, the $R$-module $I(A)$ has rank $1$ and is therefore indecomposable. Moreover, the assignment \[ A \mapsto [I(A)] \] -defines a bijection between $\Min(\SGR)$ and the set of isomorphism classes of homogeneous $R$-submodules of $\bar{R} = \kk[x]$\todo{this also holds for $\kk[x,x^{-1}]$, interesting?}. +defines a bijection between $\Min(\SGR)$ and the set of isomorphism classes of homogeneous $R$-submodules of $\bar{R} = \kk[x]$. \end{lemma} \begin{proof} -For simplicity, set $Q \coloneqq \Frac(R) = \kk(x)$\todo{note $\Frac{R} = \kk(x)$} and fix $A \in \Min(\SGR)$. Observe that $Q \otimes_R I(A)$ is naturally isomorphic, as a $Q$-vector space, to $I(A)_{(0)} \subseteq Q$. Since $I(A)$ contains $1$, it follows that $I(A)_{(0)} = Q$ and $I(A)$ has rank $1$. +For simplicity, set $Q \coloneqq \Frac(R) = \kk(x)$ and fix $A \in \Min(\SGR)$. Observe that $Q \otimes_R I(A)$ is naturally isomorphic, as a $Q$-vector space, to $I(A)_{(0)} \subseteq Q$. Since $I(A)$ contains $1$, it follows that $I(A)_{(0)} = Q$ and $I(A)$ has rank $1$. For the second part, it is clear that $I(A)$, $A \in \Min(\SGR)$, are homogeneous fractional ideals of $R$, so the map is well-defined. To show injectivity, suppose we are given an $R$-module isomorphism $\varphi \colon I(A) \to I(B)$ for some $A$, $B \in \Min(\SGR)$. This induces an isomorphism of $Q$-vector spaces $\varphi_{(0)} \colon I(A)_{(0)} \to I(B)_{(0)}$, where both sides equal~$Q$. %and $\varphi_{(0)}(f/1) = \varphi(f)$ for all $f \in I(A)$. @@ -208,41 +233,126 @@ $I(A)_\mathfrak{m}$ is a non-zero $R_\mathfrak{m}$-submodule of $\kk(x)$, hence \todo{add direct sums are Cohen--Macaulay (Bruns/Herzog, 1993, Prop. 1.2.9)} \begin{lemma} -It holds that $\qpower{\bar{R}} = \bar{R}^{\otimes q}$ as $R$-module. +For $q \in \NN$, it holds that $\qpower{\bar{R}} = \bar{R}^{\oplus q}$ as $R$-module. \end{lemma} \begin{proof} -Note that $\bar{R}^{1/q} = \kk[x^{1/q}] = \bigoplus_{r \in [0,q]} x^{r/q} \cdot \kk[x]$ +First, note that +\[ \bar{R}^{1/q} = \kk[x^{1/q}] = \bigoplus_{r \in [0,q-1]} x^{r/q} \cdot \kk[x] \cong \bar{R}^{\oplus q} \] +as $R$-modules. By Lemma~\ref{lemma:cannonc-isom}, $\bar{R}^{1/q}$ and $\qpower{\bar{R}}$ are isomorphic as $\bar{R}$-modules, and hence, by restriction of scalars along $R \subseteq \bar{R}$, also as $R$-modules. \end{proof} -%\begin{example} -%\end{example} +\begin{example}\label{ex:large-powers} +Let $\SGR = \langle 3,4 \rangle \subseteq \NN_0$ and $R = \kk[\SGR] = \kk[x^3,x^4]$. -\begin{proposition} -For sufficiently large $q$, all indecomposable summands of $R^{1/q}$ are isomorphic to the normalization $\bar{R} = \kk[x]$ of $R$. That is, $R^{1/q} \cong \bar{R}^{\oplus q}$. +\begin{center} +\begin{figure}[htbp] +\begin{tikzpicture}[x=0.8cm] + \draw[->] (0,0) -- (9.5,0); + + % labels + \foreach \x in {0,1,2,5,6,7,8,9} + \draw (\x,0.08) -- (\x,-0.08) + node[below=3pt] {$\x$}; + + % generators + \foreach \x in {3,4} + \draw[red] (\x,0.08) -- (\x,-0.08) + node[below=3pt,red] {$\x$}; + + % black dots for semigroups elements + \foreach \x in {0,6,7,8,9} + \fill (\x,0) circle (2.5pt); + + % red dots for generators + \foreach \x in {3,4} + \fill[red] (\x,0) circle (2.5pt); + + % bracket for frobenius number + \draw[thick] (6,-0.18) -- (6,0.18); + \draw[thick] (6,0.18) -- (6.15,0.18); + \draw[thick] (6,-0.18) -- (6.15,-0.18); +\end{tikzpicture} +\caption{The numerical semigroup $\SGR = \langle 3, 4\rangle$} +\end{figure} +\end{center} + +The set +\[ \{ x^{(3/q)e_1 + (4/q)e_2} : e_i \in [0,q-1] \} \] +generates $R^{1/q} = \kk[x^{3/q},x^{4/q}]$ as an $R$-module. For $q =2$, this generating set is $\{ 1, x^{3/2}, x^2, x^{7/2} \}$, and grouping the numerators of the exponents by residue modulo $2$ yields the $R$-module decomposition +\[ R^{1/2} = \langle 1, x^2 \rangle_R \oplus x^{3/2} \cdot \langle 1, x^2 \rangle_R \cong I(\{2 \})^{\oplus 2} \] +For $q = 4$, a direct computation shows $R^{1/4} \cong \bar{R}^{\oplus 4}$. Keeping in mind that $R^{1/q}$ is isomorphic to $\qpower{R}$ and that the construction of the pseudo-Frobenius twist respects direct sums, we conclude that +\[ R^{1/8} \cong \qpower[2]{(\qpower[4]{R})} = \qpower[2]{(\bar{R}^{\oplus 4})} = (\qpower[2]{\bar{R}})^{\oplus 4} \cong \bar{R}^{\oplus 8}, \] +where the last isomorphism follows from the previous lemma. By induction, we obtain that $R^{1/2^e} \cong \bar{R}^{\oplus 2^e}$ for all $e \geq 2$. Consequently, there exist summands appearing in the decomposition of $R^{1/2^e}$ for some $e \in \NN$ that do \emph{not} appear in the decomposition of $R^{1/2^\ell}$ for large $\ell$. This stands in contrast to the case where $\SGR$ is a normal affine semigroup and $R=\kk[\SGR]$ is a toric algebra~\cite{Bruns2005}. +\end{example} + +% NOTE: next proposition holds when q \in \SGR + +\begin{proposition}\label{prop:asy-behav} +For $q \in \SGR$, every indecomposable summands of $R^{1/q}$ is isomorphic to the normalization $\bar{R} = \kk[x]$ of $R$. That is, $R^{1/q} \cong \bar{R}^{\oplus q}$. In particular, this holds for all sufficiently large $q$. \end{proposition} \begin{proof} -Let $q$ be such that $\ZZ_{\geq q} \subseteq S$, and fix $r \in [0,q-1]$. For simplicity, set -\[ \supp(I_{q,r}) \coloneqq \{ a/q \mid a \in S, \, a \equiv r \pmod{q} \} \] -and observe that $\supp(I_{q,r}) \subseteq \{ m_{q,r}/q + n \mid n \in \NN_0\}$, where $m_{q,r}/q$ is the minimum of $\supp(I_{q,r})$. Conversely, for each $n \in \NN_0$, the choice of $q$ ensures $nq \in S$, and hence $m_{q,r} + nq \in S$. Since $m_{q,r}+nq \equiv r \pmod q$, we +Fix $r \in [0,q-1]$. For simplicity, set +\[ \supp(I_{q,r}) \coloneqq \{ a/q \mid a \in \SGR, \, a \equiv r \pmod{q} \} \] +and observe that $\supp(I_{q,r}) \subseteq \{ m_{q,r}/q + n \mid n \in \NN_0\}$, where $m_{q,r}/q$ is the minimum of $\supp(I_{q,r})$. Conversely, for each $n \in \NN_0$, the choice of $q$ ensures $nq \in \SGR$, and hence $m_{q,r} + nq \in \SGR$. Since $m_{q,r}+nq \equiv r \pmod q$, we obtain $m_{q,r}/q + n \in \supp(I_{q,r})$, and altogether \[ \supp(I_{q,r}) = \{ m_{q,r}/q + n \mid n \in \NN_0\}. \] It now follows from~\eqref{eq:ind-summand-as-vector-space} that \[ I_{q,r} = \spn_\kk \{ x^{m_{q,r}/q + n} \mid n \in \NN_0 \} = x^{m_{q,r}/q} \cdot \spn_\kk\{ x^n \mid n \in \NN_0 \} \cong k[x] . \] \end{proof} -%\begin{proposition} -% R is not q-split -%\end{proposition} +\begin{example} +Revisiting Example~\ref{ex:large-powers}, it remains to determine the decomposition of $R^{1/5}$ in order to obtain a complete description of the decompositions of $R^{1/q}$ for $q\in\NN$. A direct computation yields +\[ R^{1/5}\cong I(\{2\}) \oplus \bar{R}^{\oplus 4}. \] +Since $\bar{R} = \kk[x]$ corresponds to $I(\{1,2\})$, the indecomposable summands that occur in $R^{1/q}$ for some $q \in \NN$ are precisely $I(\{2\})$ and $I(\{1,2\})$. On the other hand, +\[ \Min(\SGR) = \{ \emptyset,\{1\}, \{2\}, \{5\}, \{1,2\} \}, \] +so not every potential candidate $I(A)$ with $A \in \Min(\SGR)$ occurs as a direct summand. +\end{example} -%\begin{proposition} -% if two consecutive powers have same indecomposable summands, then so do all larger powers -%\end{proposition} +\begin{corollary} +For each $q \geq 2$, $R$ is not $q$-split, meaning that $R$ does not occur as a direct summand of $R^{1/q}$. +\end{corollary} + +\todo{\emph{note:} this can also be easily proven directly!} + +\begin{proof} +Fix $q \geq 2$, and let $p$ be a prime divisor of $q$. Suppose, for contradiction, that $R$ is $q$-split. By Lemma~\ref{lemma:q-split}, $R$ is then $p^e$-split for every $e \in \NN$. This contradicts Proposition~\ref{prop:asy-behav}, which shows that, for all sufficiently large $e$, $R^{1/p^e}$ decomposes as a direct sum of copies of $\bar{R}$, and therefore cannot contain $R$ as a direct summand. +\end{proof} + +\begin{proposition} +Let $p$ be a prime divisor of $q$. If $R^{1/q}$ and $R^{1/(pq)}$ have the same indecomposable summands up to isomorphism, then the same holds for $R^{1/(p^e q)}$ for every $e \in \NN$. +\end{proposition} + +\begin{proof} +Decomposing $R^{1/q} \cong \bigoplus_{i=1}^k I(A_i)^{\oplus n_i}$ for pairwise distinct $A_i \in \Min(\SGR)$, it holds that +\[ R^{1/(pq)} \cong \qpower[p]{(R^{1/q})} \cong \bigoplus_{i=1}^k \bigl(\qpower[p]{I(A_i)}\bigr)^{\oplus n_i}. \] +Thus, by assumption, the indecomposable summands occuring in the $R$-modules $\qpower[p]{I(A_i)}$, $i \in [1,k]$, are, up to isomorphism, precisely $I(A_1)$, \dots $I(A_k)$. Repeated application of this observation shows that $R^{1/(p^e q)}$ has the same indecomposable summands, up to isomorphism, for every $e \in \NN$. +\end{proof} \section{Global dimension of endomorphism algebras} -\todo[inline]{For $\A \coloneqq \oplus_{A \in \Min(S)} I(A)$, think about $\gldim(\End_R(\A))$ and/or $\gldim(\End_R(R^{1/q}))$ for small $q$.} +Throughout this section, let $\SGR$ be a numerical semigroup and $R = \kk[\SGR]$ the corresponding monomial curve. + +Let $\Lambda$ be a commutative Noetherian ring and $M$ a finitely generated faithful $\Lambda$-module. The endomorphism algebra $\End_\Lambda(M)$ is called a \textbf{noncommutative resolution of singularities} of $\Lambda$ if it has finite global dimension. + +\begin{theorem} +For $q \in \SGR$, the global dimension of $\End_R(R^{1/q})$ is equal to the Krull dimension of $R$. In particular, this is true for all sufficiently large $q$. +\end{theorem} + +\begin{proof} +For $q \in \SGR$, Proposition~\ref{prop:asy-behav} gives $R^{1/q}\cong \bar R^{\oplus q}$. By~\cite[Cor. 2.3]{Leuschke2012}, multiplicities of direct summands do not affect the endomorphism ring up to Morita equivalence; hence $\End_R(R^{1/q})$ is Morita equivalent to $\End_R(\bar{R})$. As in the proof of Lemma~\ref{lemma:rank-1}, since $\bar{R}$ has rank $1$, any $R$-module endomorphism of $\bar{R}$ is induced by multiplication with an element of $\bar{R}$, and thus $\End_R(\bar{R}) \cong \bar{R}$. Consequently, we obtain +\[ \gldim(\End_R(R^{1/q})) = \gldim(\bar{R}) = \gldim(\kk[x]) = 1 = \dim(R), \] +where the first equality follows from the Morita invariance of global dimension. +\end{proof} + +Following the argument of Faber, Muller, and Smith~\cite[Theorem~6.13]{Faber2019} for the corresponding result on toric rings, we obtain the following: + +\begin{theorem} +For a perfect field $\kk$ of prime characteristic, the ring $D_\kk(R)$ of differential operators on $R$ has finite global dimension. +\end{theorem} + +\todo[inline]{for $\A \coloneqq \oplus_{A \in \Min(\SGR)} I(A)$, think about $\gldim(\End_R(\A))$ and/or $\gldim(\End_R(R^{1/q}))$ for small $q$} \printbibliography diff --git a/references.bib b/references.bib index 85d4ac8..4b0e24b 100644 --- a/references.bib +++ b/references.bib @@ -56,4 +56,63 @@ author = {Bruns, Winfried and Li, Ping and Römer, Tim}, urldate = {2026-08-12}, date = {2006-08}, langid = {english}, +} + +@incollection{Bruns2005, + author = {Bruns, Winfried}, + title = {Conic divisor classes over a normal monoid algebra}, + booktitle = {Commutative Algebra and Algebraic Geometry}, + series = {Contemporary Mathematics}, + volume = {390}, + pages = {63--71}, + publisher = {American Mathematical Society}, + location = {Providence, RI}, + year = {2005}, +} + + +@incollection{Blickle2013, + location = {New York, {NY}}, + title = {$p^{-1}$-Linear Maps in Algebra and Geometry}, + isbn = {978-1-4614-5291-1 978-1-4614-5292-8}, + url = {https://link.springer.com/10.1007/978-1-4614-5292-8_5}, + doi = {10.1007/978-1-4614-5292-8_5}, + pages = {123--205}, + booktitle = {Commutative Algebra}, + publisher = {Springer New York}, + author = {Blickle, Manuel and Schwede, Karl}, + editor = {Peeva, Irena}, + urldate = {2026-09-01}, + date = {2013}, + langid = {english}, +} + +@inbook{Leuschke2012, +url = {https://doi.org/10.1515/9783110250404.293}, +title = {Non-commutative Crepant Resolutions: Scenes From Categorical Geometry}, +booktitle = {Progress in Commutative Algebra 1}, +booktitle = {Combinatorics and Homology}, +author = {Graham J. Leuschke}, +editor = {Christopher Francisco and Lee C. Klingler and Sean Sather-Wagstaff and Janet C. Vassilev}, +publisher = {De Gruyter}, +address = {Berlin, Boston}, +pages = {293--364}, +doi = {doi:10.1515/9783110250404.293}, +isbn = {9783110250404}, +year = {2012}, +lastchecked = {2026-09-02} +} + +@article{Faber2019, +title = {Non-commutative resolutions of toric varieties}, +journal = {Advances in Mathematics}, +volume = {351}, +pages = {236-274}, +year = {2019}, +issn = {0001-8708}, +doi = {https://doi.org/10.1016/j.aim.2019.04.021}, +url = {https://www.sciencedirect.com/science/article/pii/S0001870819301823}, +author = {Eleonore Faber and Greg Muller and Karen E. Smith}, +keywords = {Toric variety, Non-commutative resolution, Finite global dimension, Non-commutative crepant resolution, Rings of differential operators, Strong F-regularity}, +abstract = {Let R be the coordinate ring of an affine toric variety. We prove, using direct elementary methods, that the endomorphism ring EndR(A), where A is the (finite) direct sum of all (isomorphism classes of) conic R-modules, has finite global dimension equal to the dimension of R. This gives a precise version, and an elementary proof, of a theorem of Špenko and Van den Bergh implying that EndR(A) has finite global dimension. Furthermore, we show that EndR(A) is a non-commutative crepant resolution if and only if the toric variety is simplicial. For toric varieties over a perfect field k of prime characteristic, we show that the ring of differential operators Dk(R) has finite global dimension.} } \ No newline at end of file From 1843aa43c924c7d124ba95f66e0793ce5c61d271 Mon Sep 17 00:00:00 2001 From: Markus Maximilian Rudolf Tripp Date: Wed, 2 Sep 2026 17:51:12 +0200 Subject: [PATCH 4/8] add .pdf link --- .gitlab-ci.yml | 9 +++++ README.md | 92 +------------------------------------------------- 2 files changed, 10 insertions(+), 91 deletions(-) create mode 100644 .gitlab-ci.yml diff --git a/.gitlab-ci.yml b/.gitlab-ci.yml new file mode 100644 index 0000000..0b42565 --- /dev/null +++ b/.gitlab-ci.yml @@ -0,0 +1,9 @@ +compile_latex: + image: texlive/texlive:latest + + script: + - latexmk -pdf -interaction=nonstopmode -halt-on-error ncr-of-monomial-curves.tex + + artifacts: + paths: + - ncr-of-monomial-curves.pdf \ No newline at end of file diff --git a/README.md b/README.md index b7e25ca..3f8326d 100644 --- a/README.md +++ b/README.md @@ -1,93 +1,3 @@ # ncr-of-monomial-curves - - -## Getting started - -To make it easy for you to get started with GitLab, here's a list of recommended next steps. - -Already a pro? Just edit this README.md and make it your own. Want to make it easy? [Use the template at the bottom](#editing-this-readme)! - -## Add your files - -* [Create](https://docs.gitlab.com/user/project/repository/web_editor/#create-a-file) or [upload](https://docs.gitlab.com/user/project/repository/web_editor/#upload-a-file) files -* [Add files using the command line](https://docs.gitlab.com/topics/git/add_files/#add-files-to-a-git-repository) or push an existing Git repository with the following command: - -``` -cd existing_repo -git remote add origin https://git.uni-graz.at/markus.tripp/ncr-of-monomial-curves.git -git branch -M main -git push -uf origin main -``` - -## Integrate with your tools - -* [Set up project integrations](https://git.uni-graz.at/markus.tripp/ncr-of-monomial-curves/-/settings/integrations) - -## Collaborate with your team - -* [Invite team members and collaborators](https://docs.gitlab.com/user/project/members/) -* [Create a new merge request](https://docs.gitlab.com/user/project/merge_requests/creating_merge_requests/) -* [Automatically close issues from merge requests](https://docs.gitlab.com/user/project/issues/managing_issues/#closing-issues-automatically) -* [Enable merge request approvals](https://docs.gitlab.com/user/project/merge_requests/approvals/) -* [Set auto-merge](https://docs.gitlab.com/user/project/merge_requests/auto_merge/) - -## Test and Deploy - -Use the built-in continuous integration in GitLab. - -* [Get started with GitLab CI/CD](https://docs.gitlab.com/ci/quick_start/) -* [Analyze your code for known vulnerabilities with Static Application Security Testing (SAST)](https://docs.gitlab.com/user/application_security/sast/) -* [Deploy to Kubernetes, Amazon EC2, or Amazon ECS using Auto Deploy](https://docs.gitlab.com/topics/autodevops/requirements/) -* [Use pull-based deployments for improved Kubernetes management](https://docs.gitlab.com/user/clusters/agent/) -* [Set up protected environments](https://docs.gitlab.com/ci/environments/protected_environments/) - -*** - -# Editing this README - -When you're ready to make this README your own, just edit this file and use the handy template below (or feel free to structure it however you want - this is just a starting point!). Thanks to [makeareadme.com](https://www.makeareadme.com/) for this template. - -## Suggestions for a good README - -Every project is different, so consider which of these sections apply to yours. The sections used in the template are suggestions for most open source projects. Also keep in mind that while a README can be too long and detailed, too long is better than too short. If you think your README is too long, consider utilizing another form of documentation rather than cutting out information. - -## Name -Choose a self-explaining name for your project. - -## Description -Let people know what your project can do specifically. Provide context and add a link to any reference visitors might be unfamiliar with. A list of Features or a Background subsection can also be added here. If there are alternatives to your project, this is a good place to list differentiating factors. - -## Badges -On some READMEs, you may see small images that convey metadata, such as whether or not all the tests are passing for the project. You can use Shields to add some to your README. Many services also have instructions for adding a badge. - -## Visuals -Depending on what you are making, it can be a good idea to include screenshots or even a video (you'll frequently see GIFs rather than actual videos). Tools like ttygif can help, but check out Asciinema for a more sophisticated method. - -## Installation -Within a particular ecosystem, there may be a common way of installing things, such as using Yarn, NuGet, or Homebrew. However, consider the possibility that whoever is reading your README is a novice and would like more guidance. Listing specific steps helps remove ambiguity and gets people to using your project as quickly as possible. If it only runs in a specific context like a particular programming language version or operating system or has dependencies that have to be installed manually, also add a Requirements subsection. - -## Usage -Use examples liberally, and show the expected output if you can. It's helpful to have inline the smallest example of usage that you can demonstrate, while providing links to more sophisticated examples if they are too long to reasonably include in the README. - -## Support -Tell people where they can go to for help. It can be any combination of an issue tracker, a chat room, an email address, etc. - -## Roadmap -If you have ideas for releases in the future, it is a good idea to list them in the README. - -## Contributing -State if you are open to contributions and what your requirements are for accepting them. - -For people who want to make changes to your project, it's helpful to have some documentation on how to get started. Perhaps there is a script that they should run or some environment variables that they need to set. Make these steps explicit. These instructions could also be useful to your future self. - -You can also document commands to lint the code or run tests. These steps help to ensure high code quality and reduce the likelihood that the changes inadvertently break something. Having instructions for running tests is especially helpful if it requires external setup, such as starting a Selenium server for testing in a browser. - -## Authors and acknowledgment -Show your appreciation to those who have contributed to the project. - -## License -For open source projects, say how it is licensed. - -## Project status -If you have run out of energy or time for your project, put a note at the top of the README saying that development has slowed down or stopped completely. Someone may choose to fork your project or volunteer to step in as a maintainer or owner, allowing your project to keep going. You can also make an explicit request for maintainers. +[Latest compiled PDF](https://git.uni-graz.at/markus.tripp/ncr-of-monomial-curves/-/jobs/artifacts/main/raw/ncr-of-monomial-curves.pdf?job=compile_latex) From ff35da1de2d3bde05bed3a285eac52b65de135e8 Mon Sep 17 00:00:00 2001 From: Markus Maximilian Rudolf Tripp Date: Wed, 2 Sep 2026 18:52:34 +0200 Subject: [PATCH 5/8] small changes --- ncr-of-monomial-curves.tex | 40 ++++++++++++++++++++++---------------- 1 file changed, 23 insertions(+), 17 deletions(-) diff --git a/ncr-of-monomial-curves.tex b/ncr-of-monomial-curves.tex index 1077c10..ed7bbb3 100644 --- a/ncr-of-monomial-curves.tex +++ b/ncr-of-monomial-curves.tex @@ -82,9 +82,9 @@ \section{Monomial curves} \section{Frobenius maps on affine semigroups} -Let $\SGR$ be an affine semigroup, i.e., a finitely generated submonoid of $\ZZ^d$ for some $d \in \NN$, and let $R \coloneqq \kk[\SGR]$ denote the associated semigroup algebra. Fix a positive integer $q$, and write $q^{-1}\SGR \coloneqq \{ a/q \mid a \in \SGR \} \subseteq \QQ^d$ for the affine semigroup $\SGR$ refined by $q$. Consider the semigroup algebra +Let $\SGR$ be an affine semigroup, i.e., a finitely generated submonoid of $\ZZ^d$ for some $d \in \NN$, and let $R \coloneqq \kk[\SGR]$ denote its semigroup algebra, which defines a toric variety. Fix a positive integer $q$, and write $q^{-1}\SGR \coloneqq \{ a/q \mid a \in \SGR \} \subseteq \QQ^d$ for the affine semigroup $\SGR$ refined by $q$. Consider the semigroup algebra \[ R^{1/q} \coloneqq \kk[q^{-1}\SGR] = \spn_\kk\{ x^{a/q} \mid a \in \SGR \} \] -regarded as a subring of $\kk[\QQ^d] = \spn_\kk \{ x^\alpha \mid \alpha \in \QQ^d \}$, which carries a natural $\QQ^d$-grading. While the semigroup ring $R^{1/q}$ is defined for arbitrary $q \in \NN$ in any characteristic, the notation is motivated by the case $q = p^e$ with $\kk$ perfect of characteristic $p$, where $R^{1/q}$ is the ring of $q$-th roots of elements in $R = \kk[\SGR]$. Further, for $r = \sum_{a \in \QQ^d} k_a x^{a}$, we set $\qpower{r} \coloneqq \sum_{a \in \QQ^d} k_a x^{aq}$, and write $\qpower{F}$ for the resulting \emph{$q$-th pseudo-Frobenius endomorphism} on $R$, which maps $r$ to $\qpower{r}$. +regarded as a subring of $\kk[\QQ^d] = \spn_\kk \{ x^\alpha \mid \alpha \in \QQ^d \}$, which carries a natural $\QQ^d$-grading. While the semigroup ring $R^{1/q}$ is defined for arbitrary $q \in \NN$ in any characteristic, the notation is motivated by the case $q = p^e$ with $\kk$ perfect of prime characteristic $p$, where $R^{1/q}$ is the ring of $q$-th roots of elements in $R$. Further, for $r = \sum_{a \in \QQ^d} k_a x^{a}$, we set $\qpower{r} \coloneqq \sum_{a \in \QQ^d} k_a x^{aq}$, and write $\qpower{F}$ for the resulting \emph{$q$-th pseudo-Frobenius (ring) endomorphism} of $R$, which maps $r$ to $\qpower{r}$. \begin{remark} % PREVIOUS VERSION % Note that even in positive characteristic $p$ with $q = p^e$, the $q$-th pseudo-Frobenius twist need not agree with the usual $q$-th Frobenius twist, where $R$-scalar multiplication is pulled back via the $e$-times iterated Frobenius endomorphism $F^e \colon r \mapsto r^{p^e}$ on $R$. The two agree precisely when the underlying field $\kk$ is finite, in which case $\kk \subseteq R$ is fixed pointwise by $F^e$. @@ -104,9 +104,9 @@ defined by $r = \sum_{a \in \SGR} k_a x^{a/q} \mapsto \qpower{r} = \sum_{a \in \ \end{lemma} \begin{proof} -By abuse of notation, we also write $\qpower{F}$ for the map $R^{1/q} \to \qpower{R}$ above; it is clearly additive and bijective. To verify compatibility with the respective $R$-module structures, write $r = \sum_{a \in \SGR} k_a x^a \in R$ and $f = \sum_{a \in \SGR} \ell_a x^{a/q} \in R^{1/q}$. Then -\[ \qpower{F}(rf) = \qpower{F}\Bigl(\sum_{a, b \in \SGR} k_a\ell_b x^{a + b/q}\Bigr) = \sum_{a, b \in \SGR} k_a\ell_b x^{aq + b}, \] -which is precisely $r\cdot\qpower{F}(f)$ in $\qpower{R}$. +By abuse of notation, we also write $\qpower{F}$ for the map $R^{1/q} \to \qpower{R}$ above; it is clearly additive and bijective. To verify compatibility with the respective $R$-module structures, write $r = \sum_{a \in \SGR} k_a x^a \in R$ and $m = \sum_{a \in \SGR} l_a x^{a/q} \in R^{1/q}$. Then +\[ \qpower{F}(rm) = \qpower{F}\Bigl(\sum_{a, b \in \SGR} k_a l_b x^{a + b/q}\Bigr) = \sum_{a, b \in \SGR} k_a l_b x^{aq + b}, \] +which is precisely $r\cdot\qpower{F}(m)$ in $\qpower{R}$. \end{proof} \begin{definition} @@ -120,7 +120,7 @@ In semigroup-thereotic terms, $R = \kk[\SGR]$ is $q$-weakly normal if and only i In positive characteristic $p$, the Frobenius endomorphism yields $\qpower[p]{r} \in R$ if and only if $r^p \in R$, from which we obtain the following: \begin{corollary} -If $\kk$ has characteristic $p >0$, then $R$ is $p$-weakly normal +If $\kk$ has prime characteristic $p$, then $R$ is $p$-weakly normal % MORE INFORMATION % (or, equivalently, $p^e$-weakly normal for all $e \in \NN$) if and only if $R$ is weakly normal (i.e., for any $r \in \Frac(R)$, $r^p$ implies $r \in R$). In particular, $R$ is seminormal in this case. \end{corollary} @@ -132,7 +132,7 @@ $R$ is $q$-split if and only if it is $q$-weakly normal. \begin{proof} % FAKE PROOF % Suppose $R$ is $q$-split, with $\sigma \colon \qpower{R} \to R$ an $R$-module homomorphism satisfying $\sigma \circ \qpower{F} = \id_R$; in particular, $\sigma(1) = 1$. Now let $r \in \Frac(R)$ be such that $\qpower{r} \in R$. Then $r = r\sigma(1) = \sigma(\qpower{r}) \in R$. -Suppose $R$ is $q$-split, with $\sigma \colon \qpower{R} \to R$ an $R$-module homomorphism satisfying $\sigma \circ \qpower{F} = \id_R$. Let $r \in \kk[G(\SGR)]$ be such that $\qpower{r} \in R$, and choose $s \in R^\bullet$ with $sr \in R$. Then +Suppose $R$ is $q$-split, with $\sigma \colon \qpower{R} \to R$ an $R$-module homomorphism satisfying $\sigma \circ \qpower{F} = \id_R$. Let $r \in \kk[G(\SGR)]$ be such that $\qpower{r} \in R$, and choose $s \in R\setminus \{0 \}$ with $sr \in R$. Then \[ sr = \sigma\bigl( \qpower{F}(sr) \bigr) = \sigma(\qpower{s}\qpower{r}) = s\sigma(\qpower{r}), \] and cancelling $s$ yields $r = \sigma(\qpower{r}) \in R$. @@ -150,7 +150,7 @@ Let $q = p_1^{e_1} \cdots p_k^{e_k}$, where the $p_i$ are distinct primes. Then \begin{proof} Assume that $R$ is $q$-split, i.e., there exists a $R$-module homomorphism $\sigma \colon \qpower{R} \to R$ such that $\sigma \circ \qpower{F} = \id_R$. We first claim that, for each $i \in [1,k]$, $R$ is $p_i^{\ell_i}$ split for every $\ell_i \in \NN_0$. To this end, fix an $i \in [1,k]$ and consider the $R$-module homomorphism $f_i \colon \qpower[p_i]{R} \to \qpower{R}$, defined by $r \mapsto \qpower[q/p_i]{r}$. Setting $\sigma_i \coloneqq \sigma \circ f_i$, we obtain that \[ \sigma_i(\qpower[p_i]{F}(r)) = \sigma(f_i(\qpower[p_i]{r})) = \sigma(\qpower{r}) = r \] -for all $r \in R$, so $R$ is $p_i$-split. Now, an induction on $\ell_i$ shows that $\sigma_i^{\ell_i} \circ \qpower[p_i^{\ell_i}]{F} = \id_R$ for all $\ell_i \in \NN_0$, where $\sigma_i^{\ell_i}$ denotes the $\ell_i$-fold composition of $\sigma_i$. Moreover, $\sigma_i^{\ell_i}$, regarded as a map $\qpower[p_i^{\ell_i}]{R} \to R$, is readily verified to be $R$-linear, providing the claim. +for all $r \in R$, so $R$ is $p_i$-split. Now, an induction on $\ell_i$ shows that, as maps of sets, $\sigma_i^{\ell_i} \circ \qpower[p_i^{\ell_i}]{F} = \id_R$ for all $\ell_i \in \NN_0$, where $\sigma_i^{\ell_i}$ denotes the $\ell_i$-fold composition of $\sigma_i$. Moreover, $\sigma_i^{\ell_i}$, regarded as a map $\qpower[p_i^{\ell_i}]{R} \to R$, is readily verified to be $R$-linear, providing the claim. Finally, fix $(\ell_1, \dots, \ell_k) \in \NN_0^k$, and set $q^\prime \coloneqq p_1^{\ell_1} \cdots p_k^{\ell_k}$. By the first part, $R$ is $p_i^{\ell_i}$-split for each $i \in [0,k]$. Hence, there exist $R$-module homomorphisms $\sigma_i^\prime \colon \qpower[p_i^{\ell_i}]{R} \to R$ such that $\sigma_i^\prime \circ \qpower[p_i^{\ell_i}]{F} = \id_R$. Setting $\sigma^\prime \coloneqq \sigma_1^\prime \circ \cdots \circ \sigma_k^\prime$, a direct verification shows that $\sigma^\prime$ gives the required splitting of $\qpower[q^\prime]{F}$. \end{proof} @@ -158,7 +158,7 @@ Finally, fix $(\ell_1, \dots, \ell_k) \in \NN_0^k$, and set $q^\prime \coloneqq \section{$R$-module decomposition of $R^{1/q}$} We now specialize to the case where $\SGR = \langle a_1, \dots, a_n \rangle$ %= \{ c_1a_1 + \cdots + c_na_n \mid c_i \in \NN_0\} \subseteq \NN_0 \} -is a numerical semigroup. Following Shibuta~\cite{Shibuta2017}, we sketch how $R^{1/q} = \kk[x^{a_1/q}, \dots, x^{a_n/q}]$ decomposes as an $R$-module. Observe that +is a numerical semigroup and $R = \kk[\SGR]$ is the corresponding monomial curve. Following Shibuta~\cite{Shibuta2017}, we sketch how $R^{1/q} = \kk[x^{a_1/q}, \dots, x^{a_n/q}]$ decomposes as an $R$-module. Observe that \[ \Gamma_q \coloneqq \left\{ \bigl( x^{a_1/q} \bigr)^{e_1} \cdots \bigl( x^{a_n/q} \bigr)^{e_n} \mid e_i \in [0,q-1] \right\} \] @@ -167,7 +167,7 @@ $a/q$ with $a \in \NN_0$, this amounts to grouping elements by the residue of $a \[ \Gamma_q = \Gamma_{q,0} \sqcup \Gamma_{q,1} \sqcup \cdots \sqcup \Gamma_{q,q-1}, \] -where $\Gamma_{q,r} \coloneqq \{ x^{a/q} \in \Gamma_q \mid a \equiv r \pmod{q}\}$. Let $I_{q,r}$ be the $R$-submodule of $\kk[x^q \mid q \in \QQ_{\geq 0}]$ +where $\Gamma_{q,r} \coloneqq \{ x^{a/q} \in \Gamma_q \mid a \equiv r \pmod{q}\}$. Let $I_{q,r}$ be the $R$-submodule of $\kk[x^q \mid q \in \QQ]$ generated by $\Gamma_{q,r}$. As a $\kk$-vector space, \begin{equation}\label{eq:ind-summand-as-vector-space} I_{q,r} = \spn_\kk \{ x^{a/q} \mid a \in \SGR, \, a \equiv r \pmod{q} \}, @@ -207,11 +207,11 @@ defines a bijection between $\Min(\SGR)$ and the set of isomorphism classes of h \begin{proof} For simplicity, set $Q \coloneqq \Frac(R) = \kk(x)$ and fix $A \in \Min(\SGR)$. Observe that $Q \otimes_R I(A)$ is naturally isomorphic, as a $Q$-vector space, to $I(A)_{(0)} \subseteq Q$. Since $I(A)$ contains $1$, it follows that $I(A)_{(0)} = Q$ and $I(A)$ has rank $1$. -For the second part, it is clear that $I(A)$, $A \in \Min(\SGR)$, are homogeneous fractional ideals of $R$, so the map is well-defined. To show injectivity, suppose we are given an $R$-module isomorphism $\varphi \colon I(A) \to I(B)$ for some $A$, $B \in \Min(\SGR)$. This induces an isomorphism of $Q$-vector spaces $\varphi_{(0)} \colon I(A)_{(0)} \to I(B)_{(0)}$, where both sides equal~$Q$. +For the second part, it is clear that $I(A)$, $A \in \Min(\SGR)$, are homogeneous $R$-submodules of $\bar{R}$, so the map is well-defined. To show injectivity, suppose we are given an $R$-module isomorphism $\varphi \colon I(A) \to I(B)$ for some $A$, $B \in \Min(\SGR)$. This induces an isomorphism of $Q$-vector spaces $\varphi_{(0)} \colon I(A)_{(0)} \to I(B)_{(0)}$, where both sides equal~$Q$. %and $\varphi_{(0)}(f/1) = \varphi(f)$ for all $f \in I(A)$. In particular, $\varphi_{(0)}$ (and therefore $\varphi$) is given by multiplication with some $g_0 \in Q$, so $\varphi(f) = g_0 f$ for all $f \in I(A)$. Taking $f=1$ yields $g_0 = \varphi(1) \in I(B)$. Since $\varphi$ is surjective and $1 \in I(B)$, a degree argument yields $g_0 \in \kk\setminus\{ 0 \}$. Consequently, \[ I(B) = \varphi(I(A)) = g_0 \cdot I(A) = I(A), \] -and because $A$ and $B$ are both minimal with respect to $\preceq_\SGR$, it follows that~$A = B$. Now, let $I \subseteq \bar{R}$ be an arbitrary homogeneous $R$-submodule. Because $R$ is Noetherian and $\bar{R}$ is finitely generated as an $R$-module, $I$ is finitely generated, i.e, $I = \langle x^\beta \mid \beta \in B\rangle_R$ for some finite subset $B \subseteq \NN_0$. Define $m \coloneqq \min(B)$ and $A^\prime \coloneqq \{ \beta - m \mid \beta \in B \setminus \{ 0 \} \}$. Consequently, $I = x^m \cdot I(A^\prime)$, which implies that $I \cong I(A^\prime)$. The claim then follows from the fact that $I(A^\prime)=I(A)$, where $A$ denotes the set of minimal elements of $A^\prime \cap (\gapset)$ with respect to $\preceq_\SGR$. +and because $A$ and $B$ are both minimal with respect to $\preceq_\SGR$, it follows that~$A = B$. Now, let $I \subseteq \bar{R}$ be an arbitrary homogeneous $R$-submodule. Because $R$ is Noetherian and $\bar{R}$ is finitely generated as an $R$-module, $I$ is finitely generated, i.e, $I = \langle x^\beta \mid \beta \in B\rangle_R$ for some finite subset $B \subseteq \NN_0$. Define $m \coloneqq \min(B)$ and $A^\prime \coloneqq \{ \beta - m \mid \beta \in B \} \setminus \{ 0 \}$. Consequently, $I = x^m \cdot I(A^\prime)$, which implies that $I \cong I(A^\prime)$. The claim then follows from the fact that $I(A^\prime)=I(A)$, where $A$ denotes the set of minimal elements of $A^\prime \cap (\gapset)$ with respect to $\preceq_\SGR$. \end{proof} \begin{remark} @@ -282,7 +282,7 @@ The set generates $R^{1/q} = \kk[x^{3/q},x^{4/q}]$ as an $R$-module. For $q =2$, this generating set is $\{ 1, x^{3/2}, x^2, x^{7/2} \}$, and grouping the numerators of the exponents by residue modulo $2$ yields the $R$-module decomposition \[ R^{1/2} = \langle 1, x^2 \rangle_R \oplus x^{3/2} \cdot \langle 1, x^2 \rangle_R \cong I(\{2 \})^{\oplus 2} \] For $q = 4$, a direct computation shows $R^{1/4} \cong \bar{R}^{\oplus 4}$. Keeping in mind that $R^{1/q}$ is isomorphic to $\qpower{R}$ and that the construction of the pseudo-Frobenius twist respects direct sums, we conclude that -\[ R^{1/8} \cong \qpower[2]{(\qpower[4]{R})} = \qpower[2]{(\bar{R}^{\oplus 4})} = (\qpower[2]{\bar{R}})^{\oplus 4} \cong \bar{R}^{\oplus 8}, \] +\[ R^{1/8} \cong \qpower[2]{(\qpower[4]{R})} \cong \qpower[2]{(\bar{R}^{\oplus 4})} = (\qpower[2]{\bar{R}})^{\oplus 4} \cong \bar{R}^{\oplus 8}, \] where the last isomorphism follows from the previous lemma. By induction, we obtain that $R^{1/2^e} \cong \bar{R}^{\oplus 2^e}$ for all $e \geq 2$. Consequently, there exist summands appearing in the decomposition of $R^{1/2^e}$ for some $e \in \NN$ that do \emph{not} appear in the decomposition of $R^{1/2^\ell}$ for large $\ell$. This stands in contrast to the case where $\SGR$ is a normal affine semigroup and $R=\kk[\SGR]$ is a toric algebra~\cite{Bruns2005}. \end{example} @@ -302,6 +302,8 @@ It now follows from~\eqref{eq:ind-summand-as-vector-space} that \[ I_{q,r} = \spn_\kk \{ x^{m_{q,r}/q + n} \mid n \in \NN_0 \} = x^{m_{q,r}/q} \cdot \spn_\kk\{ x^n \mid n \in \NN_0 \} \cong k[x] . \] \end{proof} +\todo[inline]{think about what can be said about decompositions of $R^{1/q}$ for $q \notin \SGR$} + \begin{example} Revisiting Example~\ref{ex:large-powers}, it remains to determine the decomposition of $R^{1/5}$ in order to obtain a complete description of the decompositions of $R^{1/q}$ for $q\in\NN$. A direct computation yields \[ R^{1/5}\cong I(\{2\}) \oplus \bar{R}^{\oplus 4}. \] @@ -332,9 +334,9 @@ Thus, by assumption, the indecomposable summands occuring in the $R$-modules $\q \section{Global dimension of endomorphism algebras} -Throughout this section, let $\SGR$ be a numerical semigroup and $R = \kk[\SGR]$ the corresponding monomial curve. +Throughout this section, let $\SGR$ again be a numerical semigroup and $R = \kk[\SGR]$ the corresponding monomial curve. -Let $\Lambda$ be a commutative Noetherian ring and $M$ a finitely generated faithful $\Lambda$-module. The endomorphism algebra $\End_\Lambda(M)$ is called a \textbf{noncommutative resolution of singularities} of $\Lambda$ if it has finite global dimension. +For a commutative Noetherian ring $\Lambda$ and a finitely generated faithful $\Lambda$-module $M$, the endomorphism algebra $\End_\Lambda(M)$ is called a \textbf{noncommutative resolution of singularities} of $\Lambda$ if it has finite global dimension. \begin{theorem} For $q \in \SGR$, the global dimension of $\End_R(R^{1/q})$ is equal to the Krull dimension of $R$. In particular, this is true for all sufficiently large $q$. @@ -346,13 +348,17 @@ For $q \in \SGR$, Proposition~\ref{prop:asy-behav} gives $R^{1/q}\cong \bar R^{\ where the first equality follows from the Morita invariance of global dimension. \end{proof} -Following the argument of Faber, Muller, and Smith~\cite[Theorem~6.13]{Faber2019} for the corresponding result on toric rings, we obtain the following: +\todo[inline]{is there a notion of crepant for non-normal, non-Gorenstein rings?} + +\todo[inline]{for $\A \coloneqq \oplus_{A \in \Min(\SGR)} I(A)$, think about $\gldim(\End_R(\A))$ and/or $\gldim(\End_R(R^{1/q}))$ for small $q$} + +Following the argument of Faber, Muller, and Smith~\cite[Theorem~6.13]{Faber2019} for the corresponding result on normal toric varieties, we obtain the following: \begin{theorem} For a perfect field $\kk$ of prime characteristic, the ring $D_\kk(R)$ of differential operators on $R$ has finite global dimension. \end{theorem} -\todo[inline]{for $\A \coloneqq \oplus_{A \in \Min(\SGR)} I(A)$, think about $\gldim(\End_R(\A))$ and/or $\gldim(\End_R(R^{1/q}))$ for small $q$} +\todo[inline]{is there machinery to show equality for the global dimension of $D_\kk(R)$ and $\dim(R)$?} \printbibliography From 48ee5bcbd7c15a968af29723117f9d1ce5615c28 Mon Sep 17 00:00:00 2001 From: Markus Maximilian Rudolf Tripp Date: Tue, 15 Sep 2026 15:09:41 +0200 Subject: [PATCH 6/8] quick intermediate save --- ncr-of-monomial-curves.tex | 214 ++++++++++++++++++++++++++++--------- references.bib | 60 +++++++++++ 2 files changed, 224 insertions(+), 50 deletions(-) diff --git a/ncr-of-monomial-curves.tex b/ncr-of-monomial-curves.tex index ed7bbb3..cfb1468 100644 --- a/ncr-of-monomial-curves.tex +++ b/ncr-of-monomial-curves.tex @@ -3,9 +3,11 @@ \usepackage[T1]{fontenc} \usepackage{mathtools} \usepackage{amssymb} +\usepackage{stmaryrd} \usepackage{todonotes} \usepackage{blindtext} \usepackage{tikz} +\usepackage{graphicx} \usetikzlibrary{arrows.meta,positioning,decorations.pathreplacing} %\usepackage{parskip} @@ -60,7 +62,9 @@ \DeclareMathOperator{\gldim}{gl.dim} \DeclareMathOperator{\End}{End} \DeclareMathOperator{\id}{id} +\DeclareMathOperator{\gp}{gp} \newcommand{\qpower}[2][q]{{}^{#1}#2} +\newcommand{\qmult}[1][q]{\mathbin{\star}_#1}%\mathbin{\vcenter{\hbox{\scalebox{0.8}{$\circledast$}}}}_#1 \newcommand{\gapset}{\NN_0\setminus\SGR} \newcommand{\kk}{\mathsf{k}} \newcommand{\SGR}{\mathsf{S}} @@ -68,6 +72,9 @@ \newcommand{\ZZ}{\mathbb{Z}} \newcommand{\QQ}{\mathbb{Q}} \newcommand{\A}{\mathbb{A}} +\newcommand{\shcong}{\cong_{\mathrm{sh}}} +\newcommand{\grcong}{\cong_{\mathrm{gr}}} +\newcommand{\intval}[1]{\llbracket #1 \rrbracket} \title{Noncommutative Resolutions of Monomial Curves} \author{Markus M. R. Tripp} @@ -79,102 +86,185 @@ \begin{abstract} \end{abstract} +\section{Introduction} + +some results/terminology in section 2 are made independent of the characteristic + \section{Monomial curves} -\section{Frobenius maps on affine semigroups} -Let $\SGR$ be an affine semigroup, i.e., a finitely generated submonoid of $\ZZ^d$ for some $d \in \NN$, and let $R \coloneqq \kk[\SGR]$ denote its semigroup algebra, which defines a toric variety. Fix a positive integer $q$, and write $q^{-1}\SGR \coloneqq \{ a/q \mid a \in \SGR \} \subseteq \QQ^d$ for the affine semigroup $\SGR$ refined by $q$. Consider the semigroup algebra +We write $\NN = \{1, 2, 3, \dots \}$ for the set of natural numbers. For $a$, $b \in \ZZ$, we denote by $\intval{a,b}$ the set of all integers from $a$ to $b$, including both endpoints. + +Throughout, we fix an arbitrary ground field $\kk$. We begin by recalling some basic facts about toric algebras, focusing in particular on the one-dimensional case. See~\cite{Cox2011} for further details. + +Let $\SGR$ be an affine semigroup, that is, a finitely generated submonoid of $\ZZ^d$ for some $d \in \NN$. The \textbf{toric algebra} associated with $\SGR$ is the semigroup algebra +\[ R \coloneqq \kk[\SGR] = \spn_\kk \{ x^a \mid a \in \SGR \}. \] +The algebra $R$ is a finitely generated $\kk$-subalgebra of the Laurent polynomial ring~$\kk[x_1^{\pm 1}, \dots, x_d^{\pm 1}]$. In particular, it is a Noetherian domain and inherits the natural $\ZZ^d$-grading. Its Krull dimension is equal to the rank of $\gp(\SGR)$, where $\gp(\SGR)$ denotes the subgroup of $\ZZ^d$ generated by $\SGR$~\cite[Prop. 1.1.14]{Cox2011}. + +%For $\QQ^d$-graded $R$-modules $M$, $M_1$, \dots, $M_k$, we write $M \cong_{\textrm{sh}} M_1 \oplus \cdots \oplus M_k$ whenever there exist $q_i \in \QQ$ such that $M \cong M_1(d_1) \oplus \cdots \oplus M_k(d_k)$ as $\QQ^d$-graded $R$-modules. + +\todo{use widebar} + +\todo{adapt phrasinig in correspondence gap modules} + +\todo{use notation $\grcong$?} + +\todo{affine semigroup rings are (non-normal) toric varieties} + +\todo{1-dimensional toric varieties are monomial curves} + +\todo{define normal, seminormal, $G(S)$, weakly normal (i.e., for any $r \in \Frac(R)$, $r^p \in R$ implies $r \in R$), note that one can replace $\Frac(R)$ by $\kk[G(S)]$, and weakly normal implies seminormal} + +\todo{monomial curves are non-seminormal, non-normal, singular at 0} + +\todo{say what the normalization, fraction field is} + +\begin{lemma} +Let $M$, $M^\prime$ and $M^{\prime\prime}$ be $\QQ^d$-graded $R$-modules, and suppose that $M \shcong M^{\prime} \oplus M^{\prime\prime}$, then $\qpower{M} \shcong \qpower{M^{\prime}} \oplus \qpower{M^{\prime\prime}}$. +\end{lemma} + +\begin{proof} +By assumption, there are $d^\prime$, $d^{\prime\prime} \in \QQ$ such that $M \grcong M^\prime(d^\prime) \oplus M^{\prime\prime}(d^{\prime\prime})$. + +As the construction of the pseudo-Frobenius twist respects direct sums, we have +\[ \qpower{M} \grcong \qpower{(M^{\prime}(d^\prime))} \oplus \qpower{(M^{\prime\prime}(d^{\prime\prime}))} = (\qpower{M^{\prime}})(d^\prime/q) \oplus (\qpower{M^{\prime\prime}})(d^{\prime\prime}/q), \] +where the last equality follows from the computation +\begin{align*} +\left(\qpower{(M^{\prime}(d^\prime))} \oplus \qpower{\left(M^{\prime\prime}(d^{\prime\prime})\right)}\right)_\alpha &= \bigl(M^{\prime}(d^\prime)\bigr)_{q\alpha} \oplus \bigl(M^{\prime\prime}(d^{\prime\prime})\bigr)_{q\alpha} \\ +&= (M^{\prime})_{q\alpha + d^\prime} \oplus (M^{\prime\prime})_{q\alpha + d^{\prime\prime}} \\ +&= (\qpower{M^\prime})_{\alpha + \frac{d^\prime}{q}} \oplus (\qpower{M^{\prime\prime}})_{\alpha + \frac{d^{\prime\prime}}{q}} \\ +&= \left((\qpower{M^{\prime}})(d^\prime/q) \oplus (\qpower{M^{\prime\prime}})(d^{\prime\prime}/q)\right)_\alpha. +\end{align*} +\end{proof} + +\begin{proposition} +For the semigroup algebra $R = \kk[\SGR]$ of an affine semigroup $\SGR$, the category of finitely generated $\QQ^d$-graded $R$-modules is Krull-Schmidt. +\end{proposition} + +\section{Frobenius maps on toric algebras} +Let $\SGR$ be an affine semigroup, i.e., a finitely generated submonoid of $\ZZ^d$ for some $d \in \NN$, and let $R \coloneqq \kk[\SGR]$ denote its semigroup algebra. Fix a positive integer $q$, and write $q^{-1}\SGR = \{ a/q \mid a \in \SGR \} \subseteq \QQ^d$ for the rescaling of $\SGR$ by the factor $1/q$. Consider the semigroup algebra \[ R^{1/q} \coloneqq \kk[q^{-1}\SGR] = \spn_\kk\{ x^{a/q} \mid a \in \SGR \} \] -regarded as a subring of $\kk[\QQ^d] = \spn_\kk \{ x^\alpha \mid \alpha \in \QQ^d \}$, which carries a natural $\QQ^d$-grading. While the semigroup ring $R^{1/q}$ is defined for arbitrary $q \in \NN$ in any characteristic, the notation is motivated by the case $q = p^e$ with $\kk$ perfect of prime characteristic $p$, where $R^{1/q}$ is the ring of $q$-th roots of elements in $R$. Further, for $r = \sum_{a \in \QQ^d} k_a x^{a}$, we set $\qpower{r} \coloneqq \sum_{a \in \QQ^d} k_a x^{aq}$, and write $\qpower{F}$ for the resulting \emph{$q$-th pseudo-Frobenius (ring) endomorphism} of $R$, which maps $r$ to $\qpower{r}$. +regarded as a subring of $\kk[\QQ^d] = \spn_\kk \{ x^\alpha \mid \alpha \in \QQ^d \}$, which carries a natural $\QQ^d$-grading. While the semigroup ring $R^{1/q}$ is defined for arbitrary $q \in \NN$ in any characteristic, the notation is motivated by the case $q = p^e$ with $\kk$ perfect of prime characteristic $p$, where $R^{1/q}$ is the ring of $q$-th roots of elements in $R$. Further, for $r = \sum_{a \in \QQ^d} k_a x^{a} \in \kk[\QQ^d]$, we set $\qpower{r} \coloneqq \sum_{a \in \QQ^d} k_a x^{qa}$, and write $\qpower{F}$ for the resulting \emph{$q$-th pseudo-Frobenius (ring) endomorphism} of $R$, which maps $r$ to $\qpower{r}$. \begin{remark} % PREVIOUS VERSION % Note that even in positive characteristic $p$ with $q = p^e$, the $q$-th pseudo-Frobenius twist need not agree with the usual $q$-th Frobenius twist, where $R$-scalar multiplication is pulled back via the $e$-times iterated Frobenius endomorphism $F^e \colon r \mapsto r^{p^e}$ on $R$. The two agree precisely when the underlying field $\kk$ is finite, in which case $\kk \subseteq R$ is fixed pointwise by $F^e$. -Note that even in positive characteristic $p$ with $q = p^e$, the $q$-th pseudo-Frobenius endomorphism $\qpower{F}$ need not agree with the usual $e$-times iterated Frobenius endomorphism $F^e \colon R \to R$, mapping $r$ to $r^{q}$ . The two agree precisely when the underlying field $\kk$ is finite, in which case $\kk \subseteq R$ is fixed pointwise by $F^e$. +Note that even in positive characteristic $p$ with $q = p^e$, the $q$-th pseudo-Frobenius endomorphism $\qpower{F}$ need not agree with the usual $e$-times iterated Frobenius endomorphism $F^e \colon R \to R$, mapping $r$ to $r^{q}$. %The two agree precisely when the underlying field $\kk$ is finite of order $p^f$ with $f \mid e$, in which case $\kk \subseteq R$ is fixed pointwise by $F^e$. \end{remark} +\begin{proposition} +Assume that $\kk$ is a field of prime characteristic $p$ and that $q = p^e$. Then the $e$-times iterated Frobenius endomorphism $F^e$ and the $q$-th pseudo-Frobenius endomorphism $\qpower{F}$ agree precisely when $\kk$ is finite of order $p^f$ with $f \mid e$. +\end{proposition} + +\begin{proof} +First, note that for $r = \sum_{a \in \SGR} k_a x^a \in R$, we have $F^e(r) = \sum_{a \in \SGR} k_a^{q} x^{qa}$. Hence, $\qpower{F}$ and $F^e$ agree precisely when $F^e$ fixes $\kk\subseteq R$ pointwise. + +Suppose first that $\kk$ is finite of order $p^f$. By Lagrange's theorem, $k^{p^f} = k$ for every $k \in \kk$. If $f \mid e$, then repeated application of this identity gives $k^{p^e} = k$ for all $k \in \kk$. Now assume that $f \nmid e$. Since the multiplicative group of a finite field is cyclic~\cite[§ 9, Prop. 18]{Dummit2004}, there exists an element $k \in \kk \setminus \{ 0 \}$ of order $p^f -1$. We claim that $k^{p^e} \neq k$. Suppose, for contradiction, that $k^{p^e} = k$. Then $k^{p^e -1 } = 1$, and hence $p^f -1 \mid p^e -1$. It follows that +\[ p^f -1 = \gcd(p^e-1,p^f-1) = p^{\gcd(e,f)}-1, \] +where the last equality is a standard identity; see, for example,~\cite[Example 1.14]{Mollin2008}. Therefore, $f = \gcd(e,f)$, and hence $f \mid e$, contradicting our assumption. Finally, if $\kk$ is infinite, the polynomial $x^q - x \in \kk[x]$ cannot vanish on all of $\kk$, so there exists some $k \in \kk$ such that $k^q \neq k$. +\end{proof} + \begin{definition} -Let $M$ be an $R$-module. The \textbf{$q$-th pseudo-Frobenius twist} of $M$, denoted $\qpower{M}$, is the $R$-module obtained by restricting scalars along the \mbox{$q$-th} pseudo-Frobenius endomorphism $\qpower{F}$. That is, $\qpower{M}$ is the $R$-module with the same underlying abelian group as $M$, but with $R$-action given by~$r\cdot m \coloneqq \qpower{r}m$ +Let $M$ be a $\QQ^d$-graded $R$-module. The \textbf{$q$-th pseudo-Frobenius twist} of $M$, denoted $\qpower{M}$, is the $R$-module obtained by restricting scalars along the \mbox{$q$-th} pseudo-Frobenius endomorphism $\qpower{F}$. That is, $\qpower{M}$ is the $R$-module with the same underlying abelian group as $M$, but with $R$-action given by~$r \qmult m \coloneqq \qpower{r}m$ % EXPLICIT DECLARATION % \[ \Bigl( \sum_{a \in S} k_a x^a \Bigr) \cdot_q \; m \coloneqq \bigl( \sum_{a \in S} k_a x^{aq} \bigr) m \] for all $r \in R$ and $m \in M$. \end{definition} +\begin{remark} +The $q$-th pseudo-Frobenius twist of $M$ carries a $\QQ^d$-graded $R$-module structure when endowed with the homogeneous components $(\qpower{M})_\alpha \coloneqq M_{q \alpha}$ for $\alpha \in \QQ^d$. Indeed, +\[ R_\alpha \qmult (\qpower{M})_\beta \subseteq R_{q\alpha}M_{q \beta} \subseteq M_{q(\alpha + \beta)} = (\qpower{M})_{\alpha + \beta} \] +for all $\alpha$, $\beta \in \QQ^d$. +\end{remark} + \begin{lemma}\label{lemma:cannonc-isom} -There is a cannonical $R$-module isomorphism +There is a cannonical $\QQ^d$-graded $R$-module isomorphism \[ R^{1/q} \xrightarrow{\;\sim\;} \qpower{R}, \] defined by $r = \sum_{a \in \SGR} k_a x^{a/q} \mapsto \qpower{r} = \sum_{a \in \SGR} k_a x^a$. \end{lemma} \begin{proof} -By abuse of notation, we also write $\qpower{F}$ for the map $R^{1/q} \to \qpower{R}$ above; it is clearly additive and bijective. To verify compatibility with the respective $R$-module structures, write $r = \sum_{a \in \SGR} k_a x^a \in R$ and $m = \sum_{a \in \SGR} l_a x^{a/q} \in R^{1/q}$. Then -\[ \qpower{F}(rm) = \qpower{F}\Bigl(\sum_{a, b \in \SGR} k_a l_b x^{a + b/q}\Bigr) = \sum_{a, b \in \SGR} k_a l_b x^{aq + b}, \] -which is precisely $r\cdot\qpower{F}(m)$ in $\qpower{R}$. +By abuse of notation, we also denote by $\qpower{F}$ the map $R^{1/q} \to \qpower{R}$ defined above. This map is clearly additive, bijective and preserves the $\QQ^d$-grading. To verify that it is compatible with the respective $R$-module structures, let $r = \sum_{a \in \SGR} k_a x^a \in R$ and $m = \sum_{a \in \SGR} l_a x^{a/q} \in R^{1/q}$. Then +\[ \qpower{F}(rm) = \qpower{F}\Bigl(\sum_{a, b \in \SGR} k_a l_b x^{a + \frac{b}{q}}\Bigr) = \sum_{a, b \in \SGR} k_a l_b x^{qa + b}, \] +which is precisely $r \qmult \qpower{F}(m)$ in $\qpower{R}$. \end{proof} \begin{definition} -We say that $R$ is \textbf{$q$-split} if the Frobenius map $\qpower{F} \colon R \to \qpower{R}$ splits as a map of $R$-modules. Moreover, we call $R$ \textbf{$q$-weakly normal} if, for every $r \in \kk[G(\SGR)]$\todo{$\Frac(R)$?}, $\qpower{r} \in R$ implies $r \in R$. +We say that $R$ is \textbf{$q$-split} if the pseudo-Frobenius map $\qpower{F} \colon R \to \qpower{R}$ splits as a map of $R$-modules, that is, if there exists an $R$-module homomorphism $\sigma \colon \qpower{R} \to R$ such that $\sigma \circ \qpower{F} = \id_R$. Moreover, we call $R$ \textbf{$q$-weakly normal} if, for every $r \in \kk[\gp(\SGR)]$, $\qpower{r} \in R$ implies $r \in R$. \end{definition} \begin{remark} -In semigroup-thereotic terms, $R = \kk[\SGR]$ is $q$-weakly normal if and only if, for every $a \in G(\SGR)$, $qa \in S$ implies $a \in S$. In particular, $\SGR$ (or equivalently, $R$) is normal if and only if $R$ is $q$-weakly normal for all $q \in \NN$. +%Recall that $R$ is $q$-split if and only if $R$ is a direct summand of $\qpower{R}$; see, for example,~\cite[§ 10, Prop. 26]{Dummit2004}. +By~\cite[Prop. 2.3.1]{Nastasescu2004}, the map $\qpower{F}$ splits as an $R$-module homomorphism if and only if it splits in the category of $\QQ^d$-graded $R$-modules. In other words, whenever $\qpower{F}$ splits, the splitting map $\sigma$ may be chosen to be $\QQ^d$-graded. Hence, $R$ is $q$-split if and only if it is a graded direct summand of $\qpower{R}$, that is, if there exists a $\QQ^d$-graded $R$-module $N$ such that $\qpower{R}\cong R \oplus N$ as $\QQ^d$-graded $R$-modules. \end{remark} -In positive characteristic $p$, the Frobenius endomorphism yields $\qpower[p]{r} \in R$ if and only if $r^p \in R$, from which we obtain the following: +%\begin{remark} +%In semigroup-thereotic terms, $R = \kk[\SGR]$ is $q$-weakly normal if and only if, for every $a \in G(\SGR)$, $qa \in S$ implies $a \in S$. In particular, $\SGR$ (or equivalently, $R$) is normal if and only if $R$ is $q$-weakly normal for all $q \in \NN$. +%\end{remark} + +In prime characteristic $p$, the Frobenius endomorphism yields $\qpower[p]{r} \in R$ if and only if $r^p \in R$, from which we obtain the following. \begin{corollary} -If $\kk$ has prime characteristic $p$, then $R$ is $p$-weakly normal +Assume that $\kk$ is a field of prime characteristic $p$. Then $R$ is $p$-weakly normal % MORE INFORMATION % (or, equivalently, $p^e$-weakly normal for all $e \in \NN$) -if and only if $R$ is weakly normal (i.e., for any $r \in \Frac(R)$, $r^p$ implies $r \in R$). In particular, $R$ is seminormal in this case. +if and only if $R$ is weakly normal. In particular, $R$ is seminormal in this case. \end{corollary} -\begin{proposition}\textnormal{\cite{Hochster1976}, \cite[Prop. 5.3]{Yasuda2009}} +It is known that, in prime characteristic, the Frobenius map splits as a map of $R$-modules if and only if $R$ is weakly normal. The argument of \cite[Prop. 5.3]{Yasuda2009} extends verbatim to the characteristic-independent setting upon replacing the Frobenius map by the pseudo-Frobenius map. For the reader's convenience, we include the argument here. + +\begin{proposition}\textnormal{\cite[Prop. 5.31]{Hochster1976}, \cite[Prop. 5.3]{Yasuda2009}} $R$ is $q$-split if and only if it is $q$-weakly normal. \end{proposition} \begin{proof} % FAKE PROOF % Suppose $R$ is $q$-split, with $\sigma \colon \qpower{R} \to R$ an $R$-module homomorphism satisfying $\sigma \circ \qpower{F} = \id_R$; in particular, $\sigma(1) = 1$. Now let $r \in \Frac(R)$ be such that $\qpower{r} \in R$. Then $r = r\sigma(1) = \sigma(\qpower{r}) \in R$. -Suppose $R$ is $q$-split, with $\sigma \colon \qpower{R} \to R$ an $R$-module homomorphism satisfying $\sigma \circ \qpower{F} = \id_R$. Let $r \in \kk[G(\SGR)]$ be such that $\qpower{r} \in R$, and choose $s \in R\setminus \{0 \}$ with $sr \in R$. Then -\[ sr = \sigma\bigl( \qpower{F}(sr) \bigr) = \sigma(\qpower{s}\qpower{r}) = s\sigma(\qpower{r}), \] +Suppose $R$ is $q$-split, with $\sigma \colon \qpower{R} \to R$ an $R$-module homomorphism satisfying $\sigma \circ \qpower{F} = \id_R$. Let $r \in \kk[\gp(\SGR)]$ be such that $\qpower{r} \in R$, and choose $s \in R\setminus \{0 \}$ with $sr \in R$. Then +\[ sr = \sigma\bigl( \qpower{F}(sr) \bigr) = \sigma(\qpower{s}\qpower{r}) = \sigma(s \qmult (\qpower{r})) = s\sigma(\qpower{r}), \] and cancelling $s$ yields $r = \sigma(\qpower{r}) \in R$. -Conversely, suppose that $R$ is $q$-weakly normal. Let $\sigma \colon \qpower{R} \to R$ be the map that sends $f \in \qpower{R}$ to the unique $g \in R$ with $f = \qpower{g}$, if such a $g$ exists, and to $0$ otherwise. Clearly $\sigma \circ \qpower{F} = \id_R$, so it remains to show that $\sigma$ is a homomorphism of $R$-modules. As $\sigma$ is additive, it suffices to check that $\sigma(x^{aq+b}) = x^a \sigma(x^b)$ for $a$, $b \in \SGR$, which is straightforward using the assumption that $R$ is $q$-weakly normal. % MORE EXPLANATION % Suppose first that $b = qc$ for some $c \in \SGR$. Then $\sigma(x^{aq+b}) = \sigma(x^{q(a+c)}) = x^{a+c} = x^a \sigma(x^b)$. Now suppose there exists no $c \in \SGR$ with $b = qc$, so $\sigma(x^b) = 0$. If $aq+b = qc$ for some $c \in \SGR$, then $\qpower{x^{c-a}} = x^b \in R$, so $x^{c-a} \in R$ by assumption, giving $c-a \in \SGR$ and $b = q(c-a)$, a contradiction. Hence $\sigma(x^{aq+b}) = 0 = x^a\sigma(x^b)$. +Conversely, suppose that $R$ is $q$-weakly normal. Let $\sigma \colon \qpower{R} \to R$ be the map that sends $f \in \qpower{R}$ to the unique $g \in R$ with $f = \qpower{g}$, if such a $g$ exists, and to $0$ otherwise. Clearly $\sigma \circ \qpower{F} = \id_R$, so it remains to show that $\sigma$ is a homomorphism of $R$-modules. As $\sigma$ is additive, it suffices to check that $\sigma(x^a \qmult x^b) = x^a \sigma(x^b)$ for $a$, $b \in \SGR$. Suppose first that $b = qc$ for some $c \in \SGR$. Then +\[ \sigma(x^a \qmult x^b) = \sigma(x^{q(a+c)}) = x^{a+c} = x^a \sigma(x^b). \] +Now suppose that $b$ is not of the form $b = qc$ for some $c \in \SGR$, meaning that $\sigma(x^b) = 0$. We claim that $qa + b$ is also not of the form $qa+ b = qc$ for some $c \in \SGR$. Indeed, suppose that there exists a $c \in \SGR$ such that $qa + b = qc$. Then $\qpower{(x^{c-a})} = x^b \in R$. By assumption, this implies that $x^{c-a} \in R$, and hence $c-a \in \SGR$. Consequently, $b = q(c-a)$, contradicting our assumption on $b$. It follows that $\sigma(x^a \qmult x^b) = \sigma(x^{qa+b}) = 0$. \end{proof} \begin{corollary}\textnormal{\cite[Cor. 6.3]{Bruns2006}} -$R$ is normal if and only if $R$ is $q$-split for all $q \in \NN$. +$R$ is normal if and only if $R$ is $q$-weakly normal for all $q \in \NN$ if and only if $R$ is $q$-split for all $q \in \NN$. \end{corollary} +\begin{proof} +By~\cite[Prop. 1]{Hochster1972}, the semigroup algebra $R = \kk[\SGR]$ is normal if and only if $\SGR$ is saturated, that is, if for every $a \in \gp(\SGR)$ and $q \in \NN$, the condition $qa \in \SGR$ implies $a \in \SGR$. This is precisely the semigroup-theoretic formulation of $R$ being $q$-weakly normal for all $q \in \NN$. By the previous proposition, this is equivalent to $R$ being $q$-weakly normal for every $q \in \NN$. +\end{proof} + \begin{lemma}\textnormal{\cite[Lemma 5.0.3]{Blickle2013}}\label{lemma:q-split} -Let $q = p_1^{e_1} \cdots p_k^{e_k}$, where the $p_i$ are distinct primes. Then $R$ is $q$-split if and only if $R$ is $(p_1^{\ell_1}\cdots p_k^{\ell_k})$-split for all $(\ell_1, \dots, \ell_k) \in \NN_0^k$. +Let $q = p_1^{e_1} \cdots p_k^{e_k}$, where the $p_i$ are distinct primes and $e_i \in \NN$. Then $R$ is $q$-split if and only if $R$ is $(p_1^{\ell_1}\cdots p_k^{\ell_k})$-split for all $(\ell_1, \dots, \ell_k) \in \NN_0^k$. \end{lemma} \begin{proof} -Assume that $R$ is $q$-split, i.e., there exists a $R$-module homomorphism $\sigma \colon \qpower{R} \to R$ such that $\sigma \circ \qpower{F} = \id_R$. We first claim that, for each $i \in [1,k]$, $R$ is $p_i^{\ell_i}$ split for every $\ell_i \in \NN_0$. To this end, fix an $i \in [1,k]$ and consider the $R$-module homomorphism $f_i \colon \qpower[p_i]{R} \to \qpower{R}$, defined by $r \mapsto \qpower[q/p_i]{r}$. Setting $\sigma_i \coloneqq \sigma \circ f_i$, we obtain that +Assume that $R$ is $q$-split, i.e., there exists a $R$-module homomorphism $\sigma \colon \qpower{R} \to R$ such that $\sigma \circ \qpower{F} = \id_R$. We first claim that, for each $i \in \intval{1,k}$, $R$~is $p_i^{\ell_i}$-split for every $\ell_i \in \NN_0$. To this end, fix an $i \in \intval{1,k}$ and consider the $R$-module homomorphism $f_i \colon \qpower[p_i]{R} \to \qpower{R}$, defined by $r \mapsto \qpower[q/p_i]{r}$. Setting $\sigma_i \coloneqq \sigma \circ f_i$, we obtain that \[ \sigma_i(\qpower[p_i]{F}(r)) = \sigma(f_i(\qpower[p_i]{r})) = \sigma(\qpower{r}) = r \] for all $r \in R$, so $R$ is $p_i$-split. Now, an induction on $\ell_i$ shows that, as maps of sets, $\sigma_i^{\ell_i} \circ \qpower[p_i^{\ell_i}]{F} = \id_R$ for all $\ell_i \in \NN_0$, where $\sigma_i^{\ell_i}$ denotes the $\ell_i$-fold composition of $\sigma_i$. Moreover, $\sigma_i^{\ell_i}$, regarded as a map $\qpower[p_i^{\ell_i}]{R} \to R$, is readily verified to be $R$-linear, providing the claim. -Finally, fix $(\ell_1, \dots, \ell_k) \in \NN_0^k$, and set $q^\prime \coloneqq p_1^{\ell_1} \cdots p_k^{\ell_k}$. By the first part, $R$ is $p_i^{\ell_i}$-split for each $i \in [0,k]$. Hence, there exist $R$-module homomorphisms $\sigma_i^\prime \colon \qpower[p_i^{\ell_i}]{R} \to R$ such that $\sigma_i^\prime \circ \qpower[p_i^{\ell_i}]{F} = \id_R$. Setting $\sigma^\prime \coloneqq \sigma_1^\prime \circ \cdots \circ \sigma_k^\prime$, a direct verification shows that $\sigma^\prime$ gives the required splitting of $\qpower[q^\prime]{F}$. +Finally, fix $(\ell_1, \dots, \ell_k) \in \NN_0^k$, and set $q^\prime \coloneqq p_1^{\ell_1} \cdots p_k^{\ell_k}$. By the first part, $R$ is $p_i^{\ell_i}$-split for each $i \in \intval{1,k}$. Hence, there exist $R$-module homomorphisms $\sigma_i^\prime \colon \qpower[p_i^{\ell_i}]{R} \to R$ such that $\sigma_i^\prime \circ \qpower[p_i^{\ell_i}]{F} = \id_R$. Setting $\sigma^\prime \coloneqq \sigma_1^\prime \circ \cdots \circ \sigma_k^\prime$, a direct verification shows that $\sigma^\prime$ gives the required splitting of $\qpower[q^\prime]{F}$. \end{proof} \section{$R$-module decomposition of $R^{1/q}$} -We now specialize to the case where $\SGR = \langle a_1, \dots, a_n \rangle$ %= \{ c_1a_1 + \cdots + c_na_n \mid c_i \in \NN_0\} \subseteq \NN_0 \} -is a numerical semigroup and $R = \kk[\SGR]$ is the corresponding monomial curve. Following Shibuta~\cite{Shibuta2017}, we sketch how $R^{1/q} = \kk[x^{a_1/q}, \dots, x^{a_n/q}]$ decomposes as an $R$-module. Observe that +We now specialize to the case where $\SGR = \langle a_1, \dots, a_n \rangle \subseteq \NN_0$ %= \{ c_1a_1 + \cdots + c_na_n \mid c_i \in \NN_0\} \subseteq \NN_0 \} +is a numerical semigroup and $R = \kk[\SGR]$ is the corresponding semigroup algebra. Following Shibuta~\cite{Shibuta2017}, we sketch, for a fixed $q \in \NN$, how $R^{1/q} = \kk[x^{a_1/q}, \dots, x^{a_n/q}]$ decomposes as a~$\QQ_{\geq 0}$-graded $R$-module. Observe that \[ -\Gamma_q \coloneqq \left\{ \bigl( x^{a_1/q} \bigr)^{e_1} \cdots \bigl( x^{a_n/q} \bigr)^{e_n} \mid e_i \in [0,q-1] \right\} +\Gamma_q \coloneqq \left\{ \bigl( x^{a_1/q} \bigr)^{e_1} \cdots \bigl( x^{a_n/q} \bigr)^{e_n} \mid e_i \in \intval{0,q-1} \right\} \] is a homogeneous generating set for $R^{1/q}$. We partition $\Gamma_q$ by degree modulo $\ZZ$: writing each degree as $a/q$ with $a \in \NN_0$, this amounts to grouping elements by the residue of $a$ modulo $q$. This yields \[ \Gamma_q = \Gamma_{q,0} \sqcup \Gamma_{q,1} \sqcup \cdots \sqcup \Gamma_{q,q-1}, \] -where $\Gamma_{q,r} \coloneqq \{ x^{a/q} \in \Gamma_q \mid a \equiv r \pmod{q}\}$. Let $I_{q,r}$ be the $R$-submodule of $\kk[x^q \mid q \in \QQ]$ +where $\Gamma_{q,r} \coloneqq \{ x^{a/q} \in \Gamma_q \mid a \equiv r \pmod{q}\}$. Let $I_{q,r}$ be the $R$-submodule of $\kk[\QQ_{\geq 0}] \coloneqq \kk[x^q \mid q \in \QQ_{\geq 0}]$ generated by $\Gamma_{q,r}$. As a $\kk$-vector space, \begin{equation}\label{eq:ind-summand-as-vector-space} I_{q,r} = \spn_\kk \{ x^{a/q} \mid a \in \SGR, \, a \equiv r \pmod{q} \}, \end{equation} -and, by construction, summing over $r$ gives the $R$-module decomposition +and, by construction, summing over $r$ gives the $\QQ_{\geq 0}$-graded $R$-module decomposition \[ -R^{1/q} = \bigoplus_{r \in [0,q-1]} I_{q,r}, +R^{1/q} = \bigoplus_{r \in \intval{0,q-1}} I_{q,r}, \] where each summand is indecomposable (see Lemma~\ref{lemma:rank-1}). @@ -189,40 +279,61 @@ where each summand is indecomposable (see Lemma~\ref{lemma:rank-1}). %\todo{note $\Frac{R} = \kk(x)$} %\end{proof} -Moreover, writing $m_{q,r}/q$ for the minimal degree occurring among -elements of $\Gamma_{q,r}$, we have -\[ -I_{q,r} = x^{m_{q,r}/q} \cdot \langle x^\alpha \mid \alpha \in \NN_0, \, m_{q,r}/q + \alpha \in \Gamma_{q,r} \rangle_R, -\] -from which it follows that every indecomposable summand of $R^{1/q}$ is isomorphic to $I(A) \coloneqq \langle x^\alpha \mid \alpha \in A \cup \{ 0 \} \rangle_R$ for some $A \subseteq \gapset$\todo{name?}. We may always choose $A$ to be minimal with respect to the order $\preceq_\SGR$ induced by $\SGR$, where $\alpha \preceq_\SGR \beta$ means $\beta - \alpha \in \SGR$. Write -\[ \Min(\SGR) \coloneqq \{ A \in \gapset \mid A \text{ is minimal with respect to } \preceq_\SGR \}. \] -In particular, as $q$ ranges over $\NN$, only finitely many indecomposable summands of $R^{1/q}$ occur, up to isomorphism. +\begin{definition}\label{def:gap-module} +Let $A\subseteq \NN$. The (normalized) \textbf{gap module defined by $A$} is the $\NN_0$-graded $R$-submodule of $\bar{R}=\kk[x]$ given by +\[ I(A) \coloneqq \langle x^\alpha \mid \alpha \in A\cup \{0\} \rangle_R. \] +\end{definition} + +\begin{remark} +Clearly, in Definition~\ref{def:gap-module}, the set $A$ may always be chosen as a subset of the gap set $\gapset$. In particular, there are only finitely many gap modules. Moreover, we may assume that $A$ consists of elements that are minimal with respect to the partial order $\preceq_\SGR$ induced by $\SGR$, where $\alpha \preceq_\SGR \beta$ +means $\beta - \alpha \in \SGR$. We write +\[ \Min(\SGR) \coloneqq \{ A \subseteq \gapset \mid A \text{ is minimal with respect to } \preceq_\SGR \}. \] +\end{remark} + +\begin{corollary} +Every $\QQ_{\geq 0}$-graded indecomposable summand of $R^{1/q}$ is, up to a shift in degree, isomorphic to a gap module. +\end{corollary} + +\begin{proof} +Let $\frac{m_{q,r}}{q}$ denote the minimal degree occurring among the elements of $\Gamma_{q,r}$, and set +\[ A_{q,r} \coloneqq \Bigl\{ \alpha \in \NN \, \Bigm| \, \frac{m_{q,r}}{q} + \alpha \in \Gamma_{q,r} \Bigr\}, \] +from which it follows that $I_{q,r} = x^{\frac{m_{q,r}}{q}} \cdot I(A_{q,r})$. +%\[ +%I_{q,r} = x^{m_{q,r}/q} \cdot \Big\langle x^\alpha \mid \alpha \in \NN_0, \, \frac{m_{q,r}}{q} + \alpha \in \Gamma_{q,r} \Big\rangle_R. +%\] +\end{proof} + +In particular, as $q$ ranges over $\NN$, only finitely many isomorphism classes of $\QQ_{\geq 0}$-graded indecomposable summands of $R^{1/q}$ occur, up to shifts in degree. More precisely, for every $q \in \NN$, there exist $m_A \in \NN_0$ for $A \in \Min(\SGR)$ such that +\[ R^{1/q} \shcong \bigoplus_{A\in\Min(\SGR)} I(A)^{\oplus m_A}. \] + +%In particular, as $q$ ranges over $\NN$, only finitely many isomorphism classes of $\QQ_{\geq 0}$-graded indecomposable summands of $R^{1/q}$ occur, up to shifts in degree. More precisely, for every $q \in \NN$, there exist non-negative integers $m_A$ for $A \in \Min(\SGR)$ and non-negative rational numbers $d_{A,j}$ for $j \in [m_A]$ such that +%\[ R^{1/q} \cong \bigoplus_{A\in\Min(\SGR)} \bigoplus_{j \in [m_A]} x^{d_{A,j}} \cdot I(A) \] +%as $\QQ_{\geq 0}$-graded modules. + \begin{lemma}\label{lemma:rank-1} -For each $A \in \Min(\SGR)$, the $R$-module $I(A)$ has rank $1$ and is therefore indecomposable. Moreover, the assignment +For each $A \in \Min(\SGR)$, the gap module $I(A)$ has rank $1$ and is therefore indecomposable. Moreover, the assignment \[ A \mapsto [I(A)] \] -defines a bijection between $\Min(\SGR)$ and the set of isomorphism classes of homogeneous $R$-submodules of $\bar{R} = \kk[x]$. +defines a bijection between $\Min(\SGR)$ and the set of isomorphism classes of $\NN_0$-graded $R$-submodules of $\bar{R} = \kk[x]$. \end{lemma} \begin{proof} For simplicity, set $Q \coloneqq \Frac(R) = \kk(x)$ and fix $A \in \Min(\SGR)$. Observe that $Q \otimes_R I(A)$ is naturally isomorphic, as a $Q$-vector space, to $I(A)_{(0)} \subseteq Q$. Since $I(A)$ contains $1$, it follows that $I(A)_{(0)} = Q$ and $I(A)$ has rank $1$. -For the second part, it is clear that $I(A)$, $A \in \Min(\SGR)$, are homogeneous $R$-submodules of $\bar{R}$, so the map is well-defined. To show injectivity, suppose we are given an $R$-module isomorphism $\varphi \colon I(A) \to I(B)$ for some $A$, $B \in \Min(\SGR)$. This induces an isomorphism of $Q$-vector spaces $\varphi_{(0)} \colon I(A)_{(0)} \to I(B)_{(0)}$, where both sides equal~$Q$. -%and $\varphi_{(0)}(f/1) = \varphi(f)$ for all $f \in I(A)$. -In particular, $\varphi_{(0)}$ (and therefore $\varphi$) is given by multiplication with some $g_0 \in Q$, so $\varphi(f) = g_0 f$ for all $f \in I(A)$. Taking $f=1$ yields $g_0 = \varphi(1) \in I(B)$. Since $\varphi$ is surjective and $1 \in I(B)$, a degree argument yields $g_0 \in \kk\setminus\{ 0 \}$. Consequently, -\[ I(B) = \varphi(I(A)) = g_0 \cdot I(A) = I(A), \] -and because $A$ and $B$ are both minimal with respect to $\preceq_\SGR$, it follows that~$A = B$. Now, let $I \subseteq \bar{R}$ be an arbitrary homogeneous $R$-submodule. Because $R$ is Noetherian and $\bar{R}$ is finitely generated as an $R$-module, $I$ is finitely generated, i.e, $I = \langle x^\beta \mid \beta \in B\rangle_R$ for some finite subset $B \subseteq \NN_0$. Define $m \coloneqq \min(B)$ and $A^\prime \coloneqq \{ \beta - m \mid \beta \in B \} \setminus \{ 0 \}$. Consequently, $I = x^m \cdot I(A^\prime)$, which implies that $I \cong I(A^\prime)$. The claim then follows from the fact that $I(A^\prime)=I(A)$, where $A$ denotes the set of minimal elements of $A^\prime \cap (\gapset)$ with respect to $\preceq_\SGR$. +For the second part, each $I(A)$ with $A \in \Min(\SGR)$ is an $\NN_0$-graded $R$-submodule of $\bar{R}$, so the map is well-defined. To prove injectivity, suppose that $\varphi \colon I(A)\to I(B)$ is an $R$-module isomorphism for some $A$, $B \in \Min(\SGR)$. Localizing at $(0)$ yields an isomorphism of $Q$-vector spaces $\varphi_{(0)}\colon I(A)_{(0)} \to I(B)_{(0)}$, where both source and target are equal to $Q$. Hence $\varphi$, and thus also $\varphi_{(0)}$, is given by multiplication by some $g_0 \in Q$, so $\varphi(f) = g_0f$ for all $f \in I(A)$. A simple degree argument shows that $g_0 \in \kk \setminus \{ 0 \}$. Therefore, +\[ I(B) = \varphi(I(A)) = g_0 \cdot I(A) =I(A). \] +Since $A$ and $B$ are both minimal with respect to $\preceq_\SGR$, it follows that $A = B$. Now, let $I \subseteq \bar{R}$ be an arbitrary $\NN_0$-graded $R$-submodule. Because $R$ is Noetherian and $\bar{R}$ is finitely generated as an $R$-module, $I$ is finitely generated, i.e, $I = \langle x^\beta \mid \beta \in B\rangle_R$ for some finite subset $B \subseteq \NN_0$. Define $m \coloneqq \min(B)$ and $A^\prime \coloneqq \{ \beta - m \mid \beta \in B \} \setminus \{ 0 \}$. Consequently, $I = x^m \cdot I(A^\prime)$, which implies that $I \cong I(A^\prime)$. The claim then follows from the fact that $I(A^\prime)=I(A)$, where $A$ denotes the set of minimal elements of $A^\prime \cap (\gapset)$ with respect to $\preceq_\SGR$. \end{proof} \begin{remark} -We now present another perspective on the $R$-modules $I(A)$. For $A \in \Min(\SGR)$, the $R$-module $I(A)$ admits the $\kk$-vector space description +We now present another perspective on the gap modules $I(A)$. For $A \in \Min(\SGR)$, the $R$-module $I(A)$ admits the $\kk$-vector space description \[ I(A)=\spn_\kk \{x^\alpha \mid \alpha\in \SGR_A \}, \] where $\SGR_A\coloneqq (A \cup \{ 0 \}) +\SGR$. The set $\SGR_A$ is a normalized relative ideal of $\SGR$, that is, a subset of $\NN_0$ containing $0$ that is closed under addition by elements of $\SGR$. One readily checks that $A \mapsto \SGR_A$ defines a one-to-one correspondence between the elements of $\Min(\SGR)$ and the normalized relative ideals of $\SGR$. \todo{reference for this terminology?} \end{remark} \begin{corollary} -For each $A \in \Min(\SGR)$, the $R$-module $I(A)$ is maximal Cohen--Macaulay. +For each $A \in \Min(\SGR)$, the gap module $I(A)$ is maximal Cohen--Macaulay. \end{corollary} \begin{proof} @@ -233,7 +344,7 @@ $I(A)_\mathfrak{m}$ is a non-zero $R_\mathfrak{m}$-submodule of $\kk(x)$, hence \todo{add direct sums are Cohen--Macaulay (Bruns/Herzog, 1993, Prop. 1.2.9)} \begin{lemma} -For $q \in \NN$, it holds that $\qpower{\bar{R}} = \bar{R}^{\oplus q}$ as $R$-module. +For $q \in \NN$, it holds that $\qpower{\bar{R}} \shcong \bar{R}^{\oplus q}$. \end{lemma} \begin{proof} @@ -278,7 +389,7 @@ Let $\SGR = \langle 3,4 \rangle \subseteq \NN_0$ and $R = \kk[\SGR] = \kk[x^3,x^ \end{center} The set -\[ \{ x^{(3/q)e_1 + (4/q)e_2} : e_i \in [0,q-1] \} \] +\[ \{ x^{(3/q)e_1 + (4/q)e_2} : e_i \in \intval{0,q-1} \} \] generates $R^{1/q} = \kk[x^{3/q},x^{4/q}]$ as an $R$-module. For $q =2$, this generating set is $\{ 1, x^{3/2}, x^2, x^{7/2} \}$, and grouping the numerators of the exponents by residue modulo $2$ yields the $R$-module decomposition \[ R^{1/2} = \langle 1, x^2 \rangle_R \oplus x^{3/2} \cdot \langle 1, x^2 \rangle_R \cong I(\{2 \})^{\oplus 2} \] For $q = 4$, a direct computation shows $R^{1/4} \cong \bar{R}^{\oplus 4}$. Keeping in mind that $R^{1/q}$ is isomorphic to $\qpower{R}$ and that the construction of the pseudo-Frobenius twist respects direct sums, we conclude that @@ -289,17 +400,20 @@ where the last isomorphism follows from the previous lemma. By induction, we obt % NOTE: next proposition holds when q \in \SGR \begin{proposition}\label{prop:asy-behav} -For $q \in \SGR$, every indecomposable summands of $R^{1/q}$ is isomorphic to the normalization $\bar{R} = \kk[x]$ of $R$. That is, $R^{1/q} \cong \bar{R}^{\oplus q}$. In particular, this holds for all sufficiently large $q$. +It holds that $q \in \SGR$ if and only if every indecomposable summand of $R^{1/q}$ is isomorphic to the normalization $\bar{R} = \kk[x]$ of $R$, that is, $R^{1/q} \cong \bar{R}^{\oplus q}$. In particular, this holds for all sufficiently large $q$. +%For $q \in \SGR$, every indecomposable summands of $R^{1/q}$ is isomorphic to the normalization $\bar{R} = \kk[x]$ of $R$. That is, $R^{1/q} \cong \bar{R}^{\oplus q}$. In particular, this holds for all sufficiently large $q$. \end{proposition} \begin{proof} -Fix $r \in [0,q-1]$. For simplicity, set +Assume that $q \in SGR$, and fix $r \in \intval{0,q-1}$. For simplicity, set \[ \supp(I_{q,r}) \coloneqq \{ a/q \mid a \in \SGR, \, a \equiv r \pmod{q} \} \] and observe that $\supp(I_{q,r}) \subseteq \{ m_{q,r}/q + n \mid n \in \NN_0\}$, where $m_{q,r}/q$ is the minimum of $\supp(I_{q,r})$. Conversely, for each $n \in \NN_0$, the choice of $q$ ensures $nq \in \SGR$, and hence $m_{q,r} + nq \in \SGR$. Since $m_{q,r}+nq \equiv r \pmod q$, we obtain $m_{q,r}/q + n \in \supp(I_{q,r})$, and altogether \[ \supp(I_{q,r}) = \{ m_{q,r}/q + n \mid n \in \NN_0\}. \] It now follows from~\eqref{eq:ind-summand-as-vector-space} that \[ I_{q,r} = \spn_\kk \{ x^{m_{q,r}/q + n} \mid n \in \NN_0 \} = x^{m_{q,r}/q} \cdot \spn_\kk\{ x^n \mid n \in \NN_0 \} \cong k[x] . \] + + \end{proof} \todo[inline]{think about what can be said about decompositions of $R^{1/q}$ for $q \notin \SGR$} @@ -329,7 +443,7 @@ Let $p$ be a prime divisor of $q$. If $R^{1/q}$ and $R^{1/(pq)}$ have the same i \begin{proof} Decomposing $R^{1/q} \cong \bigoplus_{i=1}^k I(A_i)^{\oplus n_i}$ for pairwise distinct $A_i \in \Min(\SGR)$, it holds that \[ R^{1/(pq)} \cong \qpower[p]{(R^{1/q})} \cong \bigoplus_{i=1}^k \bigl(\qpower[p]{I(A_i)}\bigr)^{\oplus n_i}. \] -Thus, by assumption, the indecomposable summands occuring in the $R$-modules $\qpower[p]{I(A_i)}$, $i \in [1,k]$, are, up to isomorphism, precisely $I(A_1)$, \dots $I(A_k)$. Repeated application of this observation shows that $R^{1/(p^e q)}$ has the same indecomposable summands, up to isomorphism, for every $e \in \NN$. +Thus, by assumption, the indecomposable summands occuring in the $R$-modules $\qpower[p]{I(A_i)}$, $i \in \intval{1,k}$, are, up to isomorphism, precisely $I(A_1)$, \dots $I(A_k)$. Repeated application of this observation shows that $R^{1/(p^e q)}$ has the same indecomposable summands, up to isomorphism, for every $e \in \NN$. \end{proof} \section{Global dimension of endomorphism algebras} diff --git a/references.bib b/references.bib index 4b0e24b..e893191 100644 --- a/references.bib +++ b/references.bib @@ -115,4 +115,64 @@ url = {https://www.sciencedirect.com/science/article/pii/S0001870819301823}, author = {Eleonore Faber and Greg Muller and Karen E. Smith}, keywords = {Toric variety, Non-commutative resolution, Finite global dimension, Non-commutative crepant resolution, Rings of differential operators, Strong F-regularity}, abstract = {Let R be the coordinate ring of an affine toric variety. We prove, using direct elementary methods, that the endomorphism ring EndR(A), where A is the (finite) direct sum of all (isomorphism classes of) conic R-modules, has finite global dimension equal to the dimension of R. This gives a precise version, and an elementary proof, of a theorem of Špenko and Van den Bergh implying that EndR(A) has finite global dimension. Furthermore, we show that EndR(A) is a non-commutative crepant resolution if and only if the toric variety is simplicial. For toric varieties over a perfect field k of prime characteristic, we show that the ring of differential operators Dk(R) has finite global dimension.} +} + +@book{Dummit2004, +author = {Dummit, David S. and Foote, Richard M.}, +title = {Abstract Algebra}, +edition = {3}, +publisher = {John Wiley \& Sons}, +location = {Hoboken, NJ}, +year = {2004}, +} + +@book{Mollin2008, +location = {Boca Raton}, +edition = {2nd ed}, +title = {Fundamental number theory with applications}, +isbn = {978-1-4200-6659-3}, +series = {Discrete mathematics and its applications}, +pagetotal = {369}, +publisher = {Chapman \& Hall/{CRC}}, +author = {Mollin, Richard A.}, +date = {2008}, +} + +@book{Nastasescu2004, +location = {Berlin ; New York}, +title = {Methods of graded rings}, +isbn = {978-3-540-20746-7}, +series = {Lecture notes in mathematics}, +pagetotal = {304}, +number = {1836}, +publisher = {Springer}, +author = {Năstăsescu, Constantin and Oystaeyen, Freddy Van}, +date = {2004}, +keywords = {Associative rings, Graded rings}, +} + +@article{Hochster1972, + ISSN = {0003486X, 19398980}, + URL = {http://www.jstor.org/stable/1970791}, + author = {Melvin Hochster}, + journal = {Annals of Mathematics}, + number = {2}, + pages = {318--337}, + publisher = {[Annals of Mathematics, Trustees of Princeton University on Behalf of the Annals of Mathematics, Mathematics Department, Princeton University]}, + title = {Rings of Invariants of Tori, Cohen-Macaulay Rings Generated by Monomials, and Polytopes}, + urldate = {2026-09-10}, + volume = {96}, + year = {1972} +} + +@book{Cox2011, +author = {Cox, David A. and Little, John B. and Schenck, Henry K.}, +title = {Toric Varieties}, +series = {Graduate Studies in Mathematics}, +volume = {124}, +publisher = {American Mathematical Society}, +address = {Providence, RI}, +year = {2011}, +isbn = {978-0-8218-4819-2}, +doi = {10.1090/gsm/124} } \ No newline at end of file From 575cf4acb81a81d7f8ab1bfc99931a64362fe36f Mon Sep 17 00:00:00 2001 From: Markus Maximilian Rudolf Tripp Date: Thu, 17 Sep 2026 22:12:28 +0200 Subject: [PATCH 7/8] first part of the preliminaries --- ncr-of-monomial-curves.tex | 135 +++++++++++++++++++++---------------- references.bib | 32 +++++++++ 2 files changed, 110 insertions(+), 57 deletions(-) diff --git a/ncr-of-monomial-curves.tex b/ncr-of-monomial-curves.tex index cfb1468..a158a81 100644 --- a/ncr-of-monomial-curves.tex +++ b/ncr-of-monomial-curves.tex @@ -7,6 +7,7 @@ \usepackage{todonotes} \usepackage{blindtext} \usepackage{tikz} +\usepackage[most]{tcolorbox} \usepackage{graphicx} \usetikzlibrary{arrows.meta,positioning,decorations.pathreplacing} %\usepackage{parskip} @@ -26,6 +27,8 @@ \clearlist{language}% } +\numberwithin{equation}{section} + \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} @@ -54,6 +57,7 @@ } \DeclareMathOperator{\supp}{supp} +\DeclareMathOperator{\Spec}{Spec} \DeclareMathOperator{\Frac}{Frac} \DeclareMathOperator{\spn}{span} \DeclareMathOperator{\Min}{Min} @@ -88,9 +92,9 @@ \section{Introduction} -some results/terminology in section 2 are made independent of the characteristic +% some results/terminology in section 2 are made independent of the characteristic -\section{Monomial curves} +\section{Preliminaries} We write $\NN = \{1, 2, 3, \dots \}$ for the set of natural numbers. For $a$, $b \in \ZZ$, we denote by $\intval{a,b}$ the set of all integers from $a$ to $b$, including both endpoints. @@ -98,47 +102,64 @@ Throughout, we fix an arbitrary ground field $\kk$. We begin by recalling some b Let $\SGR$ be an affine semigroup, that is, a finitely generated submonoid of $\ZZ^d$ for some $d \in \NN$. The \textbf{toric algebra} associated with $\SGR$ is the semigroup algebra \[ R \coloneqq \kk[\SGR] = \spn_\kk \{ x^a \mid a \in \SGR \}. \] -The algebra $R$ is a finitely generated $\kk$-subalgebra of the Laurent polynomial ring~$\kk[x_1^{\pm 1}, \dots, x_d^{\pm 1}]$. In particular, it is a Noetherian domain and inherits the natural $\ZZ^d$-grading. Its Krull dimension is equal to the rank of $\gp(\SGR)$, where $\gp(\SGR)$ denotes the subgroup of $\ZZ^d$ generated by $\SGR$~\cite[Prop. 1.1.14]{Cox2011}. +The algebra $R$ is a finitely generated $\kk$-subalgebra of the Laurent polynomial ring~$\kk[x_1^{\pm 1}, \dots, x_d^{\pm 1}]$. In particular, it is a Noetherian domain and inherits the natural $\ZZ^d$-grading. Its Krull dimension is equal to the rank of $\gp(\SGR)$, where $\gp(\SGR)$ denotes the subgroup of $\ZZ^d$ generated by $\SGR$~\cite[Prop. 1.1.14]{Cox2011}. Consequently, $R$ is one-dimensional if and only if $\gp(\SGR) \cong \ZZ$, in which case we may assume without loss of generality that $\gp(\SGR) = \ZZ$. Every additive submonoid of $\ZZ$ is either a group, a submonoid of $\NN_0$, or a submonoid of $-\NN_0$~\cite[§ 2]{Geroldinger2025}. Hence, when $R$ is one-dimensional, $\SGR$ is, up to isomorphism of monoids, either $\ZZ$ or a submonoid of $\NN_0$. Accordingly, either $R\cong\kk[x^{\pm 1}]$ or $R \cong \kk[x^{a_1},\dots,x^{a_n}]$ for some $a_1$, \dots, $a_n \in \NN_0$. +When $\kk$ is algebraically closed, the algebra $R=\kk[\SGR]$ is the coordinate ring of the associated affine toric variety $X = \Spec(R)$. In dimension one, the corresponding affine toric variety is either the one-dimensional torus $\Spec(\kk[x^{\pm1}]) = \A^1_\kk \setminus \{0\}$, or an \textbf{affine monomial curve} $\Spec(\kk[x^{a_1},\dots,x^{a_n}])$, parametrized by +\begin{equation}\label{eq:para-mon-curve} + \varphi \colon \A^1_\kk \to \A^n_\kk, \qquad t \longmapsto (t^{a_1},\dots,t^{a_n}). +\end{equation} + +\subsection{Numerical monoids} + +Let $\SGR$ be a submonoid of $\NN_0$. Writing $\SGR = \gcd(\SGR) \cdot \SGR^\prime$, where $\SGR^\prime$ is a submonoid of $\NN_0$ with $\gcd(\SGR^\prime) = 1$, yields an isomorphism of monoids between $\SGR$ and $\SGR^\prime$. Thus, it suffices to consider submonoids of $\NN_0$ with greatest common divisor $1$, called \textbf{numerical monoids}. Crucially, the gap set $\gapset$ of a numerical monoid $\SGR$ is finite. For further background on numerical monoids, see~\cite{Rosales2009}. + +For a numerical monoid $\SGR=\langle a_1,\dots, a_n\rangle \subseteq \NN_0$, the parametrization $\varphi$ from~\eqref{eq:para-mon-curve} of the associated affine monomial curve $X = \Spec(R)$ restricts to an isomorphism of varieties $\A_\kk^1 \setminus \{ 0 \} \xrightarrow{\sim} X \setminus \{ 0 \}$. Hence $X$ is nonsingular away from the origin. At the origin, it is singular unless $\SGR = \NN_0$, in which case $R=\kk[x]$ and $X=\A_\kk^1$. + +\subsection{Normality, seminormality, and weak normality} + +A domain $\Lambda$ is called \textbf{normal} if it is integrally closed in its field of fractions $\Frac(\Lambda)$, and \textbf{seminormal} if, for every $\lambda \in \Frac(\Lambda)$, the conditions $\lambda^2$, $\lambda^3 \in \Lambda$ imply $\lambda \in \Lambda$. The normalization $\overline{\Lambda}$ and the seminormalization ${}^+\Lambda$ of $\Lambda$ are the intersections of all intermediate normal and seminormal rings $\Gamma$, respectively, satisfying $\Lambda \subseteq \Gamma \subseteq \Frac(\Lambda)$. If $\Lambda$ has prime characteristic $p$, then $\Lambda$ is called \textbf{weakly normal} if, for every $\lambda \in \Frac(\Lambda)$, the condition $\lambda^p \in \Lambda$ implies $\lambda\in\Lambda$. Clearly, normality implies seminormality. In prime characteristic $p$, normality implies weak normality, and weak normality in turn implies seminormality. Indeed, suppose that $\Lambda$ is weakly normal and let $\lambda \in \Frac(\Lambda)$ satisfy $\lambda^2$, $\lambda^3 \in \Lambda$. Write $p = 2n^\prime + 3n^{\prime\prime}$ for some $n^\prime$, $n^{\prime\prime} \in \NN_0$. Then +\[ +\lambda^p = (\lambda^2)^{n^\prime}(\lambda^3)^{n^{\prime\prime}} \in \Lambda, +\] +and hence $\lambda \in \Lambda$. + +For the toric algebra $R = \kk[\SGR]$, the field of fractions in the definitions above may be replaced by $\kk[\gp(\SGR)]$. Indeed, $\gp(\SGR) \cong \ZZ^{d^\prime}$ for some $d^\prime \in \NN_0$, so $\kk[\gp(\SGR)]$ is isomorphic to a Laurent polynomial ring and is therefore normal. Consequently, every element of $\Frac(R)$ that is integral over $R$ already belongs to $\kk[\gp(\SGR)]$.\todo{mention normality, ... for affine monoids?} + +In the case of a numerical semigroup $\SGR$, the corresponding toric algebra $R$ satisfies ${}^+R=\kk[x]$. Indeed, let $n_1 > \cdots > n_k$ be the elements of $\gapset$. Since $n_1$ is the largest gap of $\SGR$, we have $(x^{n_1})^2,(x^{n_1})^3\in R\subseteq{}^+R$, and hence $x^{n_1}\in{}^+R$ by seminormality. +%Since $n_1$ is the largest gap of $\SGR$, both $2n_1$ and $3n_1$ belong to $\SGR$. Hence $(x^{n_1})^2$, $(x^{n_1})^3 \in R \subseteq{}^+R$, and seminormality gives $x^{n_1}\in{}^+R$. +Repeating this argument successively for the remaining gaps shows that $x^{n_1}$, \dots, $x^{n_k} \in{}^+R$, and therefore $\kk[x] \subseteq{}^+R$. Since $\kk[x]$ is normal, hence seminormal, the reverse inclusion follows. Since $\kk[x]$ is normal, we also have $\overline{R} = \kk[x]$, and consequently $\Frac(R) = \kk(x)$. Thus, unless $\SGR=\NN_0$, the ring $R$ is neither normal nor seminormal. + +\subsection{Graded Krull–Schmidt theorem} + +The category of finitely generated $\QQ^d$-graded $R$-modules satisfies the \textbf{Krull--Schmidt property}. That is, every such module decomposes as a finite direct sum of $\QQ^d$-graded $R$-modules that are indecomposable in the graded category, and this decomposition is unique up to graded isomorphism and permutation of the summands~\cite[§ 2]{Shibuta2017}. + +% we are interested in the unique summands of such graded decomposition only up to shift of degrees so that we introduce the following notations (define \grcong) %For $\QQ^d$-graded $R$-modules $M$, $M_1$, \dots, $M_k$, we write $M \cong_{\textrm{sh}} M_1 \oplus \cdots \oplus M_k$ whenever there exist $q_i \in \QQ$ such that $M \cong M_1(d_1) \oplus \cdots \oplus M_k(d_k)$ as $\QQ^d$-graded $R$-modules. - -\todo{use widebar} +%Clearly if $M \shcong M_1 \oplus M_2$ and $M_2 \shcong M_2^\prime \oplus M_2^{\prime\prime}$, then $M \shcong M_1 \oplus M_2^\prime \oplus M_2^{\prime\prime}$. \todo{adapt phrasinig in correspondence gap modules} -\todo{use notation $\grcong$?} +%\begin{lemma}\label{lemma:qpower-under-shift} +%Let $M$, $M^\prime$ and $M^{\prime\prime}$ be $\QQ^d$-graded $R$-modules, and suppose that $M \shcong M^{\prime} \oplus M^{\prime\prime}$, then $\qpower{M} \shcong \qpower{M^{\prime}} \oplus \qpower{M^{\prime\prime}}$. +%\end{lemma} -\todo{affine semigroup rings are (non-normal) toric varieties} +%\begin{proof} +%By assumption, there are $d^\prime$, $d^{\prime\prime} \in \QQ$ such that $M \grcong M^\prime(d^\prime) \oplus M^{\prime\prime}(d^{\prime\prime})$. -\todo{1-dimensional toric varieties are monomial curves} +%As the construction of the pseudo-Frobenius twist respects direct sums, we have +%\[ \qpower{M} \grcong \qpower{(M^{\prime}(d^\prime))} \oplus \qpower{(M^{\prime\prime}(d^{\prime\prime}))} = (\qpower{M^{\prime}})(d^\prime/q) \oplus (\qpower{M^{\prime\prime}})(d^{\prime\prime}/q), \] +%where the last equality follows from the computation +%\begin{align*} +%\left(\qpower{(M^{\prime}(d^\prime))} \oplus \qpower{\left(M^{\prime\prime}(d^{\prime\prime})\right)}\right)_\alpha &= \bigl(M^{\prime}(d^\prime)\bigr)_{q\alpha} \oplus \bigl(M^{\prime\prime}(d^{\prime\prime})\bigr)_{q\alpha} \\ +%&= (M^{\prime})_{q\alpha + d^\prime} \oplus (M^{\prime\prime})_{q\alpha + d^{\prime\prime}} \\ +%&= (\qpower{M^\prime})_{\alpha + \frac{d^\prime}{q}} \oplus (\qpower{M^{\prime\prime}})_{\alpha + \frac{d^{\prime\prime}}{q}} \\ +%&= \left((\qpower{M^{\prime}})(d^\prime/q) \oplus (\qpower{M^{\prime\prime}})(d^{\prime\prime}/q)\right)_\alpha. +%\end{align*} +%\end{proof} -\todo{define normal, seminormal, $G(S)$, weakly normal (i.e., for any $r \in \Frac(R)$, $r^p \in R$ implies $r \in R$), note that one can replace $\Frac(R)$ by $\kk[G(S)]$, and weakly normal implies seminormal} - -\todo{monomial curves are non-seminormal, non-normal, singular at 0} - -\todo{say what the normalization, fraction field is} - -\begin{lemma} -Let $M$, $M^\prime$ and $M^{\prime\prime}$ be $\QQ^d$-graded $R$-modules, and suppose that $M \shcong M^{\prime} \oplus M^{\prime\prime}$, then $\qpower{M} \shcong \qpower{M^{\prime}} \oplus \qpower{M^{\prime\prime}}$. -\end{lemma} - -\begin{proof} -By assumption, there are $d^\prime$, $d^{\prime\prime} \in \QQ$ such that $M \grcong M^\prime(d^\prime) \oplus M^{\prime\prime}(d^{\prime\prime})$. - -As the construction of the pseudo-Frobenius twist respects direct sums, we have -\[ \qpower{M} \grcong \qpower{(M^{\prime}(d^\prime))} \oplus \qpower{(M^{\prime\prime}(d^{\prime\prime}))} = (\qpower{M^{\prime}})(d^\prime/q) \oplus (\qpower{M^{\prime\prime}})(d^{\prime\prime}/q), \] -where the last equality follows from the computation -\begin{align*} -\left(\qpower{(M^{\prime}(d^\prime))} \oplus \qpower{\left(M^{\prime\prime}(d^{\prime\prime})\right)}\right)_\alpha &= \bigl(M^{\prime}(d^\prime)\bigr)_{q\alpha} \oplus \bigl(M^{\prime\prime}(d^{\prime\prime})\bigr)_{q\alpha} \\ -&= (M^{\prime})_{q\alpha + d^\prime} \oplus (M^{\prime\prime})_{q\alpha + d^{\prime\prime}} \\ -&= (\qpower{M^\prime})_{\alpha + \frac{d^\prime}{q}} \oplus (\qpower{M^{\prime\prime}})_{\alpha + \frac{d^{\prime\prime}}{q}} \\ -&= \left((\qpower{M^{\prime}})(d^\prime/q) \oplus (\qpower{M^{\prime\prime}})(d^{\prime\prime}/q)\right)_\alpha. -\end{align*} -\end{proof} - -\begin{proposition} -For the semigroup algebra $R = \kk[\SGR]$ of an affine semigroup $\SGR$, the category of finitely generated $\QQ^d$-graded $R$-modules is Krull-Schmidt. -\end{proposition} +%\begin{proposition} +%For the semigroup algebra $R = \kk[\SGR]$ of an affine semigroup $\SGR$, the category of finitely generated $\QQ^d$-graded $R$-modules is Krull-Schmidt. +%\end{proposition} \section{Frobenius maps on toric algebras} Let $\SGR$ be an affine semigroup, i.e., a finitely generated submonoid of $\ZZ^d$ for some $d \in \NN$, and let $R \coloneqq \kk[\SGR]$ denote its semigroup algebra. Fix a positive integer $q$, and write $q^{-1}\SGR = \{ a/q \mid a \in \SGR \} \subseteq \QQ^d$ for the rescaling of $\SGR$ by the factor $1/q$. Consider the semigroup algebra @@ -230,7 +251,7 @@ $R$ is normal if and only if $R$ is $q$-weakly normal for all $q \in \NN$ if and \end{corollary} \begin{proof} -By~\cite[Prop. 1]{Hochster1972}, the semigroup algebra $R = \kk[\SGR]$ is normal if and only if $\SGR$ is saturated, that is, if for every $a \in \gp(\SGR)$ and $q \in \NN$, the condition $qa \in \SGR$ implies $a \in \SGR$. This is precisely the semigroup-theoretic formulation of $R$ being $q$-weakly normal for all $q \in \NN$. By the previous proposition, this is equivalent to $R$ being $q$-weakly normal for every $q \in \NN$. +By~\cite[Prop. 1]{Hochster1972}, the semigroup algebra $R = \kk[\SGR]$ is normal if and only if $\SGR$ is normal, that is, if for every $a \in \gp(\SGR)$ and $q \in \NN$, the condition $qa \in \SGR$ implies $a \in \SGR$. This is precisely the semigroup-theoretic formulation of $R$ being $q$-weakly normal for all $q \in \NN$. By the previous proposition, this is equivalent to $R$ being $q$-weakly normal for every $q \in \NN$. \end{proof} \begin{lemma}\textnormal{\cite[Lemma 5.0.3]{Blickle2013}}\label{lemma:q-split} @@ -280,7 +301,7 @@ where each summand is indecomposable (see Lemma~\ref{lemma:rank-1}). %\end{proof} \begin{definition}\label{def:gap-module} -Let $A\subseteq \NN$. The (normalized) \textbf{gap module defined by $A$} is the $\NN_0$-graded $R$-submodule of $\bar{R}=\kk[x]$ given by +Let $A\subseteq \NN$. The (normalized) \textbf{gap module defined by $A$} is the $\NN_0$-graded $R$-submodule of $\overline{R}=\kk[x]$ given by \[ I(A) \coloneqq \langle x^\alpha \mid \alpha \in A\cup \{0\} \rangle_R. \] \end{definition} @@ -314,15 +335,15 @@ In particular, as $q$ ranges over $\NN$, only finitely many isomorphism classes \begin{lemma}\label{lemma:rank-1} For each $A \in \Min(\SGR)$, the gap module $I(A)$ has rank $1$ and is therefore indecomposable. Moreover, the assignment \[ A \mapsto [I(A)] \] -defines a bijection between $\Min(\SGR)$ and the set of isomorphism classes of $\NN_0$-graded $R$-submodules of $\bar{R} = \kk[x]$. +defines a bijection between $\Min(\SGR)$ and the set of isomorphism classes of $\NN_0$-graded $R$-submodules of $\overline{R} = \kk[x]$. \end{lemma} \begin{proof} For simplicity, set $Q \coloneqq \Frac(R) = \kk(x)$ and fix $A \in \Min(\SGR)$. Observe that $Q \otimes_R I(A)$ is naturally isomorphic, as a $Q$-vector space, to $I(A)_{(0)} \subseteq Q$. Since $I(A)$ contains $1$, it follows that $I(A)_{(0)} = Q$ and $I(A)$ has rank $1$. -For the second part, each $I(A)$ with $A \in \Min(\SGR)$ is an $\NN_0$-graded $R$-submodule of $\bar{R}$, so the map is well-defined. To prove injectivity, suppose that $\varphi \colon I(A)\to I(B)$ is an $R$-module isomorphism for some $A$, $B \in \Min(\SGR)$. Localizing at $(0)$ yields an isomorphism of $Q$-vector spaces $\varphi_{(0)}\colon I(A)_{(0)} \to I(B)_{(0)}$, where both source and target are equal to $Q$. Hence $\varphi$, and thus also $\varphi_{(0)}$, is given by multiplication by some $g_0 \in Q$, so $\varphi(f) = g_0f$ for all $f \in I(A)$. A simple degree argument shows that $g_0 \in \kk \setminus \{ 0 \}$. Therefore, +For the second part, each $I(A)$ with $A \in \Min(\SGR)$ is an $\NN_0$-graded $R$-submodule of $\overline{R}$, so the map is well-defined. To prove injectivity, suppose that $\varphi \colon I(A)\to I(B)$ is an $R$-module isomorphism for some $A$, $B \in \Min(\SGR)$. Localizing at $(0)$ yields an isomorphism of $Q$-vector spaces $\varphi_{(0)}\colon I(A)_{(0)} \to I(B)_{(0)}$, where both source and target are equal to $Q$. Hence $\varphi$, and thus also $\varphi_{(0)}$, is given by multiplication by some $g_0 \in Q$, so $\varphi(f) = g_0f$ for all $f \in I(A)$. A simple degree argument shows that $g_0 \in \kk \setminus \{ 0 \}$. Therefore, \[ I(B) = \varphi(I(A)) = g_0 \cdot I(A) =I(A). \] -Since $A$ and $B$ are both minimal with respect to $\preceq_\SGR$, it follows that $A = B$. Now, let $I \subseteq \bar{R}$ be an arbitrary $\NN_0$-graded $R$-submodule. Because $R$ is Noetherian and $\bar{R}$ is finitely generated as an $R$-module, $I$ is finitely generated, i.e, $I = \langle x^\beta \mid \beta \in B\rangle_R$ for some finite subset $B \subseteq \NN_0$. Define $m \coloneqq \min(B)$ and $A^\prime \coloneqq \{ \beta - m \mid \beta \in B \} \setminus \{ 0 \}$. Consequently, $I = x^m \cdot I(A^\prime)$, which implies that $I \cong I(A^\prime)$. The claim then follows from the fact that $I(A^\prime)=I(A)$, where $A$ denotes the set of minimal elements of $A^\prime \cap (\gapset)$ with respect to $\preceq_\SGR$. +Since $A$ and $B$ are both minimal with respect to $\preceq_\SGR$, it follows that $A = B$. Now, let $I \subseteq \overline{R}$ be an arbitrary $\NN_0$-graded $R$-submodule. Because $R$ is Noetherian and $\overline{R}$ is finitely generated as an $R$-module, $I$ is finitely generated, i.e, $I = \langle x^\beta \mid \beta \in B\rangle_R$ for some finite subset $B \subseteq \NN_0$. Define $m \coloneqq \min(B)$ and $A^\prime \coloneqq \{ \beta - m \mid \beta \in B \} \setminus \{ 0 \}$. Consequently, $I = x^m \cdot I(A^\prime)$, which implies that $I \cong I(A^\prime)$. The claim then follows from the fact that $I(A^\prime)=I(A)$, where $A$ denotes the set of minimal elements of $A^\prime \cap (\gapset)$ with respect to $\preceq_\SGR$. \end{proof} \begin{remark} @@ -337,20 +358,20 @@ For each $A \in \Min(\SGR)$, the gap module $I(A)$ is maximal Cohen--Macaulay. \end{corollary} \begin{proof} -By the Cohen--Seidenberg theorems, $\dim(R) = \dim(\bar{R}) = \dim(\kk[x]) = 1$. Fix $A \in \Min(\SGR)$ and a maximal ideal $\mathfrak{m}$ of $R$. Since $R$ is a domain of Krull dimension $1$, every maximal ideal has height $1$; in particular $\dim(R_\mathfrak{m}) = \hgt(\mathfrak{m}) = 1$. The localization +By the Cohen--Seidenberg theorems, $\dim(R) = \dim(\overline{R}) = \dim(\kk[x]) = 1$. Fix $A \in \Min(\SGR)$ and a maximal ideal $\mathfrak{m}$ of $R$. Since $R$ is a domain of Krull dimension $1$, every maximal ideal has height $1$; in particular $\dim(R_\mathfrak{m}) = \hgt(\mathfrak{m}) = 1$. The localization $I(A)_\mathfrak{m}$ is a non-zero $R_\mathfrak{m}$-submodule of $\kk(x)$, hence torsion-free, so $1 \leq \depth(I(A)_\mathfrak{m}) \leq \dim(R_\mathfrak{m}) = 1$. Hence, $I(A)$ is maximal Cohen--Macaulay. \end{proof} \todo{add direct sums are Cohen--Macaulay (Bruns/Herzog, 1993, Prop. 1.2.9)} \begin{lemma} -For $q \in \NN$, it holds that $\qpower{\bar{R}} \shcong \bar{R}^{\oplus q}$. +For $q \in \NN$, it holds that $\qpower{\overline{R}} \shcong \overline{R}^{\oplus q}$. \end{lemma} \begin{proof} First, note that -\[ \bar{R}^{1/q} = \kk[x^{1/q}] = \bigoplus_{r \in [0,q-1]} x^{r/q} \cdot \kk[x] \cong \bar{R}^{\oplus q} \] -as $R$-modules. By Lemma~\ref{lemma:cannonc-isom}, $\bar{R}^{1/q}$ and $\qpower{\bar{R}}$ are isomorphic as $\bar{R}$-modules, and hence, by restriction of scalars along $R \subseteq \bar{R}$, also as $R$-modules. +\[ \overline{R}^{1/q} = \kk[x^{1/q}] = \bigoplus_{r \in [0,q-1]} x^{r/q} \cdot \kk[x] \shcong \overline{R}^{\oplus q} \] +as $R$-modules. By Lemma~\ref{lemma:cannonc-isom}, $\overline{R}^{1/q}$ and $\qpower{\overline{R}}$ are isomorphic as $\QQ_{\geq 0}$-graded modules over $\overline{R}$, and hence, by restriction of scalars along $R \subseteq \overline{R}$, also as $\QQ_{\geq 0}$-graded modules over $R$. \end{proof} \begin{example}\label{ex:large-powers} @@ -389,19 +410,19 @@ Let $\SGR = \langle 3,4 \rangle \subseteq \NN_0$ and $R = \kk[\SGR] = \kk[x^3,x^ \end{center} The set -\[ \{ x^{(3/q)e_1 + (4/q)e_2} : e_i \in \intval{0,q-1} \} \] -generates $R^{1/q} = \kk[x^{3/q},x^{4/q}]$ as an $R$-module. For $q =2$, this generating set is $\{ 1, x^{3/2}, x^2, x^{7/2} \}$, and grouping the numerators of the exponents by residue modulo $2$ yields the $R$-module decomposition -\[ R^{1/2} = \langle 1, x^2 \rangle_R \oplus x^{3/2} \cdot \langle 1, x^2 \rangle_R \cong I(\{2 \})^{\oplus 2} \] -For $q = 4$, a direct computation shows $R^{1/4} \cong \bar{R}^{\oplus 4}$. Keeping in mind that $R^{1/q}$ is isomorphic to $\qpower{R}$ and that the construction of the pseudo-Frobenius twist respects direct sums, we conclude that -\[ R^{1/8} \cong \qpower[2]{(\qpower[4]{R})} \cong \qpower[2]{(\bar{R}^{\oplus 4})} = (\qpower[2]{\bar{R}})^{\oplus 4} \cong \bar{R}^{\oplus 8}, \] -where the last isomorphism follows from the previous lemma. By induction, we obtain that $R^{1/2^e} \cong \bar{R}^{\oplus 2^e}$ for all $e \geq 2$. Consequently, there exist summands appearing in the decomposition of $R^{1/2^e}$ for some $e \in \NN$ that do \emph{not} appear in the decomposition of $R^{1/2^\ell}$ for large $\ell$. This stands in contrast to the case where $\SGR$ is a normal affine semigroup and $R=\kk[\SGR]$ is a toric algebra~\cite{Bruns2005}. +\[ \bigl\{ x^{(3/q)e_1 + (4/q)e_2} : e_i \in \intval{0,q-1} \bigr\} \] +generates $R^{1/q} = \kk[x^{3/q},x^{4/q}]$ as an $R$-module. For $q =2$, this homogeneous generating set is $\{ 1, x^{3/2}, x^2, x^{7/2} \}$, and grouping the numerators of the exponents by residue modulo $2$ yields the $\QQ_{\geq 0}$-graded $R$-module decomposition +\[ R^{1/2} = \langle 1, x^2 \rangle_R \oplus x^{3/2} \cdot \langle 1, x^2 \rangle_R \shcong I(\{2 \})^{\oplus 2} \] +For $q = 4$, a direct computation shows $R^{1/4} \shcong \overline{R}^{\oplus 4}$. Using the $\QQ_{\geq 0}$-graded isomorphism between $R^{1/q}$ and $\qpower{R}$, we compute +\[ R^{1/8} \grcong \qpower[8]{R} = \qpower[2]{(\qpower[4]{R})} \grcong \qpower[2]{(R^{1/4})}. \] +Hence, by Lemma~\ref{lemma:qpower-under-shift}, it holds that $R^{1/8} \shcong (\qpower[2]{\overline{R}})^{\oplus 4}$. The preceding lemma then yields $R^{1/8} \shcong \overline{R}^{\oplus 8}$. By induction, we obtain that $R^{1/2^e} \shcong \overline{R}^{\oplus 2^e}$ for all $e \geq 2$. Consequently, there exist summands appearing in the $\QQ_{\geq 0}$-graded direct sum decomposition of $R^{1/2^e}$ for some $e \in \NN$ that do \emph{not} appear in the decomposition of $R^{1/2^\ell}$ for large $\ell$. This stands in contrast to the case where $\SGR$ is a normal affine semigroup and $R=\kk[\SGR]$ is a normal toric algebra~\cite{Bruns2005}. \end{example} % NOTE: next proposition holds when q \in \SGR \begin{proposition}\label{prop:asy-behav} -It holds that $q \in \SGR$ if and only if every indecomposable summand of $R^{1/q}$ is isomorphic to the normalization $\bar{R} = \kk[x]$ of $R$, that is, $R^{1/q} \cong \bar{R}^{\oplus q}$. In particular, this holds for all sufficiently large $q$. -%For $q \in \SGR$, every indecomposable summands of $R^{1/q}$ is isomorphic to the normalization $\bar{R} = \kk[x]$ of $R$. That is, $R^{1/q} \cong \bar{R}^{\oplus q}$. In particular, this holds for all sufficiently large $q$. +It holds that $q \in \SGR$ if and only if every indecomposable summand of $R^{1/q}$ is isomorphic to the normalization $\overline{R} = \kk[x]$ of $R$, that is, $R^{1/q} \cong \overline{R}^{\oplus q}$. In particular, this holds for all sufficiently large $q$. +%For $q \in \SGR$, every indecomposable summands of $R^{1/q}$ is isomorphic to the normalization $\overline{R} = \kk[x]$ of $R$. That is, $R^{1/q} \cong \overline{R}^{\oplus q}$. In particular, this holds for all sufficiently large $q$. \end{proposition} \begin{proof} @@ -420,8 +441,8 @@ It now follows from~\eqref{eq:ind-summand-as-vector-space} that \begin{example} Revisiting Example~\ref{ex:large-powers}, it remains to determine the decomposition of $R^{1/5}$ in order to obtain a complete description of the decompositions of $R^{1/q}$ for $q\in\NN$. A direct computation yields -\[ R^{1/5}\cong I(\{2\}) \oplus \bar{R}^{\oplus 4}. \] -Since $\bar{R} = \kk[x]$ corresponds to $I(\{1,2\})$, the indecomposable summands that occur in $R^{1/q}$ for some $q \in \NN$ are precisely $I(\{2\})$ and $I(\{1,2\})$. On the other hand, +\[ R^{1/5}\cong I(\{2\}) \oplus \overline{R}^{\oplus 4}. \] +Since $\overline{R} = \kk[x]$ corresponds to $I(\{1,2\})$, the indecomposable summands that occur in $R^{1/q}$ for some $q \in \NN$ are precisely $I(\{2\})$ and $I(\{1,2\})$. On the other hand, \[ \Min(\SGR) = \{ \emptyset,\{1\}, \{2\}, \{5\}, \{1,2\} \}, \] so not every potential candidate $I(A)$ with $A \in \Min(\SGR)$ occurs as a direct summand. \end{example} @@ -433,7 +454,7 @@ For each $q \geq 2$, $R$ is not $q$-split, meaning that $R$ does not occur as a \todo{\emph{note:} this can also be easily proven directly!} \begin{proof} -Fix $q \geq 2$, and let $p$ be a prime divisor of $q$. Suppose, for contradiction, that $R$ is $q$-split. By Lemma~\ref{lemma:q-split}, $R$ is then $p^e$-split for every $e \in \NN$. This contradicts Proposition~\ref{prop:asy-behav}, which shows that, for all sufficiently large $e$, $R^{1/p^e}$ decomposes as a direct sum of copies of $\bar{R}$, and therefore cannot contain $R$ as a direct summand. +Fix $q \geq 2$, and let $p$ be a prime divisor of $q$. Suppose, for contradiction, that $R$ is $q$-split. By Lemma~\ref{lemma:q-split}, $R$ is then $p^e$-split for every $e \in \NN$. This contradicts Proposition~\ref{prop:asy-behav}, which shows that, for all sufficiently large $e$, $R^{1/p^e}$ decomposes as a direct sum of copies of $\overline{R}$, and therefore cannot contain $R$ as a direct summand. \end{proof} \begin{proposition} @@ -457,8 +478,8 @@ For $q \in \SGR$, the global dimension of $\End_R(R^{1/q})$ is equal to the Krul \end{theorem} \begin{proof} -For $q \in \SGR$, Proposition~\ref{prop:asy-behav} gives $R^{1/q}\cong \bar R^{\oplus q}$. By~\cite[Cor. 2.3]{Leuschke2012}, multiplicities of direct summands do not affect the endomorphism ring up to Morita equivalence; hence $\End_R(R^{1/q})$ is Morita equivalent to $\End_R(\bar{R})$. As in the proof of Lemma~\ref{lemma:rank-1}, since $\bar{R}$ has rank $1$, any $R$-module endomorphism of $\bar{R}$ is induced by multiplication with an element of $\bar{R}$, and thus $\End_R(\bar{R}) \cong \bar{R}$. Consequently, we obtain -\[ \gldim(\End_R(R^{1/q})) = \gldim(\bar{R}) = \gldim(\kk[x]) = 1 = \dim(R), \] +For $q \in \SGR$, Proposition~\ref{prop:asy-behav} gives $R^{1/q}\cong \overline R^{\oplus q}$. By~\cite[Cor. 2.3]{Leuschke2012}, multiplicities of direct summands do not affect the endomorphism ring up to Morita equivalence; hence $\End_R(R^{1/q})$ is Morita equivalent to $\End_R(\overline{R})$. As in the proof of Lemma~\ref{lemma:rank-1}, since $\overline{R}$ has rank $1$, any $R$-module endomorphism of $\overline{R}$ is induced by multiplication with an element of $\overline{R}$, and thus $\End_R(\overline{R}) \cong \overline{R}$. Consequently, we obtain +\[ \gldim(\End_R(R^{1/q})) = \gldim(\overline{R}) = \gldim(\kk[x]) = 1 = \dim(R), \] where the first equality follows from the Morita invariance of global dimension. \end{proof} diff --git a/references.bib b/references.bib index e893191..74f11af 100644 --- a/references.bib +++ b/references.bib @@ -175,4 +175,36 @@ address = {Providence, RI}, year = {2011}, isbn = {978-0-8218-4819-2}, doi = {10.1090/gsm/124} +} + + +@article{Geroldinger2025, +title = {On algebraic properties of power monoids of numerical monoids}, +volume = {265}, +issn = {0021-2172, 1565-8511}, +url = {https://link.springer.com/10.1007/s11856-024-2683-0}, +doi = {10.1007/s11856-024-2683-0}, +pages = {867--900}, +number = {2}, +journaltitle = {Israel Journal of Mathematics}, +shortjournal = {Isr. J. Math.}, +author = {Bienvenu, Pierre-Yves and Geroldinger, Alfred}, +urldate = {2026-09-16}, +date = {2025-02}, +} + +@book{Rosales2009, +location = {New York, {NY}}, +title = {Numerical Semigroups}, +volume = {20}, +rights = {https://www.springernature.com/gp/researchers/text-and-data-mining}, +isbn = {978-1-4419-0159-0 978-1-4419-0160-6}, +url = {https://link.springer.com/10.1007/978-1-4419-0160-6}, +doi = {10.1007/978-1-4419-0160-6}, +series = {Developments in Mathematics}, +publisher = {Springer New York}, +author = {Rosales, J.C. and García-Sánchez, P. A.}, +urldate = {2026-09-16}, +date = {2009}, +langid = {english}, } \ No newline at end of file From 844ca7a2514ae9a39e38d8a4cd77529d558befa6 Mon Sep 17 00:00:00 2001 From: Markus Maximilian Rudolf Tripp Date: Mon, 21 Sep 2026 16:44:13 +0200 Subject: [PATCH 8/8] repairing some stuff --- ncr-of-monomial-curves.tex | 116 ++++++++++++++++++------------------- references.bib | 13 +++++ 2 files changed, 68 insertions(+), 61 deletions(-) diff --git a/ncr-of-monomial-curves.tex b/ncr-of-monomial-curves.tex index a158a81..da55c05 100644 --- a/ncr-of-monomial-curves.tex +++ b/ncr-of-monomial-curves.tex @@ -102,9 +102,9 @@ Throughout, we fix an arbitrary ground field $\kk$. We begin by recalling some b Let $\SGR$ be an affine semigroup, that is, a finitely generated submonoid of $\ZZ^d$ for some $d \in \NN$. The \textbf{toric algebra} associated with $\SGR$ is the semigroup algebra \[ R \coloneqq \kk[\SGR] = \spn_\kk \{ x^a \mid a \in \SGR \}. \] -The algebra $R$ is a finitely generated $\kk$-subalgebra of the Laurent polynomial ring~$\kk[x_1^{\pm 1}, \dots, x_d^{\pm 1}]$. In particular, it is a Noetherian domain and inherits the natural $\ZZ^d$-grading. Its Krull dimension is equal to the rank of $\gp(\SGR)$, where $\gp(\SGR)$ denotes the subgroup of $\ZZ^d$ generated by $\SGR$~\cite[Prop. 1.1.14]{Cox2011}. Consequently, $R$ is one-dimensional if and only if $\gp(\SGR) \cong \ZZ$, in which case we may assume without loss of generality that $\gp(\SGR) = \ZZ$. Every additive submonoid of $\ZZ$ is either a group, a submonoid of $\NN_0$, or a submonoid of $-\NN_0$~\cite[§ 2]{Geroldinger2025}. Hence, when $R$ is one-dimensional, $\SGR$ is, up to isomorphism of monoids, either $\ZZ$ or a submonoid of $\NN_0$. Accordingly, either $R\cong\kk[x^{\pm 1}]$ or $R \cong \kk[x^{a_1},\dots,x^{a_n}]$ for some $a_1$, \dots, $a_n \in \NN_0$. +The algebra $R$ is a finitely generated $\kk$-subalgebra of the Laurent polynomial ring~$\kk[x_1^{\pm 1}, \dots, x_d^{\pm 1}]$. In particular, it is a Noetherian domain and inherits the natural $\ZZ^d$-grading. Its Krull dimension is equal to the rank of $\gp(\SGR)$, where $\gp(\SGR)$ denotes the subgroup of $\ZZ^d$ generated by $\SGR$~\cite[Prop. 1.1.14]{Cox2011}. Consequently, $R$ is one-dimensional if and only if $\gp(\SGR) \cong \ZZ$, in which case we may assume without loss of generality that $\gp(\SGR) = \ZZ$. Every submonoid of $\ZZ$ is either a group, a submonoid of $\NN_0$, or a submonoid of $-\NN_0$~\cite[§ 2]{Geroldinger2025}. Hence, when $R$ is one-dimensional, $\SGR$ is, up to isomorphism of monoids, either $\ZZ$ or a submonoid of $\NN_0$. Accordingly, either $R\cong\kk[x^{\pm 1}]$ or $R \cong \kk[x^{a_1},\dots,x^{a_n}]$ for some $a_1$, \dots, $a_n \in \NN_0$. -When $\kk$ is algebraically closed, the algebra $R=\kk[\SGR]$ is the coordinate ring of the associated affine toric variety $X = \Spec(R)$. In dimension one, the corresponding affine toric variety is either the one-dimensional torus $\Spec(\kk[x^{\pm1}]) = \A^1_\kk \setminus \{0\}$, or an \textbf{affine monomial curve} $\Spec(\kk[x^{a_1},\dots,x^{a_n}])$, parametrized by +When $\kk$ is algebraically closed, the algebra $R=\kk[\SGR]$ is the coordinate ring of the associated affine toric variety $X = \Spec(R)$. In dimension one, the corresponding affine toric variety is either the one-dimensional torus $\Spec(\kk[x^{\pm1}]) = \A^1_\kk \setminus \{ 0 \}$, or an \textbf{affine monomial curve} $\Spec(\kk[x^{a_1},\dots,x^{a_n}])$, parametrized by \begin{equation}\label{eq:para-mon-curve} \varphi \colon \A^1_\kk \to \A^n_\kk, \qquad t \longmapsto (t^{a_1},\dots,t^{a_n}). \end{equation} @@ -117,62 +117,41 @@ For a numerical monoid $\SGR=\langle a_1,\dots, a_n\rangle \subseteq \NN_0$, the \subsection{Normality, seminormality, and weak normality} -A domain $\Lambda$ is called \textbf{normal} if it is integrally closed in its field of fractions $\Frac(\Lambda)$, and \textbf{seminormal} if, for every $\lambda \in \Frac(\Lambda)$, the conditions $\lambda^2$, $\lambda^3 \in \Lambda$ imply $\lambda \in \Lambda$. The normalization $\overline{\Lambda}$ and the seminormalization ${}^+\Lambda$ of $\Lambda$ are the intersections of all intermediate normal and seminormal rings $\Gamma$, respectively, satisfying $\Lambda \subseteq \Gamma \subseteq \Frac(\Lambda)$. If $\Lambda$ has prime characteristic $p$, then $\Lambda$ is called \textbf{weakly normal} if, for every $\lambda \in \Frac(\Lambda)$, the condition $\lambda^p \in \Lambda$ implies $\lambda\in\Lambda$. Clearly, normality implies seminormality. In prime characteristic $p$, normality implies weak normality, and weak normality in turn implies seminormality. Indeed, suppose that $\Lambda$ is weakly normal and let $\lambda \in \Frac(\Lambda)$ satisfy $\lambda^2$, $\lambda^3 \in \Lambda$. Write $p = 2n^\prime + 3n^{\prime\prime}$ for some $n^\prime$, $n^{\prime\prime} \in \NN_0$. Then +A domain $\Lambda$ is called \textbf{normal} if it is integrally closed in its field o.f fractions $\Frac(\Lambda)$, and \textbf{seminormal} if, for every $\lambda \in \Frac(\Lambda)$, $\lambda^2 \in \Lambda$ and $\lambda^3 \in \Lambda$ imply that $\lambda \in \Lambda$. The normalization~$\overline{\Lambda}$ and the seminormalization ${}^+\Lambda$ of $\Lambda$ are, respectively, the intersections of all intermediate normal and seminormal rings $\Gamma$ satisfying $\Lambda \subseteq \Gamma \subseteq \Frac(\Lambda)$ If $\Lambda$ has prime characteristic $p$, then $\Lambda$ is called \textbf{weakly normal} if, for every $\lambda \in \Frac(\Lambda)$, the condition $\lambda^p \in \Lambda$ implies $\lambda\in\Lambda$. Clearly, normality implies seminormality. In prime characteristic $p$, normality implies weak normality, and weak normality in turn implies seminormality. Indeed, suppose that $\Lambda$ is weakly normal and let $\lambda \in \Frac(\Lambda)$ satisfy $\lambda^2$, $\lambda^3 \in \Lambda$. Write $p = 2k + 3\ell$ for some $k$, $\ell \in \NN_0$. Then \[ -\lambda^p = (\lambda^2)^{n^\prime}(\lambda^3)^{n^{\prime\prime}} \in \Lambda, +\lambda^p = (\lambda^2)^k(\lambda^3)^\ell \in \Lambda, \] and hence $\lambda \in \Lambda$. -For the toric algebra $R = \kk[\SGR]$, the field of fractions in the definitions above may be replaced by $\kk[\gp(\SGR)]$. Indeed, $\gp(\SGR) \cong \ZZ^{d^\prime}$ for some $d^\prime \in \NN_0$, so $\kk[\gp(\SGR)]$ is isomorphic to a Laurent polynomial ring and is therefore normal. Consequently, every element of $\Frac(R)$ that is integral over $R$ already belongs to $\kk[\gp(\SGR)]$.\todo{mention normality, ... for affine monoids?} +For a toric algebra $R = \kk[\SGR]$, the field of fractions in the definitions above may be replaced by $\kk[\gp(\SGR)]$. Indeed, $\gp(\SGR) \cong \ZZ^{d^\prime}$ for some $d^\prime \in \NN_0$, so $\kk[\gp(\SGR)]$ is isomorphic to a Laurent polynomial ring and is therefore normal. Consequently, every element of $\Frac(R)$ that is integral over $R$ already belongs to $\kk[\gp(\SGR)]$.\todo{mention normality, ... for affine monoids?} -In the case of a numerical semigroup $\SGR$, the corresponding toric algebra $R$ satisfies ${}^+R=\kk[x]$. Indeed, let $n_1 > \cdots > n_k$ be the elements of $\gapset$. Since $n_1$ is the largest gap of $\SGR$, we have $(x^{n_1})^2,(x^{n_1})^3\in R\subseteq{}^+R$, and hence $x^{n_1}\in{}^+R$ by seminormality. +In the case of a numerical monoid $\SGR$, the corresponding toric algebra $R$ satisfies ${}^+R=\kk[x]$. Indeed, let $n_1 > \cdots > n_k$ be the elements of $\gapset$. Since $n_1$ is the largest gap of $\SGR$, we have $(x^{n_1})^2$, $(x^{n_1})^3 \in R \subseteq{}^+R$, and hence $x^{n_1}\in{}^+R$ by seminormality. %Since $n_1$ is the largest gap of $\SGR$, both $2n_1$ and $3n_1$ belong to $\SGR$. Hence $(x^{n_1})^2$, $(x^{n_1})^3 \in R \subseteq{}^+R$, and seminormality gives $x^{n_1}\in{}^+R$. -Repeating this argument successively for the remaining gaps shows that $x^{n_1}$, \dots, $x^{n_k} \in{}^+R$, and therefore $\kk[x] \subseteq{}^+R$. Since $\kk[x]$ is normal, hence seminormal, the reverse inclusion follows. Since $\kk[x]$ is normal, we also have $\overline{R} = \kk[x]$, and consequently $\Frac(R) = \kk(x)$. Thus, unless $\SGR=\NN_0$, the ring $R$ is neither normal nor seminormal. +Repeating this argument successively for the remaining gaps shows that $x^{n_1}$, \dots, $x^{n_k} \in{}^+R$, and therefore $\kk[x] \subseteq{}^+R$. Since $\kk[x]$ is normal, hence seminormal, the reverse inclusion follows. As $\kk[x]$ is normal, it also follows that $\overline{R} = \kk[x]$, and consequently $\Frac(R) = \kk(x)$. Thus, unless $\SGR=\NN_0$, the ring $R$ is neither normal nor seminormal. \subsection{Graded Krull–Schmidt theorem} -The category of finitely generated $\QQ^d$-graded $R$-modules satisfies the \textbf{Krull--Schmidt property}. That is, every such module decomposes as a finite direct sum of $\QQ^d$-graded $R$-modules that are indecomposable in the graded category, and this decomposition is unique up to graded isomorphism and permutation of the summands~\cite[§ 2]{Shibuta2017}. +The category of finitely generated $\QQ^d$-graded $R$-modules satisfies the \textbf{Krull--Schmidt property}. That is, in the graded category, every such module decomposes as a finite direct sum of indecomposable modules, uniquely up to graded isomorphism and permutation of the summands~\cite[§ 2]{Shibuta2017}. -% we are interested in the unique summands of such graded decomposition only up to shift of degrees so that we introduce the following notations (define \grcong) -%For $\QQ^d$-graded $R$-modules $M$, $M_1$, \dots, $M_k$, we write $M \cong_{\textrm{sh}} M_1 \oplus \cdots \oplus M_k$ whenever there exist $q_i \in \QQ$ such that $M \cong M_1(d_1) \oplus \cdots \oplus M_k(d_k)$ as $\QQ^d$-graded $R$-modules. -%Clearly if $M \shcong M_1 \oplus M_2$ and $M_2 \shcong M_2^\prime \oplus M_2^{\prime\prime}$, then $M \shcong M_1 \oplus M_2^\prime \oplus M_2^{\prime\prime}$. - -\todo{adapt phrasinig in correspondence gap modules} - -%\begin{lemma}\label{lemma:qpower-under-shift} -%Let $M$, $M^\prime$ and $M^{\prime\prime}$ be $\QQ^d$-graded $R$-modules, and suppose that $M \shcong M^{\prime} \oplus M^{\prime\prime}$, then $\qpower{M} \shcong \qpower{M^{\prime}} \oplus \qpower{M^{\prime\prime}}$. -%\end{lemma} - -%\begin{proof} -%By assumption, there are $d^\prime$, $d^{\prime\prime} \in \QQ$ such that $M \grcong M^\prime(d^\prime) \oplus M^{\prime\prime}(d^{\prime\prime})$. - -%As the construction of the pseudo-Frobenius twist respects direct sums, we have -%\[ \qpower{M} \grcong \qpower{(M^{\prime}(d^\prime))} \oplus \qpower{(M^{\prime\prime}(d^{\prime\prime}))} = (\qpower{M^{\prime}})(d^\prime/q) \oplus (\qpower{M^{\prime\prime}})(d^{\prime\prime}/q), \] -%where the last equality follows from the computation -%\begin{align*} -%\left(\qpower{(M^{\prime}(d^\prime))} \oplus \qpower{\left(M^{\prime\prime}(d^{\prime\prime})\right)}\right)_\alpha &= \bigl(M^{\prime}(d^\prime)\bigr)_{q\alpha} \oplus \bigl(M^{\prime\prime}(d^{\prime\prime})\bigr)_{q\alpha} \\ -%&= (M^{\prime})_{q\alpha + d^\prime} \oplus (M^{\prime\prime})_{q\alpha + d^{\prime\prime}} \\ -%&= (\qpower{M^\prime})_{\alpha + \frac{d^\prime}{q}} \oplus (\qpower{M^{\prime\prime}})_{\alpha + \frac{d^{\prime\prime}}{q}} \\ -%&= \left((\qpower{M^{\prime}})(d^\prime/q) \oplus (\qpower{M^{\prime\prime}})(d^{\prime\prime}/q)\right)_\alpha. -%\end{align*} -%\end{proof} - -%\begin{proposition} -%For the semigroup algebra $R = \kk[\SGR]$ of an affine semigroup $\SGR$, the category of finitely generated $\QQ^d$-graded $R$-modules is Krull-Schmidt. -%\end{proposition} +We are interested in the summands of such graded decompositions only up to degree shifts, and therefore introduce the following notation. Let $M$, $N$, $M_1$, \dots,~$M_k$ be $\QQ^d$-graded $R$-modules. We write $M \grcong N$ if $M$ and $N$ are isomorphic as $\QQ^d$-graded $R$-modules. Moreover, we write +\[ M \shcong M_1 \oplus \cdots \oplus M_k \] +if there exist $d_i \in \QQ^d$ such that $M\grcong M_1(d_1)\oplus\cdots\oplus M_k(d_k)$, where $M_i(d_i)$ denotes the degree shift of $M_i$ by $d_i$. The notation is compatible with refinements of decompositions. If $M \shcong M_1 \oplus \cdots \oplus M_k$ and $M_1 \shcong M_1^\prime \oplus \cdots \oplus M_\ell^\prime$, then $M \shcong M_1^\prime \oplus \cdots \oplus M_\ell^\prime \oplus M_2 \oplus \cdots \oplus M_k$. \section{Frobenius maps on toric algebras} -Let $\SGR$ be an affine semigroup, i.e., a finitely generated submonoid of $\ZZ^d$ for some $d \in \NN$, and let $R \coloneqq \kk[\SGR]$ denote its semigroup algebra. Fix a positive integer $q$, and write $q^{-1}\SGR = \{ a/q \mid a \in \SGR \} \subseteq \QQ^d$ for the rescaling of $\SGR$ by the factor $1/q$. Consider the semigroup algebra +Let $R = \kk[\SGR]$ be the toric algebra associated with an affine monoid $\SGR \subseteq \ZZ^d$. Fix a positive integer $q$, and write $q^{-1}\SGR = \{ a/q \mid a \in \SGR \} \subseteq \QQ^d$ for the rescaling of $\SGR$ by the factor $1/q$. Consider the semigroup algebra \[ R^{1/q} \coloneqq \kk[q^{-1}\SGR] = \spn_\kk\{ x^{a/q} \mid a \in \SGR \} \] -regarded as a subring of $\kk[\QQ^d] = \spn_\kk \{ x^\alpha \mid \alpha \in \QQ^d \}$, which carries a natural $\QQ^d$-grading. While the semigroup ring $R^{1/q}$ is defined for arbitrary $q \in \NN$ in any characteristic, the notation is motivated by the case $q = p^e$ with $\kk$ perfect of prime characteristic $p$, where $R^{1/q}$ is the ring of $q$-th roots of elements in $R$. Further, for $r = \sum_{a \in \QQ^d} k_a x^{a} \in \kk[\QQ^d]$, we set $\qpower{r} \coloneqq \sum_{a \in \QQ^d} k_a x^{qa}$, and write $\qpower{F}$ for the resulting \emph{$q$-th pseudo-Frobenius (ring) endomorphism} of $R$, which maps $r$ to $\qpower{r}$. +regarded as a $\kk$-subalgebra of $\kk[\QQ^d] = \spn_\kk \{ x^\alpha \mid \alpha \in \QQ^d \}$, which carries a natural $\QQ^d$-grading. While the algebra $R^{1/q}$ is defined for arbitrary $q \in \NN$ in any characteristic, the notation is motivated by the case $q = p^e$ with $\kk$ perfect of prime characteristic $p$, where $R^{1/q}$ is the ring of $q$-th roots of elements in $R$. Further, for $r = \sum_{a \in \QQ^d} k_a x^{a} \in \kk[\QQ^d]$, we set $\qpower{r} \coloneqq \sum_{a \in \QQ^d} k_a x^{qa}$, and denote by $\qpower{F}$ the resulting $\kk$-algebra endomorphism +\[ \qpower{F} \colon R \to R, \qquad r \longmapsto \qpower{r}, \] +%of $R$ mapping $r$ to $\qpower{r}$, +referred to as the \emph{$q$-th pseudo-Frobenius endomorphism} of $R$. \begin{remark} % PREVIOUS VERSION % Note that even in positive characteristic $p$ with $q = p^e$, the $q$-th pseudo-Frobenius twist need not agree with the usual $q$-th Frobenius twist, where $R$-scalar multiplication is pulled back via the $e$-times iterated Frobenius endomorphism $F^e \colon r \mapsto r^{p^e}$ on $R$. The two agree precisely when the underlying field $\kk$ is finite, in which case $\kk \subseteq R$ is fixed pointwise by $F^e$. -Note that even in positive characteristic $p$ with $q = p^e$, the $q$-th pseudo-Frobenius endomorphism $\qpower{F}$ need not agree with the usual $e$-times iterated Frobenius endomorphism $F^e \colon R \to R$, mapping $r$ to $r^{q}$. %The two agree precisely when the underlying field $\kk$ is finite of order $p^f$ with $f \mid e$, in which case $\kk \subseteq R$ is fixed pointwise by $F^e$. +Note that even in prime characteristic $p$ with $q = p^e$, the $q$-th pseudo-Frobenius endomorphism $\qpower{F}$ need not agree with the usual $e$-times iterated Frobenius endomorphism $F^e \colon R \to R$, mapping $r$ to $r^{q}$. %The two agree precisely when the underlying field $\kk$ is finite of order $p^f$ with $f \mid e$, in which case $\kk \subseteq R$ is fixed pointwise by $F^e$. \end{remark} \begin{proposition} -Assume that $\kk$ is a field of prime characteristic $p$ and that $q = p^e$. Then the $e$-times iterated Frobenius endomorphism $F^e$ and the $q$-th pseudo-Frobenius endomorphism $\qpower{F}$ agree precisely when $\kk$ is finite of order $p^f$ with $f \mid e$. +Assume that $\kk$ is a field of prime characteristic $p$ and that \mbox{$q = p^e$}. Then the $e$-times iterated Frobenius endomorphism $F^e$ and the $q$-th pseudo-Frobenius endomorphism $\qpower{F}$ agree precisely when $\kk$ is finite of order $p^f$ with $f \mid e$. \end{proposition} \begin{proof} @@ -184,7 +163,7 @@ where the last equality is a standard identity; see, for example,~\cite[Example \end{proof} \begin{definition} -Let $M$ be a $\QQ^d$-graded $R$-module. The \textbf{$q$-th pseudo-Frobenius twist} of $M$, denoted $\qpower{M}$, is the $R$-module obtained by restricting scalars along the \mbox{$q$-th} pseudo-Frobenius endomorphism $\qpower{F}$. That is, $\qpower{M}$ is the $R$-module with the same underlying abelian group as $M$, but with $R$-action given by~$r \qmult m \coloneqq \qpower{r}m$ +Let $M$ be a $\QQ^d$-graded $R$-module. The \textbf{$q$-th pseudo-Frobenius twist} of $M$, denoted $\qpower{M}$, is the $R$-module obtained by restricting scalars along the \mbox{$q$-th} pseudo-Frobenius endomorphism $\qpower{F}$. That is, $\qpower{M}$ is the $R$-module with the same underlying abelian group as $M$, but with $R$-action given by~$r \qmult m \coloneqq \qpower{r} \cdot m$ % EXPLICIT DECLARATION % \[ \Bigl( \sum_{a \in S} k_a x^a \Bigr) \cdot_q \; m \coloneqq \bigl( \sum_{a \in S} k_a x^{aq} \bigr) m \] for all $r \in R$ and $m \in M$. \end{definition} @@ -202,18 +181,34 @@ defined by $r = \sum_{a \in \SGR} k_a x^{a/q} \mapsto \qpower{r} = \sum_{a \in \ \end{lemma} \begin{proof} -By abuse of notation, we also denote by $\qpower{F}$ the map $R^{1/q} \to \qpower{R}$ defined above. This map is clearly additive, bijective and preserves the $\QQ^d$-grading. To verify that it is compatible with the respective $R$-module structures, let $r = \sum_{a \in \SGR} k_a x^a \in R$ and $m = \sum_{a \in \SGR} l_a x^{a/q} \in R^{1/q}$. Then -\[ \qpower{F}(rm) = \qpower{F}\Bigl(\sum_{a, b \in \SGR} k_a l_b x^{a + \frac{b}{q}}\Bigr) = \sum_{a, b \in \SGR} k_a l_b x^{qa + b}, \] +By abuse of notation, we also denote by $\qpower{F}$ the map $R^{1/q} \to \qpower{R}$ defined above. This map is clearly additive, bijective and preserves the $\QQ^d$-grading. To verify that it is compatible with the respective $R$-module structures, let $r = \sum_{a \in \SGR} k_a x^a \in R$ and $m = \sum_{a \in \SGR} \ell_a x^{a/q} \in R^{1/q}$. Then +\[ \qpower{F}(rm) = \qpower{F}\Bigl(\sum_{a, b \in \SGR} k_a \ell_b x^{a + \frac{b}{q}}\Bigr) = \sum_{a, b \in \SGR} k_a \ell_b x^{qa + b}, \] which is precisely $r \qmult \qpower{F}(m)$ in $\qpower{R}$. \end{proof} +\begin{lemma}\label{lemma:qpower-under-shift} +Let $M$, $M_1$, \dots, $M_k$ be $\QQ^d$-graded $R$-modules. If it holds that \mbox{$M \shcong M_1 \oplus \cdots \oplus M_k$}, then $\qpower{M} \shcong \qpower{M_1} \oplus \cdots \oplus \qpower{M_k}$. +\end{lemma} + +\begin{proof} +For simplicity of notation, we prove the result only for $k =2$. By assumption, there are $d_1$, $d_2 \in \QQ^d$ such that $M \grcong M_1(d_1) \oplus M_2(d_2)$. As the construction of the pseudo-Frobenius twist respects direct sums, we have +\[ \qpower{M} \grcong \qpower{(M_1(d_1))} \oplus \qpower{(M_2(d_2))} = (\qpower{M_1})(d_1/q) \oplus (\qpower{M_2})(d_2/q), \] +where the last equality follows from the computation +\[ \setlength{\arraycolsep}{2pt} \begin{array}{rcccc} +\left(\qpower{(M_1(d_1))} \oplus \qpower{(M_2(d_2))}\right)_\alpha & = & \bigl(M_1(d_1)\bigr)_{q\alpha} & \oplus & \bigl(M_2(d_2)\bigr)_{q\alpha} \\ +& = & (M_1)_{q\alpha+d_1} & \oplus & (M_2)_{q\alpha+d_2} \\ +& = & (\qpower{M_1})_{\alpha+\frac{d_1}{q}} & \oplus & (\qpower{M_2})_{\alpha+\frac{d_2}{q}} \\ +& = & \left((\qpower{M_1})(d_1/q)\right)_\alpha & \oplus & \left((\qpower{M_2})(d_2/q)\right)_\alpha. +\end{array} \] +\end{proof} + \begin{definition} We say that $R$ is \textbf{$q$-split} if the pseudo-Frobenius map $\qpower{F} \colon R \to \qpower{R}$ splits as a map of $R$-modules, that is, if there exists an $R$-module homomorphism $\sigma \colon \qpower{R} \to R$ such that $\sigma \circ \qpower{F} = \id_R$. Moreover, we call $R$ \textbf{$q$-weakly normal} if, for every $r \in \kk[\gp(\SGR)]$, $\qpower{r} \in R$ implies $r \in R$. \end{definition} \begin{remark} %Recall that $R$ is $q$-split if and only if $R$ is a direct summand of $\qpower{R}$; see, for example,~\cite[§ 10, Prop. 26]{Dummit2004}. -By~\cite[Prop. 2.3.1]{Nastasescu2004}, the map $\qpower{F}$ splits as an $R$-module homomorphism if and only if it splits in the category of $\QQ^d$-graded $R$-modules. In other words, whenever $\qpower{F}$ splits, the splitting map $\sigma$ may be chosen to be $\QQ^d$-graded. Hence, $R$ is $q$-split if and only if it is a graded direct summand of $\qpower{R}$, that is, if there exists a $\QQ^d$-graded $R$-module $N$ such that $\qpower{R}\cong R \oplus N$ as $\QQ^d$-graded $R$-modules. +By~\cite[Prop. 2.3.1]{Nastasescu2004}, the map $\qpower{F}$ splits as an $R$-module homomorphism if and only if it splits in the category of $\QQ^d$-graded $R$-modules. In other words, whenever $\qpower{F}$ splits, the splitting map $\sigma$ may be chosen to be $\QQ^d$-graded. Hence, $R$ is $q$-split if and only if it is a graded direct summand of $\qpower{R}$, that is, if there exists a $\QQ^d$-graded $R$-module $N$ such that $\qpower{R}\grcong R \oplus N$. \end{remark} %\begin{remark} @@ -251,7 +246,7 @@ $R$ is normal if and only if $R$ is $q$-weakly normal for all $q \in \NN$ if and \end{corollary} \begin{proof} -By~\cite[Prop. 1]{Hochster1972}, the semigroup algebra $R = \kk[\SGR]$ is normal if and only if $\SGR$ is normal, that is, if for every $a \in \gp(\SGR)$ and $q \in \NN$, the condition $qa \in \SGR$ implies $a \in \SGR$. This is precisely the semigroup-theoretic formulation of $R$ being $q$-weakly normal for all $q \in \NN$. By the previous proposition, this is equivalent to $R$ being $q$-weakly normal for every $q \in \NN$. +By~\cite[Prop. 1]{Hochster1972}, the toric algebra $R = \kk[\SGR]$ is normal if and only if $\SGR$ is normal, that is, if for every $a \in \gp(\SGR)$ and $q \in \NN$, the condition $qa \in \SGR$ implies $a \in \SGR$. This is precisely the monoid-theoretic formulation of $R$ being $q$-weakly normal for all $q \in \NN$. By the previous proposition, this is equivalent to $R$ being $q$-weakly normal for every $q \in \NN$. \end{proof} \begin{lemma}\textnormal{\cite[Lemma 5.0.3]{Blickle2013}}\label{lemma:q-split} @@ -269,7 +264,7 @@ Finally, fix $(\ell_1, \dots, \ell_k) \in \NN_0^k$, and set $q^\prime \coloneqq \section{$R$-module decomposition of $R^{1/q}$} We now specialize to the case where $\SGR = \langle a_1, \dots, a_n \rangle \subseteq \NN_0$ %= \{ c_1a_1 + \cdots + c_na_n \mid c_i \in \NN_0\} \subseteq \NN_0 \} -is a numerical semigroup and $R = \kk[\SGR]$ is the corresponding semigroup algebra. Following Shibuta~\cite{Shibuta2017}, we sketch, for a fixed $q \in \NN$, how $R^{1/q} = \kk[x^{a_1/q}, \dots, x^{a_n/q}]$ decomposes as a~$\QQ_{\geq 0}$-graded $R$-module. Observe that +is a numerical monoid and $R = \kk[\SGR]$ is the corresponding toric algebra. Following Shibuta~\cite{Shibuta2017}, we sketch, for a fixed $q \in \NN$, how $R^{1/q} = \kk[x^{a_1/q}, \dots, x^{a_n/q}]$ decomposes as a~$\QQ_{\geq 0}$-graded $R$-module. Observe that \[ \Gamma_q \coloneqq \left\{ \bigl( x^{a_1/q} \bigr)^{e_1} \cdots \bigl( x^{a_n/q} \bigr)^{e_n} \mid e_i \in \intval{0,q-1} \right\} \] @@ -278,7 +273,7 @@ $a/q$ with $a \in \NN_0$, this amounts to grouping elements by the residue of $a \[ \Gamma_q = \Gamma_{q,0} \sqcup \Gamma_{q,1} \sqcup \cdots \sqcup \Gamma_{q,q-1}, \] -where $\Gamma_{q,r} \coloneqq \{ x^{a/q} \in \Gamma_q \mid a \equiv r \pmod{q}\}$. Let $I_{q,r}$ be the $R$-submodule of $\kk[\QQ_{\geq 0}] \coloneqq \kk[x^q \mid q \in \QQ_{\geq 0}]$ +where $\Gamma_{q,r} \coloneqq \{ x^{a/q} \in \Gamma_q \mid a \equiv r \pmod{q}\}$. Let $I_{q,r}$ be the $R$-submodule of $\kk[\QQ_{\geq 0}] = \kk[x^q \mid q \in \QQ_{\geq 0}]$ generated by $\Gamma_{q,r}$. As a $\kk$-vector space, \begin{equation}\label{eq:ind-summand-as-vector-space} I_{q,r} = \spn_\kk \{ x^{a/q} \mid a \in \SGR, \, a \equiv r \pmod{q} \}, @@ -311,7 +306,7 @@ means $\beta - \alpha \in \SGR$. We write \[ \Min(\SGR) \coloneqq \{ A \subseteq \gapset \mid A \text{ is minimal with respect to } \preceq_\SGR \}. \] \end{remark} -\begin{corollary} +\begin{corollary}\label{cor:indec-summ} Every $\QQ_{\geq 0}$-graded indecomposable summand of $R^{1/q}$ is, up to a shift in degree, isomorphic to a gap module. \end{corollary} @@ -341,28 +336,27 @@ defines a bijection between $\Min(\SGR)$ and the set of isomorphism classes of $ \begin{proof} For simplicity, set $Q \coloneqq \Frac(R) = \kk(x)$ and fix $A \in \Min(\SGR)$. Observe that $Q \otimes_R I(A)$ is naturally isomorphic, as a $Q$-vector space, to $I(A)_{(0)} \subseteq Q$. Since $I(A)$ contains $1$, it follows that $I(A)_{(0)} = Q$ and $I(A)$ has rank $1$. -For the second part, each $I(A)$ with $A \in \Min(\SGR)$ is an $\NN_0$-graded $R$-submodule of $\overline{R}$, so the map is well-defined. To prove injectivity, suppose that $\varphi \colon I(A)\to I(B)$ is an $R$-module isomorphism for some $A$, $B \in \Min(\SGR)$. Localizing at $(0)$ yields an isomorphism of $Q$-vector spaces $\varphi_{(0)}\colon I(A)_{(0)} \to I(B)_{(0)}$, where both source and target are equal to $Q$. Hence $\varphi$, and thus also $\varphi_{(0)}$, is given by multiplication by some $g_0 \in Q$, so $\varphi(f) = g_0f$ for all $f \in I(A)$. A simple degree argument shows that $g_0 \in \kk \setminus \{ 0 \}$. Therefore, +For the second part, each $I(A)$ with $A \in \Min(\SGR)$ is an $\NN_0$-graded $R$-submodule of $\overline{R}$, so the map is well-defined. To prove injectivity, suppose that $\varphi \colon I(A) \to I(B)$ is an $R$-module isomorphism for some $A$, $B \in \Min(\SGR)$. Localizing at $(0)$ yields an isomorphism of $Q$-vector spaces $\varphi_{(0)}\colon I(A)_{(0)} \to I(B)_{(0)}$, where both source and target are equal to $Q$. Hence $\varphi$, and thus also $\varphi_{(0)}$, is given by multiplication by some $g_0 \in Q$, so $\varphi(f) = g_0f$ for all $f \in I(A)$. A simple degree argument shows that $g_0 \in \kk \setminus \{ 0 \}$. Therefore, \[ I(B) = \varphi(I(A)) = g_0 \cdot I(A) =I(A). \] -Since $A$ and $B$ are both minimal with respect to $\preceq_\SGR$, it follows that $A = B$. Now, let $I \subseteq \overline{R}$ be an arbitrary $\NN_0$-graded $R$-submodule. Because $R$ is Noetherian and $\overline{R}$ is finitely generated as an $R$-module, $I$ is finitely generated, i.e, $I = \langle x^\beta \mid \beta \in B\rangle_R$ for some finite subset $B \subseteq \NN_0$. Define $m \coloneqq \min(B)$ and $A^\prime \coloneqq \{ \beta - m \mid \beta \in B \} \setminus \{ 0 \}$. Consequently, $I = x^m \cdot I(A^\prime)$, which implies that $I \cong I(A^\prime)$. The claim then follows from the fact that $I(A^\prime)=I(A)$, where $A$ denotes the set of minimal elements of $A^\prime \cap (\gapset)$ with respect to $\preceq_\SGR$. +Since $A$ and $B$ are both minimal with respect to $\preceq_\SGR$, it follows that $A = B$. We now prove that the assignment is surjective. Let $I \subseteq \overline{R}$ be an arbitrary $\NN_0$-graded $R$-submodule. Because $R$ is Noetherian and $\overline{R}$ is finitely generated as an $R$-module, $I$ is finitely generated, i.e, $I = \langle x^\beta \mid \beta \in B\rangle_R$ for some finite subset $B \subseteq \NN_0$. Define $m \coloneqq \min(B)$ and $A^\prime \coloneqq \{ \beta - m \mid \beta \in B \} \setminus \{ 0 \}$. Consequently, $I = x^m \cdot I(A^\prime)$, which implies that $I \cong I(A^\prime)$. Surjectivity then follows from the equality $I(A^\prime)=I(A)$, where $A$ is the set of elements of $A^\prime \cap \gapset$ that are minimal with respect to $\preceq_\SGR$. \end{proof} \begin{remark} We now present another perspective on the gap modules $I(A)$. For $A \in \Min(\SGR)$, the $R$-module $I(A)$ admits the $\kk$-vector space description \[ I(A)=\spn_\kk \{x^\alpha \mid \alpha\in \SGR_A \}, \] where $\SGR_A\coloneqq (A \cup \{ 0 \}) +\SGR$. The set $\SGR_A$ is a normalized relative ideal of $\SGR$, that is, a subset of $\NN_0$ containing $0$ that is closed under addition by elements of $\SGR$. One readily checks that $A \mapsto \SGR_A$ defines a one-to-one correspondence between the elements of $\Min(\SGR)$ and the normalized relative ideals of $\SGR$. -\todo{reference for this terminology?} \end{remark} \begin{corollary} -For each $A \in \Min(\SGR)$, the gap module $I(A)$ is maximal Cohen--Macaulay. +For each $A \in \Min(\SGR)$, the gap module $I(A)$ is maximal Cohen--Macaulay. In particular, $R^{1/q}$ is maximal Cohen--Macaulay. \end{corollary} \begin{proof} By the Cohen--Seidenberg theorems, $\dim(R) = \dim(\overline{R}) = \dim(\kk[x]) = 1$. Fix $A \in \Min(\SGR)$ and a maximal ideal $\mathfrak{m}$ of $R$. Since $R$ is a domain of Krull dimension $1$, every maximal ideal has height $1$; in particular $\dim(R_\mathfrak{m}) = \hgt(\mathfrak{m}) = 1$. The localization $I(A)_\mathfrak{m}$ is a non-zero $R_\mathfrak{m}$-submodule of $\kk(x)$, hence torsion-free, so $1 \leq \depth(I(A)_\mathfrak{m}) \leq \dim(R_\mathfrak{m}) = 1$. Hence, $I(A)$ is maximal Cohen--Macaulay. -\end{proof} -\todo{add direct sums are Cohen--Macaulay (Bruns/Herzog, 1993, Prop. 1.2.9)} +By the depth lemma~\cite[Prop.~1.2.9]{Bruns1993}, the depth of a finite direct sum is the minimum of the depths of its summands. Hence, a finite direct sum of maximal Cohen--Macaulay modules is again maximal Cohen--Macaulay. By Corollary~\ref{cor:indec-summ}, the module $R^{1/q}$ decomposes as a finite direct sum of gap modules $I(A)$, and the claim follows. +\end{proof} \begin{lemma} For $q \in \NN$, it holds that $\qpower{\overline{R}} \shcong \overline{R}^{\oplus q}$. @@ -370,7 +364,7 @@ For $q \in \NN$, it holds that $\qpower{\overline{R}} \shcong \overline{R}^{\opl \begin{proof} First, note that -\[ \overline{R}^{1/q} = \kk[x^{1/q}] = \bigoplus_{r \in [0,q-1]} x^{r/q} \cdot \kk[x] \shcong \overline{R}^{\oplus q} \] +\[ \overline{R}^{1/q} = \kk[x^{1/q}] = \bigoplus_{r \in \intval{0,q-1}} x^{r/q} \cdot \kk[x] \shcong \overline{R}^{\oplus q} \] as $R$-modules. By Lemma~\ref{lemma:cannonc-isom}, $\overline{R}^{1/q}$ and $\qpower{\overline{R}}$ are isomorphic as $\QQ_{\geq 0}$-graded modules over $\overline{R}$, and hence, by restriction of scalars along $R \subseteq \overline{R}$, also as $\QQ_{\geq 0}$-graded modules over $R$. \end{proof} @@ -415,24 +409,24 @@ generates $R^{1/q} = \kk[x^{3/q},x^{4/q}]$ as an $R$-module. For $q =2$, this ho \[ R^{1/2} = \langle 1, x^2 \rangle_R \oplus x^{3/2} \cdot \langle 1, x^2 \rangle_R \shcong I(\{2 \})^{\oplus 2} \] For $q = 4$, a direct computation shows $R^{1/4} \shcong \overline{R}^{\oplus 4}$. Using the $\QQ_{\geq 0}$-graded isomorphism between $R^{1/q}$ and $\qpower{R}$, we compute \[ R^{1/8} \grcong \qpower[8]{R} = \qpower[2]{(\qpower[4]{R})} \grcong \qpower[2]{(R^{1/4})}. \] -Hence, by Lemma~\ref{lemma:qpower-under-shift}, it holds that $R^{1/8} \shcong (\qpower[2]{\overline{R}})^{\oplus 4}$. The preceding lemma then yields $R^{1/8} \shcong \overline{R}^{\oplus 8}$. By induction, we obtain that $R^{1/2^e} \shcong \overline{R}^{\oplus 2^e}$ for all $e \geq 2$. Consequently, there exist summands appearing in the $\QQ_{\geq 0}$-graded direct sum decomposition of $R^{1/2^e}$ for some $e \in \NN$ that do \emph{not} appear in the decomposition of $R^{1/2^\ell}$ for large $\ell$. This stands in contrast to the case where $\SGR$ is a normal affine semigroup and $R=\kk[\SGR]$ is a normal toric algebra~\cite{Bruns2005}. +Hence, by Lemma~\ref{lemma:qpower-under-shift}, it holds that $R^{1/8} \shcong (\qpower[2]{\overline{R}})^{\oplus 4}$. The preceding lemma then yields $R^{1/8} \shcong \overline{R}^{\oplus 8}$. By induction, we obtain that $R^{1/2^e} \shcong \overline{R}^{\oplus 2^e}$ for all \mbox{$e \geq 2$}. Consequently, there exist summands appearing in the $\QQ_{\geq 0}$-graded direct sum decomposition of $R^{1/2^e}$ for some $e \in \NN$ that do \emph{not} appear in the decomposition of $R^{1/2^\ell}$ for large $\ell$. This stands in contrast to the case where $\SGR$ is a normal affine monoid and $R=\kk[\SGR]$ is a normal toric algebra~\cite{Bruns2005}. \end{example} % NOTE: next proposition holds when q \in \SGR \begin{proposition}\label{prop:asy-behav} -It holds that $q \in \SGR$ if and only if every indecomposable summand of $R^{1/q}$ is isomorphic to the normalization $\overline{R} = \kk[x]$ of $R$, that is, $R^{1/q} \cong \overline{R}^{\oplus q}$. In particular, this holds for all sufficiently large $q$. +It holds that $q \in \SGR$ if and only if every $\QQ^d$-graded indecomposable summand of $R^{1/q}$ is, up to a degree shift, isomorphic to the normalization $\overline{R} = \kk[x]$ of $R$, that is, if $R^{1/q} \shcong \overline{R}^{\oplus q}$. In particular, this holds for all sufficiently large $q$. %For $q \in \SGR$, every indecomposable summands of $R^{1/q}$ is isomorphic to the normalization $\overline{R} = \kk[x]$ of $R$. That is, $R^{1/q} \cong \overline{R}^{\oplus q}$. In particular, this holds for all sufficiently large $q$. \end{proposition} \begin{proof} -Assume that $q \in SGR$, and fix $r \in \intval{0,q-1}$. For simplicity, set +Assume that $q \in \SGR$, and fix $r \in \intval{0,q-1}$. For simplicity, set \[ \supp(I_{q,r}) \coloneqq \{ a/q \mid a \in \SGR, \, a \equiv r \pmod{q} \} \] -and observe that $\supp(I_{q,r}) \subseteq \{ m_{q,r}/q + n \mid n \in \NN_0\}$, where $m_{q,r}/q$ is the minimum of $\supp(I_{q,r})$. Conversely, for each $n \in \NN_0$, the choice of $q$ ensures $nq \in \SGR$, and hence $m_{q,r} + nq \in \SGR$. Since $m_{q,r}+nq \equiv r \pmod q$, we -obtain $m_{q,r}/q + n \in \supp(I_{q,r})$, and altogether -\[ \supp(I_{q,r}) = \{ m_{q,r}/q + n \mid n \in \NN_0\}. \] +and observe that $\supp(I_{q,r}) \subseteq \{ \frac{m_{q,r}}{q} + \alpha \mid \alpha \in \NN_0\}$, where $\frac{m_{q,r}}{q}$ is the minimum of $\supp(I_{q,r})$. Conversely, for each $\alpha \in \NN_0$, the choice of $q$ ensures $\alpha q \in \SGR$, and hence $m_{q,r} + \alpha q \in \SGR$. Since $m_{q,r}+ \alpha q \equiv r \pmod q$, we +obtain $\frac{m_{q,r}}{q} + \alpha \in \supp(I_{q,r})$, and altogether +\[ \supp(I_{q,r}) = \{ \frac{m_{q,r}}{q} + \alpha \mid \alpha \in \NN_0\}. \] It now follows from~\eqref{eq:ind-summand-as-vector-space} that -\[ I_{q,r} = \spn_\kk \{ x^{m_{q,r}/q + n} \mid n \in \NN_0 \} = x^{m_{q,r}/q} \cdot \spn_\kk\{ x^n \mid n \in \NN_0 \} \cong k[x] . \] +\[ I_{q,r} = \spn_\kk \{ x^{m_{q,r}/q + \alpha} \mid \alpha \in \NN_0 \} = x^{\frac{m_{q,r}}{q}} \cdot \spn_\kk\{ x^\alpha \mid \alpha \in \NN_0 \} \cong k[x] . \] \end{proof} diff --git a/references.bib b/references.bib index 74f11af..0ee9951 100644 --- a/references.bib +++ b/references.bib @@ -207,4 +207,17 @@ author = {Rosales, J.C. and García-Sánchez, P. A.}, urldate = {2026-09-16}, date = {2009}, langid = {english}, +} + +@book{Bruns1993, +location = {Cambridge [England] ; New York}, +title = {Cohen-Macaulay rings}, +isbn = {978-0-521-41068-7}, +series = {Cambridge studies in advanced mathematics}, +pagetotal = {403}, +number = {39}, +publisher = {Cambridge University Press}, +author = {Bruns, Winfried and Herzog, Jürgen}, +date = {1993}, +keywords = {Cohen-Macaulay rings}, } \ No newline at end of file