first part of the preliminaries
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Markus Maximilian Rudolf Tripp
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\usepackage{todonotes}
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\usepackage{blindtext}
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\usepackage{tikz}
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\usepackage[most]{tcolorbox}
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\usepackage{graphicx}
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\usetikzlibrary{arrows.meta,positioning,decorations.pathreplacing}
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%\usepackage{parskip}
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\clearlist{language}%
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}
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\numberwithin{equation}{section}
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\newtheorem{theorem}{Theorem}[section]
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\newtheorem{lemma}[theorem]{Lemma}
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\newtheorem{proposition}[theorem]{Proposition}
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}
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\DeclareMathOperator{\supp}{supp}
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\DeclareMathOperator{\Spec}{Spec}
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\DeclareMathOperator{\Frac}{Frac}
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\DeclareMathOperator{\spn}{span}
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\DeclareMathOperator{\Min}{Min}
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@ -88,9 +92,9 @@
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\section{Introduction}
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some results/terminology in section 2 are made independent of the characteristic
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% some results/terminology in section 2 are made independent of the characteristic
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\section{Monomial curves}
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\section{Preliminaries}
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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.
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@ -98,47 +102,64 @@ Throughout, we fix an arbitrary ground field $\kk$. We begin by recalling some b
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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
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\[ R \coloneqq \kk[\SGR] = \spn_\kk \{ x^a \mid a \in \SGR \}. \]
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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}.
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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, $a_n \in \NN_0$.
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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
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\begin{equation}\label{eq:para-mon-curve}
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\varphi \colon \A^1_\kk \to \A^n_\kk, \qquad t \longmapsto (t^{a_1},\dots,t^{a_n}).
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\end{equation}
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\subsection{Numerical monoids}
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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}.
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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 $X=\A_\kk^1$.
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\subsection{Normality, seminormality, and weak normality}
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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
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\[
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\lambda^p = (\lambda^2)^{n^\prime}(\lambda^3)^{n^{\prime\prime}} \in \Lambda,
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\]
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and hence $\lambda \in \Lambda$.
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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?}
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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.
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%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 $x^{n_1}\in{}^+R$.
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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.
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\subsection{Graded Krull–Schmidt theorem}
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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}.
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% 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)
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%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.
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\todo{use widebar}
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%Clearly if $M \shcong M_1 \oplus M_2$ and $M_2 \shcong M_2^\prime \oplus M_2^{\prime\prime}$, then $M \shcong M_1 \oplus M_2^\prime \oplus M_2^{\prime\prime}$.
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\todo{adapt phrasinig in correspondence gap modules}
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\todo{use notation $\grcong$?}
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%\begin{lemma}\label{lemma:qpower-under-shift}
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%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 $\qpower{M} \shcong \qpower{M^{\prime}} \oplus \qpower{M^{\prime\prime}}$.
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%\end{lemma}
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\todo{affine semigroup rings are (non-normal) toric varieties}
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%\begin{proof}
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%By assumption, there are $d^\prime$, $d^{\prime\prime} \in \QQ$ such that $M \grcong M^\prime(d^\prime) \oplus M^{\prime\prime}(d^{\prime\prime})$.
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\todo{1-dimensional toric varieties are monomial curves}
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%As the construction of the pseudo-Frobenius twist respects direct sums, we have
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%\[ \qpower{M} \grcong \qpower{(M^{\prime}(d^\prime))} \oplus \qpower{(M^{\prime\prime}(d^{\prime\prime}))} = (\qpower{M^{\prime}})(d^\prime/q) \oplus (\qpower{M^{\prime\prime}})(d^{\prime\prime}/q), \]
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%where the last equality follows from the computation
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%\begin{align*}
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%\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} \\
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%&= (M^{\prime})_{q\alpha + d^\prime} \oplus (M^{\prime\prime})_{q\alpha + d^{\prime\prime}} \\
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%&= (\qpower{M^\prime})_{\alpha + \frac{d^\prime}{q}} \oplus (\qpower{M^{\prime\prime}})_{\alpha + \frac{d^{\prime\prime}}{q}} \\
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%&= \left((\qpower{M^{\prime}})(d^\prime/q) \oplus (\qpower{M^{\prime\prime}})(d^{\prime\prime}/q)\right)_\alpha.
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%\end{align*}
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%\end{proof}
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\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}
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\todo{monomial curves are non-seminormal, non-normal, singular at 0}
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\todo{say what the normalization, fraction field is}
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\begin{lemma}
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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 $\qpower{M} \shcong \qpower{M^{\prime}} \oplus \qpower{M^{\prime\prime}}$.
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\end{lemma}
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\begin{proof}
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By assumption, there are $d^\prime$, $d^{\prime\prime} \in \QQ$ such that $M \grcong M^\prime(d^\prime) \oplus M^{\prime\prime}(d^{\prime\prime})$.
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As the construction of the pseudo-Frobenius twist respects direct sums, we have
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\[ \qpower{M} \grcong \qpower{(M^{\prime}(d^\prime))} \oplus \qpower{(M^{\prime\prime}(d^{\prime\prime}))} = (\qpower{M^{\prime}})(d^\prime/q) \oplus (\qpower{M^{\prime\prime}})(d^{\prime\prime}/q), \]
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where the last equality follows from the computation
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\begin{align*}
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\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} \\
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&= (M^{\prime})_{q\alpha + d^\prime} \oplus (M^{\prime\prime})_{q\alpha + d^{\prime\prime}} \\
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&= (\qpower{M^\prime})_{\alpha + \frac{d^\prime}{q}} \oplus (\qpower{M^{\prime\prime}})_{\alpha + \frac{d^{\prime\prime}}{q}} \\
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&= \left((\qpower{M^{\prime}})(d^\prime/q) \oplus (\qpower{M^{\prime\prime}})(d^{\prime\prime}/q)\right)_\alpha.
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\end{align*}
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\end{proof}
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\begin{proposition}
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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.
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\end{proposition}
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%\begin{proposition}
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%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.
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%\end{proposition}
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\section{Frobenius maps on toric algebras}
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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
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@ -230,7 +251,7 @@ $R$ is normal if and only if $R$ is $q$-weakly normal for all $q \in \NN$ if and
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\end{corollary}
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\begin{proof}
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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 $q \in \NN$.
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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 $q \in \NN$.
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\end{proof}
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\begin{lemma}\textnormal{\cite[Lemma 5.0.3]{Blickle2013}}\label{lemma:q-split}
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@ -280,7 +301,7 @@ where each summand is indecomposable (see Lemma~\ref{lemma:rank-1}).
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%\end{proof}
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\begin{definition}\label{def:gap-module}
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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
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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
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\[ I(A) \coloneqq \langle x^\alpha \mid \alpha \in A\cup \{0\} \rangle_R. \]
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\end{definition}
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\begin{lemma}\label{lemma:rank-1}
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For each $A \in \Min(\SGR)$, the gap module $I(A)$ has rank $1$ and is therefore indecomposable. Moreover, the assignment
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\[ A \mapsto [I(A)] \]
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defines a bijection between $\Min(\SGR)$ and the set of isomorphism classes of $\NN_0$-graded $R$-submodules of $\bar{R} = \kk[x]$.
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defines a bijection between $\Min(\SGR)$ and the set of isomorphism classes of $\NN_0$-graded $R$-submodules of $\overline{R} = \kk[x]$.
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\end{lemma}
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\begin{proof}
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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 $1$.
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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,
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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,
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\[ I(B) = \varphi(I(A)) = g_0 \cdot I(A) =I(A). \]
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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 $\preceq_\SGR$.
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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 $\preceq_\SGR$.
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\end{proof}
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\begin{remark}
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@ -337,20 +358,20 @@ For each $A \in \Min(\SGR)$, the gap module $I(A)$ is maximal Cohen--Macaulay.
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\end{corollary}
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\begin{proof}
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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
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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
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$I(A)_\mathfrak{m}$ 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.
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\end{proof}
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\todo{add direct sums are Cohen--Macaulay (Bruns/Herzog, 1993, Prop. 1.2.9)}
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\begin{lemma}
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For $q \in \NN$, it holds that $\qpower{\bar{R}} \shcong \bar{R}^{\oplus q}$.
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For $q \in \NN$, it holds that $\qpower{\overline{R}} \shcong \overline{R}^{\oplus q}$.
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\end{lemma}
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\begin{proof}
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First, note that
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\[ \bar{R}^{1/q} = \kk[x^{1/q}] = \bigoplus_{r \in [0,q-1]} x^{r/q} \cdot \kk[x] \cong \bar{R}^{\oplus q} \]
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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.
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\[ \overline{R}^{1/q} = \kk[x^{1/q}] = \bigoplus_{r \in [0,q-1]} x^{r/q} \cdot \kk[x] \shcong \overline{R}^{\oplus q} \]
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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 $R$.
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\end{proof}
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\begin{example}\label{ex:large-powers}
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@ -389,19 +410,19 @@ Let $\SGR = \langle 3,4 \rangle \subseteq \NN_0$ and $R = \kk[\SGR] = \kk[x^3,x^
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\end{center}
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The set
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\[ \{ x^{(3/q)e_1 + (4/q)e_2} : e_i \in \intval{0,q-1} \} \]
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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
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\[ R^{1/2} = \langle 1, x^2 \rangle_R \oplus x^{3/2} \cdot \langle 1, x^2 \rangle_R \cong I(\{2 \})^{\oplus 2} \]
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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
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\[ R^{1/8} \cong \qpower[2]{(\qpower[4]{R})} \cong \qpower[2]{(\bar{R}^{\oplus 4})} = (\qpower[2]{\bar{R}})^{\oplus 4} \cong \bar{R}^{\oplus 8}, \]
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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}.
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\[ \bigl\{ x^{(3/q)e_1 + (4/q)e_2} : e_i \in \intval{0,q-1} \bigr\} \]
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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
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\[ R^{1/2} = \langle 1, x^2 \rangle_R \oplus x^{3/2} \cdot \langle 1, x^2 \rangle_R \shcong I(\{2 \})^{\oplus 2} \]
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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
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\[ R^{1/8} \grcong \qpower[8]{R} = \qpower[2]{(\qpower[4]{R})} \grcong \qpower[2]{(R^{1/4})}. \]
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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}.
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\end{example}
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% NOTE: next proposition holds when q \in \SGR
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\begin{proposition}\label{prop:asy-behav}
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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 $q$.
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%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 $q$.
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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 $q$.
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%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 $q$.
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\end{proposition}
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\begin{proof}
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@ -420,8 +441,8 @@ It now follows from~\eqref{eq:ind-summand-as-vector-space} that
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\begin{example}
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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
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\[ R^{1/5}\cong I(\{2\}) \oplus \bar{R}^{\oplus 4}. \]
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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,
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\[ R^{1/5}\cong I(\{2\}) \oplus \overline{R}^{\oplus 4}. \]
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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,
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\[ \Min(\SGR) = \{ \emptyset,\{1\}, \{2\}, \{5\}, \{1,2\} \}, \]
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so not every potential candidate $I(A)$ with $A \in \Min(\SGR)$ occurs as a direct summand.
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\end{example}
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@ -433,7 +454,7 @@ For each $q \geq 2$, $R$ is not $q$-split, meaning that $R$ does not occur as a
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\todo{\emph{note:} this can also be easily proven directly!}
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\begin{proof}
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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.
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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.
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\end{proof}
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|
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\begin{proposition}
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|
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@ -457,8 +478,8 @@ For $q \in \SGR$, the global dimension of $\End_R(R^{1/q})$ is equal to the Krul
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\end{theorem}
|
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\begin{proof}
|
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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
|
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\[ \gldim(\End_R(R^{1/q})) = \gldim(\bar{R}) = \gldim(\kk[x]) = 1 = \dim(R), \]
|
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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
|
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\[ \gldim(\End_R(R^{1/q})) = \gldim(\overline{R}) = \gldim(\kk[x]) = 1 = \dim(R), \]
|
||||
where the first equality follows from the Morita invariance of global dimension.
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\end{proof}
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Reference in a new issue