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