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ncr-of-monomial-curves.tex
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@ -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

59
references.bib
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@ -57,3 +57,62 @@ 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.}
}