repairing some stuff
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Markus Maximilian Rudolf Tripp
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2 changed files with 68 additions and 61 deletions
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@ -102,9 +102,9 @@ 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}. 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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The algebra $R$ is a finitely generated $\kk$-subalgebra of the Laurent polynomial ring~$\kk[x_1^{\pm 1}, \dots, x_d^{\pm 1}]$. In particular, it is a Noetherian domain and inherits the natural $\ZZ^d$-grading. Its Krull dimension is equal to the rank of $\gp(\SGR)$, where $\gp(\SGR)$ denotes the subgroup of $\ZZ^d$ generated by $\SGR$~\cite[Prop. 1.1.14]{Cox2011}. Consequently, $R$ is one-dimensional if and only if $\gp(\SGR) \cong \ZZ$, in which case we may assume without loss of generality that $\gp(\SGR) = \ZZ$. Every submonoid of $\ZZ$ is either a group, a submonoid of $\NN_0$, or a submonoid of $-\NN_0$~\cite[§ 2]{Geroldinger2025}. Hence, when $R$ is one-dimensional, $\SGR$ is, up to isomorphism of monoids, either $\ZZ$ or a submonoid of $\NN_0$. Accordingly, either $R\cong\kk[x^{\pm 1}]$ or $R \cong \kk[x^{a_1},\dots,x^{a_n}]$ for some $a_1$, \dots, $a_n \in \NN_0$.
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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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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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@ -117,62 +117,41 @@ For a numerical monoid $\SGR=\langle a_1,\dots, a_n\rangle \subseteq \NN_0$, the
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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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A domain $\Lambda$ is called \textbf{normal} if it is integrally closed in its field o.f fractions $\Frac(\Lambda)$, and \textbf{seminormal} if, for every $\lambda \in \Frac(\Lambda)$, $\lambda^2 \in \Lambda$ and $\lambda^3 \in \Lambda$ imply that $\lambda \in \Lambda$. The normalization~$\overline{\Lambda}$ and the seminormalization ${}^+\Lambda$ of $\Lambda$ are, respectively, the intersections of all intermediate normal and seminormal rings $\Gamma$ satisfying $\Lambda \subseteq \Gamma \subseteq \Frac(\Lambda)$ If $\Lambda$ has prime characteristic $p$, then $\Lambda$ is called \textbf{weakly normal} if, for every $\lambda \in \Frac(\Lambda)$, the condition $\lambda^p \in \Lambda$ implies $\lambda\in\Lambda$. Clearly, normality implies seminormality. In prime characteristic $p$, normality implies weak normality, and weak normality in turn implies seminormality. Indeed, suppose that $\Lambda$ is weakly normal and let $\lambda \in \Frac(\Lambda)$ satisfy $\lambda^2$, $\lambda^3 \in \Lambda$. Write $p = 2k + 3\ell$ for some $k$, $\ell \in \NN_0$. Then
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\[
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\lambda^p = (\lambda^2)^{n^\prime}(\lambda^3)^{n^{\prime\prime}} \in \Lambda,
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\lambda^p = (\lambda^2)^k(\lambda^3)^\ell \in \Lambda,
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\]
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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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For a toric algebra $R = \kk[\SGR]$, the field of fractions in the definitions above may be replaced by $\kk[\gp(\SGR)]$. Indeed, $\gp(\SGR) \cong \ZZ^{d^\prime}$ for some $d^\prime \in \NN_0$, so $\kk[\gp(\SGR)]$ is isomorphic to a Laurent polynomial ring and is therefore normal. Consequently, every element of $\Frac(R)$ that is integral over $R$ already belongs to $\kk[\gp(\SGR)]$.\todo{mention normality, ... for affine monoids?}
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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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In the case of a numerical monoid $\SGR$, the corresponding toric algebra $R$ satisfies ${}^+R=\kk[x]$. Indeed, let $n_1 > \cdots > n_k$ be the elements of $\gapset$. Since $n_1$ is the largest gap of $\SGR$, we have $(x^{n_1})^2$, $(x^{n_1})^3 \in R \subseteq{}^+R$, and hence $x^{n_1}\in{}^+R$ by seminormality.
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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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Repeating this argument successively for the remaining gaps shows that $x^{n_1}$, \dots, $x^{n_k} \in{}^+R$, and therefore $\kk[x] \subseteq{}^+R$. Since $\kk[x]$ is normal, hence seminormal, the reverse inclusion follows. As $\kk[x]$ is normal, it also follows that $\overline{R} = \kk[x]$, and consequently $\Frac(R) = \kk(x)$. Thus, unless $\SGR=\NN_0$, the ring $R$ is neither normal nor seminormal.
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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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The category of finitely generated $\QQ^d$-graded $R$-modules satisfies the \textbf{Krull--Schmidt property}. That is, in the graded category, every such module decomposes as a finite direct sum of indecomposable modules, uniquely up to graded isomorphism and permutation of the summands~\cite[§ 2]{Shibuta2017}.
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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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%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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%\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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%\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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We are interested in the summands of such graded decompositions only up to degree shifts, and therefore introduce the following notation. Let $M$, $N$, $M_1$, \dots,~$M_k$ be $\QQ^d$-graded $R$-modules. We write $M \grcong N$ if $M$ and $N$ are isomorphic as $\QQ^d$-graded $R$-modules. Moreover, we write
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\[ M \shcong M_1 \oplus \cdots \oplus M_k \]
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if there exist $d_i \in \QQ^d$ such that $M\grcong M_1(d_1)\oplus\cdots\oplus M_k(d_k)$, where $M_i(d_i)$ denotes the degree shift of $M_i$ by $d_i$. The notation is compatible with refinements of decompositions. If $M \shcong M_1 \oplus \cdots \oplus M_k$ and $M_1 \shcong M_1^\prime \oplus \cdots \oplus M_\ell^\prime$, then $M \shcong M_1^\prime \oplus \cdots \oplus M_\ell^\prime \oplus M_2 \oplus \cdots \oplus M_k$.
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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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Let $R = \kk[\SGR]$ be the toric algebra associated with an affine monoid $\SGR \subseteq \ZZ^d$. Fix a positive integer $q$, and write $q^{-1}\SGR = \{ a/q \mid a \in \SGR \} \subseteq \QQ^d$ for the rescaling of $\SGR$ by the factor $1/q$. Consider the semigroup algebra
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\[ R^{1/q} \coloneqq \kk[q^{-1}\SGR] = \spn_\kk\{ x^{a/q} \mid a \in \SGR \} \]
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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 $\qpower{r}$.
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regarded as a $\kk$-subalgebra of $\kk[\QQ^d] = \spn_\kk \{ x^\alpha \mid \alpha \in \QQ^d \}$, which carries a natural $\QQ^d$-grading. While the algebra $R^{1/q}$ is defined for arbitrary $q \in \NN$ in any characteristic, the notation is motivated by the case $q = p^e$ with $\kk$ perfect of prime characteristic $p$, where $R^{1/q}$ is the ring of $q$-th roots of elements in $R$. Further, for $r = \sum_{a \in \QQ^d} k_a x^{a} \in \kk[\QQ^d]$, we set $\qpower{r} \coloneqq \sum_{a \in \QQ^d} k_a x^{qa}$, and denote by $\qpower{F}$ the resulting $\kk$-algebra endomorphism
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\[ \qpower{F} \colon R \to R, \qquad r \longmapsto \qpower{r}, \]
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%of $R$ mapping $r$ to $\qpower{r}$,
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referred to as the \emph{$q$-th pseudo-Frobenius endomorphism} of $R$.
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\begin{remark}
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% 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 $F^e$.
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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 $F^e$.
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Note that even in prime characteristic $p$ with $q = p^e$, the $q$-th pseudo-Frobenius endomorphism $\qpower{F}$ need not agree with the usual $e$-times iterated Frobenius endomorphism $F^e \colon R \to R$, mapping $r$ to $r^{q}$. %The two agree precisely when the underlying field $\kk$ is finite of order $p^f$ with $f \mid e$, in which case $\kk \subseteq R$ is fixed pointwise by $F^e$.
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\end{remark}
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\begin{proposition}
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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 $f \mid e$.
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Assume that $\kk$ is a field of prime characteristic $p$ and that \mbox{$q = p^e$}. Then the $e$-times iterated Frobenius endomorphism $F^e$ and the $q$-th pseudo-Frobenius endomorphism $\qpower{F}$ agree precisely when $\kk$ is finite of order $p^f$ with $f \mid e$.
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\end{proposition}
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\begin{proof}
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@ -184,7 +163,7 @@ where the last equality is a standard identity; see, for example,~\cite[Example
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\end{proof}
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\begin{definition}
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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~$r \qmult m \coloneqq \qpower{r}m$
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Let $M$ be a $\QQ^d$-graded $R$-module. The \textbf{$q$-th pseudo-Frobenius twist} of $M$, denoted $\qpower{M}$, is the $R$-module obtained by restricting scalars along the \mbox{$q$-th} pseudo-Frobenius endomorphism $\qpower{F}$. That is, $\qpower{M}$ is the $R$-module with the same underlying abelian group as $M$, but with $R$-action given by~$r \qmult m \coloneqq \qpower{r} \cdot m$
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% EXPLICIT DECLARATION % \[ \Bigl( \sum_{a \in S} k_a x^a \Bigr) \cdot_q \; m \coloneqq \bigl( \sum_{a \in S} k_a x^{aq} \bigr) m \]
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for all $r \in R$ and $m \in M$.
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\end{definition}
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@ -202,18 +181,34 @@ defined by $r = \sum_{a \in \SGR} k_a x^{a/q} \mapsto \qpower{r} = \sum_{a \in \
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\end{lemma}
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\begin{proof}
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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
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\[ \qpower{F}(rm) = \qpower{F}\Bigl(\sum_{a, b \in \SGR} k_a l_b x^{a + \frac{b}{q}}\Bigr) = \sum_{a, b \in \SGR} k_a l_b x^{qa + b}, \]
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By abuse of notation, we also denote by $\qpower{F}$ the map $R^{1/q} \to \qpower{R}$ defined above. This map is clearly additive, bijective and preserves the $\QQ^d$-grading. To verify that it is compatible with the respective $R$-module structures, let $r = \sum_{a \in \SGR} k_a x^a \in R$ and $m = \sum_{a \in \SGR} \ell_a x^{a/q} \in R^{1/q}$. Then
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\[ \qpower{F}(rm) = \qpower{F}\Bigl(\sum_{a, b \in \SGR} k_a \ell_b x^{a + \frac{b}{q}}\Bigr) = \sum_{a, b \in \SGR} k_a \ell_b x^{qa + b}, \]
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which is precisely $r \qmult \qpower{F}(m)$ in $\qpower{R}$.
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\end{proof}
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\begin{lemma}\label{lemma:qpower-under-shift}
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Let $M$, $M_1$, \dots, $M_k$ be $\QQ^d$-graded $R$-modules. If it holds that \mbox{$M \shcong M_1 \oplus \cdots \oplus M_k$}, then $\qpower{M} \shcong \qpower{M_1} \oplus \cdots \oplus \qpower{M_k}$.
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\end{lemma}
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\begin{proof}
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For simplicity of notation, we prove the result only for $k =2$. By assumption, there are $d_1$, $d_2 \in \QQ^d$ such that $M \grcong M_1(d_1) \oplus M_2(d_2)$. As the construction of the pseudo-Frobenius twist respects direct sums, we have
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\[ \qpower{M} \grcong \qpower{(M_1(d_1))} \oplus \qpower{(M_2(d_2))} = (\qpower{M_1})(d_1/q) \oplus (\qpower{M_2})(d_2/q), \]
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where the last equality follows from the computation
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\[ \setlength{\arraycolsep}{2pt} \begin{array}{rcccc}
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\left(\qpower{(M_1(d_1))} \oplus \qpower{(M_2(d_2))}\right)_\alpha & = & \bigl(M_1(d_1)\bigr)_{q\alpha} & \oplus & \bigl(M_2(d_2)\bigr)_{q\alpha} \\
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& = & (M_1)_{q\alpha+d_1} & \oplus & (M_2)_{q\alpha+d_2} \\
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& = & (\qpower{M_1})_{\alpha+\frac{d_1}{q}} & \oplus & (\qpower{M_2})_{\alpha+\frac{d_2}{q}} \\
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& = & \left((\qpower{M_1})(d_1/q)\right)_\alpha & \oplus & \left((\qpower{M_2})(d_2/q)\right)_\alpha.
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\end{array} \]
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\end{proof}
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\begin{definition}
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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 $r \in R$.
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\end{definition}
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\begin{remark}
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%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}.
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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.
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By~\cite[Prop. 2.3.1]{Nastasescu2004}, the map $\qpower{F}$ splits as an $R$-module homomorphism if and only if it splits in the category of $\QQ^d$-graded $R$-modules. In other words, whenever $\qpower{F}$ splits, the splitting map $\sigma$ may be chosen to be $\QQ^d$-graded. Hence, $R$ is $q$-split if and only if it is a graded direct summand of $\qpower{R}$, that is, if there exists a $\QQ^d$-graded $R$-module $N$ such that $\qpower{R}\grcong R \oplus N$.
|
||||
\end{remark}
|
||||
|
||||
%\begin{remark}
|
||||
|
|
@ -251,7 +246,7 @@ $R$ is normal if and only if $R$ is $q$-weakly normal for all $q \in \NN$ if and
|
|||
\end{corollary}
|
||||
|
||||
\begin{proof}
|
||||
By~\cite[Prop. 1]{Hochster1972}, the semigroup algebra $R = \kk[\SGR]$ is normal if and only if $\SGR$ is normal, that is, if for every $a \in \gp(\SGR)$ and $q \in \NN$, the condition $qa \in \SGR$ implies $a \in \SGR$. This is precisely the semigroup-theoretic formulation of $R$ being $q$-weakly normal for all $q \in \NN$. By the previous proposition, this is equivalent to $R$ being $q$-weakly normal for every $q \in \NN$.
|
||||
By~\cite[Prop. 1]{Hochster1972}, the toric algebra $R = \kk[\SGR]$ is normal if and only if $\SGR$ is normal, that is, if for every $a \in \gp(\SGR)$ and $q \in \NN$, the condition $qa \in \SGR$ implies $a \in \SGR$. This is precisely the monoid-theoretic formulation of $R$ being $q$-weakly normal for all $q \in \NN$. By the previous proposition, this is equivalent to $R$ being $q$-weakly normal for every $q \in \NN$.
|
||||
\end{proof}
|
||||
|
||||
\begin{lemma}\textnormal{\cite[Lemma 5.0.3]{Blickle2013}}\label{lemma:q-split}
|
||||
|
|
@ -269,7 +264,7 @@ Finally, fix $(\ell_1, \dots, \ell_k) \in \NN_0^k$, and set $q^\prime \coloneqq
|
|||
\section{$R$-module decomposition of $R^{1/q}$}
|
||||
|
||||
We now specialize to the case where $\SGR = \langle a_1, \dots, a_n \rangle \subseteq \NN_0$ %= \{ c_1a_1 + \cdots + c_na_n \mid c_i \in \NN_0\} \subseteq \NN_0 \}
|
||||
is a numerical semigroup and $R = \kk[\SGR]$ is the corresponding semigroup algebra. Following Shibuta~\cite{Shibuta2017}, we sketch, for a fixed $q \in \NN$, how $R^{1/q} = \kk[x^{a_1/q}, \dots, x^{a_n/q}]$ decomposes as a~$\QQ_{\geq 0}$-graded $R$-module. Observe that
|
||||
is a numerical monoid and $R = \kk[\SGR]$ is the corresponding toric algebra. Following Shibuta~\cite{Shibuta2017}, we sketch, for a fixed $q \in \NN$, how $R^{1/q} = \kk[x^{a_1/q}, \dots, x^{a_n/q}]$ decomposes as a~$\QQ_{\geq 0}$-graded $R$-module. Observe that
|
||||
\[
|
||||
\Gamma_q \coloneqq \left\{ \bigl( x^{a_1/q} \bigr)^{e_1} \cdots \bigl( x^{a_n/q} \bigr)^{e_n} \mid e_i \in \intval{0,q-1} \right\}
|
||||
\]
|
||||
|
|
@ -278,7 +273,7 @@ $a/q$ with $a \in \NN_0$, this amounts to grouping elements by the residue of $a
|
|||
\[
|
||||
\Gamma_q = \Gamma_{q,0} \sqcup \Gamma_{q,1} \sqcup \cdots \sqcup \Gamma_{q,q-1},
|
||||
\]
|
||||
where $\Gamma_{q,r} \coloneqq \{ x^{a/q} \in \Gamma_q \mid a \equiv r \pmod{q}\}$. Let $I_{q,r}$ be the $R$-submodule of $\kk[\QQ_{\geq 0}] \coloneqq \kk[x^q \mid q \in \QQ_{\geq 0}]$
|
||||
where $\Gamma_{q,r} \coloneqq \{ x^{a/q} \in \Gamma_q \mid a \equiv r \pmod{q}\}$. Let $I_{q,r}$ be the $R$-submodule of $\kk[\QQ_{\geq 0}] = \kk[x^q \mid q \in \QQ_{\geq 0}]$
|
||||
generated by $\Gamma_{q,r}$. As a $\kk$-vector space,
|
||||
\begin{equation}\label{eq:ind-summand-as-vector-space}
|
||||
I_{q,r} = \spn_\kk \{ x^{a/q} \mid a \in \SGR, \, a \equiv r \pmod{q} \},
|
||||
|
|
@ -311,7 +306,7 @@ means $\beta - \alpha \in \SGR$. We write
|
|||
\[ \Min(\SGR) \coloneqq \{ A \subseteq \gapset \mid A \text{ is minimal with respect to } \preceq_\SGR \}. \]
|
||||
\end{remark}
|
||||
|
||||
\begin{corollary}
|
||||
\begin{corollary}\label{cor:indec-summ}
|
||||
Every $\QQ_{\geq 0}$-graded indecomposable summand of $R^{1/q}$ is, up to a shift in degree, isomorphic to a gap module.
|
||||
\end{corollary}
|
||||
|
||||
|
|
@ -341,28 +336,27 @@ defines a bijection between $\Min(\SGR)$ and the set of isomorphism classes of $
|
|||
\begin{proof}
|
||||
For simplicity, set $Q \coloneqq \Frac(R) = \kk(x)$ and fix $A \in \Min(\SGR)$. Observe that $Q \otimes_R I(A)$ is naturally isomorphic, as a $Q$-vector space, to $I(A)_{(0)} \subseteq Q$. Since $I(A)$ contains $1$, it follows that $I(A)_{(0)} = Q$ and $I(A)$ has rank $1$.
|
||||
|
||||
For the second part, each $I(A)$ with $A \in \Min(\SGR)$ is an $\NN_0$-graded $R$-submodule of $\overline{R}$, so the map is well-defined. To prove injectivity, suppose that $\varphi \colon I(A)\to I(B)$ is an $R$-module isomorphism for some $A$, $B \in \Min(\SGR)$. Localizing at $(0)$ yields an isomorphism of $Q$-vector spaces $\varphi_{(0)}\colon I(A)_{(0)} \to I(B)_{(0)}$, where both source and target are equal to $Q$. Hence $\varphi$, and thus also $\varphi_{(0)}$, is given by multiplication by some $g_0 \in Q$, so $\varphi(f) = g_0f$ for all $f \in I(A)$. A simple degree argument shows that $g_0 \in \kk \setminus \{ 0 \}$. Therefore,
|
||||
For the second part, each $I(A)$ with $A \in \Min(\SGR)$ is an $\NN_0$-graded $R$-submodule of $\overline{R}$, so the map is well-defined. To prove injectivity, suppose that $\varphi \colon I(A) \to I(B)$ is an $R$-module isomorphism for some $A$, $B \in \Min(\SGR)$. Localizing at $(0)$ yields an isomorphism of $Q$-vector spaces $\varphi_{(0)}\colon I(A)_{(0)} \to I(B)_{(0)}$, where both source and target are equal to $Q$. Hence $\varphi$, and thus also $\varphi_{(0)}$, is given by multiplication by some $g_0 \in Q$, so $\varphi(f) = g_0f$ for all $f \in I(A)$. A simple degree argument shows that $g_0 \in \kk \setminus \{ 0 \}$. Therefore,
|
||||
\[ I(B) = \varphi(I(A)) = g_0 \cdot I(A) =I(A). \]
|
||||
Since $A$ and $B$ are both minimal with respect to $\preceq_\SGR$, it follows that $A = B$. Now, let $I \subseteq \overline{R}$ be an arbitrary $\NN_0$-graded $R$-submodule. Because $R$ is Noetherian and $\overline{R}$ is finitely generated as an $R$-module, $I$ is finitely generated, i.e, $I = \langle x^\beta \mid \beta \in B\rangle_R$ for some finite subset $B \subseteq \NN_0$. Define $m \coloneqq \min(B)$ and $A^\prime \coloneqq \{ \beta - m \mid \beta \in B \} \setminus \{ 0 \}$. Consequently, $I = x^m \cdot I(A^\prime)$, which implies that $I \cong I(A^\prime)$. The claim then follows from the fact that $I(A^\prime)=I(A)$, where $A$ denotes the set of minimal elements of $A^\prime \cap (\gapset)$ with respect to $\preceq_\SGR$.
|
||||
Since $A$ and $B$ are both minimal with respect to $\preceq_\SGR$, it follows that $A = B$. We now prove that the assignment is surjective. Let $I \subseteq \overline{R}$ be an arbitrary $\NN_0$-graded $R$-submodule. Because $R$ is Noetherian and $\overline{R}$ is finitely generated as an $R$-module, $I$ is finitely generated, i.e, $I = \langle x^\beta \mid \beta \in B\rangle_R$ for some finite subset $B \subseteq \NN_0$. Define $m \coloneqq \min(B)$ and $A^\prime \coloneqq \{ \beta - m \mid \beta \in B \} \setminus \{ 0 \}$. Consequently, $I = x^m \cdot I(A^\prime)$, which implies that $I \cong I(A^\prime)$. Surjectivity then follows from the equality $I(A^\prime)=I(A)$, where $A$ is the set of elements of $A^\prime \cap \gapset$ that are minimal with respect to $\preceq_\SGR$.
|
||||
\end{proof}
|
||||
|
||||
\begin{remark}
|
||||
We now present another perspective on the gap modules $I(A)$. For $A \in \Min(\SGR)$, the $R$-module $I(A)$ admits the $\kk$-vector space description
|
||||
\[ I(A)=\spn_\kk \{x^\alpha \mid \alpha\in \SGR_A \}, \]
|
||||
where $\SGR_A\coloneqq (A \cup \{ 0 \}) +\SGR$. The set $\SGR_A$ is a normalized relative ideal of $\SGR$, that is, a subset of $\NN_0$ containing $0$ that is closed under addition by elements of $\SGR$. One readily checks that $A \mapsto \SGR_A$ defines a one-to-one correspondence between the elements of $\Min(\SGR)$ and the normalized relative ideals of $\SGR$.
|
||||
\todo{reference for this terminology?}
|
||||
\end{remark}
|
||||
|
||||
\begin{corollary}
|
||||
For each $A \in \Min(\SGR)$, the gap module $I(A)$ is maximal Cohen--Macaulay.
|
||||
For each $A \in \Min(\SGR)$, the gap module $I(A)$ is maximal Cohen--Macaulay. In particular, $R^{1/q}$ is maximal Cohen--Macaulay.
|
||||
\end{corollary}
|
||||
|
||||
\begin{proof}
|
||||
By the Cohen--Seidenberg theorems, $\dim(R) = \dim(\overline{R}) = \dim(\kk[x]) = 1$. Fix $A \in \Min(\SGR)$ and a maximal ideal $\mathfrak{m}$ of $R$. Since $R$ is a domain of Krull dimension $1$, every maximal ideal has height $1$; in particular $\dim(R_\mathfrak{m}) = \hgt(\mathfrak{m}) = 1$. The localization
|
||||
$I(A)_\mathfrak{m}$ is a non-zero $R_\mathfrak{m}$-submodule of $\kk(x)$, hence torsion-free, so $1 \leq \depth(I(A)_\mathfrak{m}) \leq \dim(R_\mathfrak{m}) = 1$. Hence, $I(A)$ is maximal Cohen--Macaulay.
|
||||
\end{proof}
|
||||
|
||||
\todo{add direct sums are Cohen--Macaulay (Bruns/Herzog, 1993, Prop. 1.2.9)}
|
||||
By the depth lemma~\cite[Prop.~1.2.9]{Bruns1993}, the depth of a finite direct sum is the minimum of the depths of its summands. Hence, a finite direct sum of maximal Cohen--Macaulay modules is again maximal Cohen--Macaulay. By Corollary~\ref{cor:indec-summ}, the module $R^{1/q}$ decomposes as a finite direct sum of gap modules $I(A)$, and the claim follows.
|
||||
\end{proof}
|
||||
|
||||
\begin{lemma}
|
||||
For $q \in \NN$, it holds that $\qpower{\overline{R}} \shcong \overline{R}^{\oplus q}$.
|
||||
|
|
@ -370,7 +364,7 @@ For $q \in \NN$, it holds that $\qpower{\overline{R}} \shcong \overline{R}^{\opl
|
|||
|
||||
\begin{proof}
|
||||
First, note that
|
||||
\[ \overline{R}^{1/q} = \kk[x^{1/q}] = \bigoplus_{r \in [0,q-1]} x^{r/q} \cdot \kk[x] \shcong \overline{R}^{\oplus q} \]
|
||||
\[ \overline{R}^{1/q} = \kk[x^{1/q}] = \bigoplus_{r \in \intval{0,q-1}} x^{r/q} \cdot \kk[x] \shcong \overline{R}^{\oplus q} \]
|
||||
as $R$-modules. By Lemma~\ref{lemma:cannonc-isom}, $\overline{R}^{1/q}$ and $\qpower{\overline{R}}$ are isomorphic as $\QQ_{\geq 0}$-graded modules over $\overline{R}$, and hence, by restriction of scalars along $R \subseteq \overline{R}$, also as $\QQ_{\geq 0}$-graded modules over $R$.
|
||||
\end{proof}
|
||||
|
||||
|
|
@ -415,24 +409,24 @@ generates $R^{1/q} = \kk[x^{3/q},x^{4/q}]$ as an $R$-module. For $q =2$, this ho
|
|||
\[ R^{1/2} = \langle 1, x^2 \rangle_R \oplus x^{3/2} \cdot \langle 1, x^2 \rangle_R \shcong I(\{2 \})^{\oplus 2} \]
|
||||
For $q = 4$, a direct computation shows $R^{1/4} \shcong \overline{R}^{\oplus 4}$. Using the $\QQ_{\geq 0}$-graded isomorphism between $R^{1/q}$ and $\qpower{R}$, we compute
|
||||
\[ R^{1/8} \grcong \qpower[8]{R} = \qpower[2]{(\qpower[4]{R})} \grcong \qpower[2]{(R^{1/4})}. \]
|
||||
Hence, by Lemma~\ref{lemma:qpower-under-shift}, it holds that $R^{1/8} \shcong (\qpower[2]{\overline{R}})^{\oplus 4}$. The preceding lemma then yields $R^{1/8} \shcong \overline{R}^{\oplus 8}$. By induction, we obtain that $R^{1/2^e} \shcong \overline{R}^{\oplus 2^e}$ for all $e \geq 2$. Consequently, there exist summands appearing in the $\QQ_{\geq 0}$-graded direct sum decomposition of $R^{1/2^e}$ for some $e \in \NN$ that do \emph{not} appear in the decomposition of $R^{1/2^\ell}$ for large $\ell$. This stands in contrast to the case where $\SGR$ is a normal affine semigroup and $R=\kk[\SGR]$ is a normal toric algebra~\cite{Bruns2005}.
|
||||
Hence, by Lemma~\ref{lemma:qpower-under-shift}, it holds that $R^{1/8} \shcong (\qpower[2]{\overline{R}})^{\oplus 4}$. The preceding lemma then yields $R^{1/8} \shcong \overline{R}^{\oplus 8}$. By induction, we obtain that $R^{1/2^e} \shcong \overline{R}^{\oplus 2^e}$ for all \mbox{$e \geq 2$}. Consequently, there exist summands appearing in the $\QQ_{\geq 0}$-graded direct sum decomposition of $R^{1/2^e}$ for some $e \in \NN$ that do \emph{not} appear in the decomposition of $R^{1/2^\ell}$ for large $\ell$. This stands in contrast to the case where $\SGR$ is a normal affine monoid and $R=\kk[\SGR]$ is a normal toric algebra~\cite{Bruns2005}.
|
||||
\end{example}
|
||||
|
||||
% NOTE: next proposition holds when q \in \SGR
|
||||
|
||||
\begin{proposition}\label{prop:asy-behav}
|
||||
It holds that $q \in \SGR$ if and only if every indecomposable summand of $R^{1/q}$ is isomorphic to the normalization $\overline{R} = \kk[x]$ of $R$, that is, $R^{1/q} \cong \overline{R}^{\oplus q}$. In particular, this holds for all sufficiently large $q$.
|
||||
It holds that $q \in \SGR$ if and only if every $\QQ^d$-graded indecomposable summand of $R^{1/q}$ is, up to a degree shift, isomorphic to the normalization $\overline{R} = \kk[x]$ of $R$, that is, if $R^{1/q} \shcong \overline{R}^{\oplus q}$. In particular, this holds for all sufficiently large $q$.
|
||||
%For $q \in \SGR$, every indecomposable summands of $R^{1/q}$ is isomorphic to the normalization $\overline{R} = \kk[x]$ of $R$. That is, $R^{1/q} \cong \overline{R}^{\oplus q}$. In particular, this holds for all sufficiently large $q$.
|
||||
\end{proposition}
|
||||
|
||||
\begin{proof}
|
||||
Assume that $q \in SGR$, and fix $r \in \intval{0,q-1}$. For simplicity, set
|
||||
Assume that $q \in \SGR$, and fix $r \in \intval{0,q-1}$. For simplicity, set
|
||||
\[ \supp(I_{q,r}) \coloneqq \{ a/q \mid a \in \SGR, \, a \equiv r \pmod{q} \} \]
|
||||
and observe that $\supp(I_{q,r}) \subseteq \{ m_{q,r}/q + n \mid n \in \NN_0\}$, where $m_{q,r}/q$ is the minimum of $\supp(I_{q,r})$. Conversely, for each $n \in \NN_0$, the choice of $q$ ensures $nq \in \SGR$, and hence $m_{q,r} + nq \in \SGR$. Since $m_{q,r}+nq \equiv r \pmod q$, we
|
||||
obtain $m_{q,r}/q + n \in \supp(I_{q,r})$, and altogether
|
||||
\[ \supp(I_{q,r}) = \{ m_{q,r}/q + n \mid n \in \NN_0\}. \]
|
||||
and observe that $\supp(I_{q,r}) \subseteq \{ \frac{m_{q,r}}{q} + \alpha \mid \alpha \in \NN_0\}$, where $\frac{m_{q,r}}{q}$ is the minimum of $\supp(I_{q,r})$. Conversely, for each $\alpha \in \NN_0$, the choice of $q$ ensures $\alpha q \in \SGR$, and hence $m_{q,r} + \alpha q \in \SGR$. Since $m_{q,r}+ \alpha q \equiv r \pmod q$, we
|
||||
obtain $\frac{m_{q,r}}{q} + \alpha \in \supp(I_{q,r})$, and altogether
|
||||
\[ \supp(I_{q,r}) = \{ \frac{m_{q,r}}{q} + \alpha \mid \alpha \in \NN_0\}. \]
|
||||
It now follows from~\eqref{eq:ind-summand-as-vector-space} that
|
||||
\[ I_{q,r} = \spn_\kk \{ x^{m_{q,r}/q + n} \mid n \in \NN_0 \} = x^{m_{q,r}/q} \cdot \spn_\kk\{ x^n \mid n \in \NN_0 \} \cong k[x] . \]
|
||||
\[ I_{q,r} = \spn_\kk \{ x^{m_{q,r}/q + \alpha} \mid \alpha \in \NN_0 \} = x^{\frac{m_{q,r}}{q}} \cdot \spn_\kk\{ x^\alpha \mid \alpha \in \NN_0 \} \cong k[x] . \]
|
||||
|
||||
|
||||
\end{proof}
|
||||
|
|
|
|||
|
|
@ -207,4 +207,17 @@ author = {Rosales, J.C. and García-Sánchez, P. A.},
|
|||
urldate = {2026-09-16},
|
||||
date = {2009},
|
||||
langid = {english},
|
||||
}
|
||||
|
||||
@book{Bruns1993,
|
||||
location = {Cambridge [England] ; New York},
|
||||
title = {Cohen-Macaulay rings},
|
||||
isbn = {978-0-521-41068-7},
|
||||
series = {Cambridge studies in advanced mathematics},
|
||||
pagetotal = {403},
|
||||
number = {39},
|
||||
publisher = {Cambridge University Press},
|
||||
author = {Bruns, Winfried and Herzog, Jürgen},
|
||||
date = {1993},
|
||||
keywords = {Cohen-Macaulay rings},
|
||||
}
|
||||
Loading…
Reference in a new issue