first part of the preliminaries

ncr-of-monomial-curves.tex

@@ -7,6 +7,7 @@
 \usepackage{todonotes}
 \usepackage{blindtext}
 \usepackage{tikz}
+\usepackage[most]{tcolorbox}
 \usepackage{graphicx}
 \usetikzlibrary{arrows.meta,positioning,decorations.pathreplacing}
 %\usepackage{parskip}
@@ -26,6 +27,8 @@
   \clearlist{language}%
 }
 
+\numberwithin{equation}{section}
+
 \newtheorem{theorem}{Theorem}[section]
 \newtheorem{lemma}[theorem]{Lemma}
 \newtheorem{proposition}[theorem]{Proposition}
@@ -54,6 +57,7 @@
 }
 
 \DeclareMathOperator{\supp}{supp}
+\DeclareMathOperator{\Spec}{Spec}
 \DeclareMathOperator{\Frac}{Frac}
 \DeclareMathOperator{\spn}{span}
 \DeclareMathOperator{\Min}{Min}
@@ -88,9 +92,9 @@
 
 \section{Introduction}
 
-some results/terminology in section 2 are made independent of the characteristic
+% some results/terminology in section 2 are made independent of the characteristic
 
-\section{Monomial curves}
+\section{Preliminaries}
 
 We write $\NN = \{1, 2, 3, \dots \}$ for the set of natural numbers. For $a$, $b \in \ZZ$, we denote by $\intval{a,b}$ the set of all integers from $a$ to $b$, including both endpoints.
 
@@ -98,47 +102,64 @@ Throughout, we fix an arbitrary ground field $\kk$. We begin by recalling some b
 
 Let $\SGR$ be an affine semigroup, that is, a finitely generated submonoid of $\ZZ^d$ for some $d \in \NN$. The \textbf{toric algebra} associated with $\SGR$ is the semigroup algebra
 \[ R \coloneqq \kk[\SGR] = \spn_\kk \{ x^a \mid a \in \SGR \}. \]
-The algebra $R$ is a finitely generated $\kk$-subalgebra of the Laurent polynomial ring~$\kk[x_1^{\pm 1}, \dots, x_d^{\pm 1}]$. In particular, it is a Noetherian domain and inherits the natural $\ZZ^d$-grading. Its Krull dimension is equal to the rank of $\gp(\SGR)$, where $\gp(\SGR)$ denotes the subgroup of $\ZZ^d$ generated by $\SGR$~\cite[Prop. 1.1.14]{Cox2011}.
+The algebra $R$ is a finitely generated $\kk$-subalgebra of the Laurent polynomial ring~$\kk[x_1^{\pm 1}, \dots, x_d^{\pm 1}]$. In particular, it is a Noetherian domain and inherits the natural $\ZZ^d$-grading. Its Krull dimension is equal to the rank of $\gp(\SGR)$, where $\gp(\SGR)$ denotes the subgroup of $\ZZ^d$ generated by $\SGR$~\cite[Prop. 1.1.14]{Cox2011}. Consequently, $R$ is one-dimensional if and only if $\gp(\SGR) \cong \ZZ$, in which case we may assume without loss of generality that $\gp(\SGR) = \ZZ$. Every additive submonoid of $\ZZ$ is either a group, a submonoid of $\NN_0$, or a submonoid of $-\NN_0$~\cite[§ 2]{Geroldinger2025}. Hence, when $R$ is one-dimensional, $\SGR$ is, up to isomorphism of monoids, either $\ZZ$ or a submonoid of $\NN_0$. Accordingly, either $R\cong\kk[x^{\pm 1}]$ or $R \cong \kk[x^{a_1},\dots,x^{a_n}]$ for some $a_1$, \dots, $a_n \in \NN_0$. 
 
+When $\kk$ is algebraically closed, the algebra $R=\kk[\SGR]$ is the coordinate ring of the associated affine toric variety $X = \Spec(R)$. In dimension one, the corresponding affine toric variety is either the one-dimensional torus $\Spec(\kk[x^{\pm1}]) = \A^1_\kk \setminus \{0\}$, or an \textbf{affine monomial curve} $\Spec(\kk[x^{a_1},\dots,x^{a_n}])$, parametrized by
+\begin{equation}\label{eq:para-mon-curve}
+ \varphi \colon \A^1_\kk \to \A^n_\kk, \qquad t \longmapsto (t^{a_1},\dots,t^{a_n}).
+\end{equation}
+
+\subsection{Numerical monoids} 
+
+Let $\SGR$ be a submonoid of $\NN_0$. Writing $\SGR = \gcd(\SGR) \cdot \SGR^\prime$, where $\SGR^\prime$ is a submonoid of $\NN_0$ with $\gcd(\SGR^\prime) = 1$, yields an isomorphism of monoids between $\SGR$ and $\SGR^\prime$. Thus, it suffices to consider submonoids of $\NN_0$ with greatest common divisor $1$, called \textbf{numerical monoids}. Crucially, the gap set $\gapset$ of a numerical monoid $\SGR$ is finite. For further background on numerical monoids, see~\cite{Rosales2009}. 
+
+For a numerical monoid $\SGR=\langle a_1,\dots, a_n\rangle \subseteq \NN_0$, the parametrization $\varphi$ from~\eqref{eq:para-mon-curve} of the associated affine monomial curve $X = \Spec(R)$ restricts to an isomorphism of varieties $\A_\kk^1 \setminus \{ 0 \} \xrightarrow{\sim} X \setminus \{ 0 \}$. Hence $X$ is nonsingular away from the origin. At the origin, it is singular unless $\SGR = \NN_0$, in which case $R=\kk[x]$ and $X=\A_\kk^1$.
+
+\subsection{Normality, seminormality, and weak normality}
+
+A domain $\Lambda$ is called \textbf{normal} if it is integrally closed in its field of fractions $\Frac(\Lambda)$, and \textbf{seminormal} if, for every $\lambda \in \Frac(\Lambda)$, the conditions $\lambda^2$, $\lambda^3 \in \Lambda$ imply $\lambda \in \Lambda$. The normalization $\overline{\Lambda}$ and the seminormalization ${}^+\Lambda$ of $\Lambda$ are the intersections of all intermediate normal and seminormal rings $\Gamma$, respectively, satisfying $\Lambda \subseteq \Gamma \subseteq \Frac(\Lambda)$. If $\Lambda$ has prime characteristic $p$, then $\Lambda$ is called \textbf{weakly normal} if, for every $\lambda \in \Frac(\Lambda)$, the condition $\lambda^p \in \Lambda$ implies $\lambda\in\Lambda$. Clearly, normality implies seminormality. In prime characteristic $p$, normality implies weak normality, and weak normality in turn implies seminormality. Indeed, suppose that $\Lambda$ is weakly normal and let $\lambda \in \Frac(\Lambda)$ satisfy $\lambda^2$, $\lambda^3 \in \Lambda$. Write $p = 2n^\prime + 3n^{\prime\prime}$ for some $n^\prime$, $n^{\prime\prime} \in \NN_0$. Then
+\[
+\lambda^p = (\lambda^2)^{n^\prime}(\lambda^3)^{n^{\prime\prime}} \in \Lambda,
+\]
+and hence $\lambda \in \Lambda$.
+
+For the toric algebra $R = \kk[\SGR]$, the field of fractions in the definitions above may be replaced by $\kk[\gp(\SGR)]$. Indeed, $\gp(\SGR) \cong \ZZ^{d^\prime}$ for some $d^\prime \in \NN_0$, so $\kk[\gp(\SGR)]$ is isomorphic to a Laurent polynomial ring and is therefore normal. Consequently, every element of $\Frac(R)$ that is integral over $R$ already belongs to $\kk[\gp(\SGR)]$.\todo{mention normality, ... for affine monoids?}
+
+In the case of a numerical semigroup $\SGR$, the corresponding toric algebra $R$ satisfies ${}^+R=\kk[x]$. Indeed, let $n_1 > \cdots > n_k$ be the elements of $\gapset$. Since $n_1$ is the largest gap of $\SGR$, we have $(x^{n_1})^2,(x^{n_1})^3\in R\subseteq{}^+R$, and hence $x^{n_1}\in{}^+R$ by seminormality.
+%Since $n_1$ is the largest gap of $\SGR$, both $2n_1$ and $3n_1$ belong to $\SGR$. Hence $(x^{n_1})^2$, $(x^{n_1})^3 \in R \subseteq{}^+R$, and seminormality gives $x^{n_1}\in{}^+R$. 
+Repeating this argument successively for the remaining gaps shows that $x^{n_1}$, \dots, $x^{n_k} \in{}^+R$, and therefore $\kk[x] \subseteq{}^+R$. Since $\kk[x]$ is normal, hence seminormal, the reverse inclusion follows. Since $\kk[x]$ is normal, we also have $\overline{R} = \kk[x]$, and consequently $\Frac(R) = \kk(x)$. Thus, unless $\SGR=\NN_0$, the ring $R$ is neither normal nor seminormal.
+
+\subsection{Graded Krull–Schmidt theorem}  
+
+The category of finitely generated $\QQ^d$-graded $R$-modules satisfies the \textbf{Krull--Schmidt property}. That is, every such module decomposes as a finite direct sum of $\QQ^d$-graded $R$-modules that are indecomposable in the graded category, and this decomposition is unique up to graded isomorphism and permutation of the summands~\cite[§ 2]{Shibuta2017}.
+
+% we are interested in the unique summands of such graded decomposition only up to shift of degrees so that we introduce the following notations (define \grcong)
 %For $\QQ^d$-graded $R$-modules $M$, $M_1$, \dots, $M_k$, we write $M \cong_{\textrm{sh}} M_1 \oplus \cdots \oplus M_k$ whenever there exist $q_i \in \QQ$ such that $M \cong M_1(d_1) \oplus \cdots \oplus M_k(d_k)$ as $\QQ^d$-graded $R$-modules.
-
+%Clearly if $M \shcong M_1 \oplus M_2$ and $M_2 \shcong M_2^\prime \oplus M_2^{\prime\prime}$, then $M \shcong M_1 \oplus M_2^\prime \oplus M_2^{\prime\prime}$.
-\todo{use widebar}
 
 \todo{adapt phrasinig in correspondence gap modules}
 
-\todo{use notation $\grcong$?}
+%\begin{lemma}\label{lemma:qpower-under-shift}
+%Let $M$, $M^\prime$ and $M^{\prime\prime}$ be $\QQ^d$-graded $R$-modules, and suppose that $M \shcong M^{\prime} \oplus M^{\prime\prime}$, then $\qpower{M} \shcong \qpower{M^{\prime}} \oplus \qpower{M^{\prime\prime}}$. 
+%\end{lemma}
 
-\todo{affine semigroup rings are (non-normal) toric varieties}
+%\begin{proof}
+%By assumption, there are $d^\prime$, $d^{\prime\prime} \in \QQ$ such that $M \grcong M^\prime(d^\prime) \oplus M^{\prime\prime}(d^{\prime\prime})$.
 
-\todo{1-dimensional toric varieties are monomial curves}
+%As the construction of the pseudo-Frobenius twist respects direct sums, we have 
+%\[ \qpower{M} \grcong \qpower{(M^{\prime}(d^\prime))} \oplus \qpower{(M^{\prime\prime}(d^{\prime\prime}))} = (\qpower{M^{\prime}})(d^\prime/q) \oplus (\qpower{M^{\prime\prime}})(d^{\prime\prime}/q), \]
+%where the last equality follows from the computation
+%\begin{align*}
+%\left(\qpower{(M^{\prime}(d^\prime))} \oplus \qpower{\left(M^{\prime\prime}(d^{\prime\prime})\right)}\right)_\alpha &= \bigl(M^{\prime}(d^\prime)\bigr)_{q\alpha} \oplus \bigl(M^{\prime\prime}(d^{\prime\prime})\bigr)_{q\alpha} \\
+%&= (M^{\prime})_{q\alpha + d^\prime} \oplus (M^{\prime\prime})_{q\alpha + d^{\prime\prime}} \\
+%&= (\qpower{M^\prime})_{\alpha + \frac{d^\prime}{q}} \oplus (\qpower{M^{\prime\prime}})_{\alpha + \frac{d^{\prime\prime}}{q}} \\
+%&= \left((\qpower{M^{\prime}})(d^\prime/q) \oplus (\qpower{M^{\prime\prime}})(d^{\prime\prime}/q)\right)_\alpha.
+%\end{align*}
+%\end{proof}
 
-\todo{define normal, seminormal, $G(S)$, weakly normal (i.e., for any $r \in \Frac(R)$, $r^p \in R$ implies $r \in R$), note that one can replace $\Frac(R)$ by $\kk[G(S)]$, and weakly normal implies seminormal}
+%\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.
-\todo{monomial curves are non-seminormal, non-normal, singular at 0}
+%\end{proposition}
-
-\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
@@ -230,7 +251,7 @@ $R$ is normal if and only if $R$ is $q$-weakly normal for all $q \in \NN$ if and
 \end{corollary}
 
 \begin{proof}
-By~\cite[Prop. 1]{Hochster1972}, the semigroup algebra $R = \kk[\SGR]$ is normal if and only if $\SGR$ is saturated, that is, if for every $a \in \gp(\SGR)$ and $q \in \NN$, the condition $qa \in \SGR$ implies $a \in \SGR$. This is precisely the semigroup-theoretic formulation of $R$ being $q$-weakly normal for all $q \in \NN$. By the previous proposition, this is equivalent to $R$ being $q$-weakly normal for every $q \in \NN$.
+By~\cite[Prop. 1]{Hochster1972}, the semigroup algebra $R = \kk[\SGR]$ is normal if and only if $\SGR$ is normal, that is, if for every $a \in \gp(\SGR)$ and $q \in \NN$, the condition $qa \in \SGR$ implies $a \in \SGR$. This is precisely the semigroup-theoretic formulation of $R$ being $q$-weakly normal for all $q \in \NN$. By the previous proposition, this is equivalent to $R$ being $q$-weakly normal for every $q \in \NN$.
 \end{proof}
 
 \begin{lemma}\textnormal{\cite[Lemma 5.0.3]{Blickle2013}}\label{lemma:q-split}
@@ -280,7 +301,7 @@ where each summand is indecomposable (see Lemma~\ref{lemma:rank-1}).
 %\end{proof}
 
 \begin{definition}\label{def:gap-module}
-Let $A\subseteq \NN$. The (normalized) \textbf{gap module defined by $A$} is the $\NN_0$-graded $R$-submodule of $\bar{R}=\kk[x]$ given by
+Let $A\subseteq \NN$. The (normalized) \textbf{gap module defined by $A$} is the $\NN_0$-graded $R$-submodule of $\overline{R}=\kk[x]$ given by
 \[ I(A) \coloneqq \langle x^\alpha \mid \alpha \in A\cup \{0\} \rangle_R. \]
 \end{definition}
 
@@ -314,15 +335,15 @@ In particular, as $q$ ranges over $\NN$, only finitely many isomorphism classes
 \begin{lemma}\label{lemma:rank-1}
 For each $A \in \Min(\SGR)$, the gap module $I(A)$ has rank $1$ and is therefore indecomposable. Moreover, the assignment 
 \[ A \mapsto [I(A)] \]
-defines a bijection between $\Min(\SGR)$ and the set of isomorphism classes of $\NN_0$-graded $R$-submodules of $\bar{R} = \kk[x]$.
+defines a bijection between $\Min(\SGR)$ and the set of isomorphism classes of $\NN_0$-graded $R$-submodules of $\overline{R} = \kk[x]$.
 \end{lemma}
 
 \begin{proof}
 For simplicity, set $Q \coloneqq \Frac(R) = \kk(x)$  and fix $A \in \Min(\SGR)$. Observe that $Q \otimes_R I(A)$ is naturally isomorphic, as a $Q$-vector space, to $I(A)_{(0)} \subseteq Q$. Since $I(A)$ contains $1$, it follows that $I(A)_{(0)} = Q$ and $I(A)$ has rank $1$.
 
-For the second part, each $I(A)$ with $A \in \Min(\SGR)$ is an $\NN_0$-graded $R$-submodule of $\bar{R}$, so the map is well-defined. To prove injectivity, suppose that $\varphi \colon I(A)\to I(B)$ is an $R$-module isomorphism for some $A$, $B \in \Min(\SGR)$. Localizing at $(0)$ yields an isomorphism of $Q$-vector spaces $\varphi_{(0)}\colon I(A)_{(0)} \to I(B)_{(0)}$, where both source and target are equal to $Q$. Hence $\varphi$, and thus also $\varphi_{(0)}$, is given by multiplication by some $g_0 \in Q$, so $\varphi(f) = g_0f$ for all $f \in I(A)$. A simple degree argument shows that $g_0 \in \kk \setminus \{ 0 \}$. Therefore,
+For the second part, each $I(A)$ with $A \in \Min(\SGR)$ is an $\NN_0$-graded $R$-submodule of $\overline{R}$, so the map is well-defined. To prove injectivity, suppose that $\varphi \colon I(A)\to I(B)$ is an $R$-module isomorphism for some $A$, $B \in \Min(\SGR)$. Localizing at $(0)$ yields an isomorphism of $Q$-vector spaces $\varphi_{(0)}\colon I(A)_{(0)} \to I(B)_{(0)}$, where both source and target are equal to $Q$. Hence $\varphi$, and thus also $\varphi_{(0)}$, is given by multiplication by some $g_0 \in Q$, so $\varphi(f) = g_0f$ for all $f \in I(A)$. A simple degree argument shows that $g_0 \in \kk \setminus \{ 0 \}$. Therefore,
 \[ I(B) = \varphi(I(A)) = g_0 \cdot I(A) =I(A). \]
-Since $A$ and $B$ are both minimal with respect to $\preceq_\SGR$, it follows that $A = B$. Now, let $I \subseteq \bar{R}$ be an arbitrary $\NN_0$-graded $R$-submodule. Because $R$ is Noetherian and $\bar{R}$ is finitely generated as an $R$-module, $I$ is finitely generated, i.e, $I = \langle x^\beta \mid \beta \in B\rangle_R$ for some finite subset $B \subseteq \NN_0$. Define $m \coloneqq \min(B)$ and $A^\prime \coloneqq \{ \beta - m \mid \beta \in B \} \setminus \{ 0 \}$. Consequently, $I = x^m \cdot I(A^\prime)$, which implies that $I \cong I(A^\prime)$. The claim then follows from the fact that $I(A^\prime)=I(A)$, where $A$ denotes the set of minimal elements of $A^\prime \cap (\gapset)$ with respect to $\preceq_\SGR$.
+Since $A$ and $B$ are both minimal with respect to $\preceq_\SGR$, it follows that $A = B$. Now, let $I \subseteq \overline{R}$ be an arbitrary $\NN_0$-graded $R$-submodule. Because $R$ is Noetherian and $\overline{R}$ is finitely generated as an $R$-module, $I$ is finitely generated, i.e, $I = \langle x^\beta \mid \beta \in B\rangle_R$ for some finite subset $B \subseteq \NN_0$. Define $m \coloneqq \min(B)$ and $A^\prime \coloneqq \{ \beta - m \mid \beta \in B \} \setminus \{ 0 \}$. Consequently, $I = x^m \cdot I(A^\prime)$, which implies that $I \cong I(A^\prime)$. The claim then follows from the fact that $I(A^\prime)=I(A)$, where $A$ denotes the set of minimal elements of $A^\prime \cap (\gapset)$ with respect to $\preceq_\SGR$.
 \end{proof}
 
 \begin{remark}
@@ -337,20 +358,20 @@ For each $A \in \Min(\SGR)$, the gap module $I(A)$ is maximal Cohen--Macaulay.
 \end{corollary}
 
 \begin{proof}
-By the Cohen--Seidenberg theorems, $\dim(R) = \dim(\bar{R}) = \dim(\kk[x]) = 1$. Fix $A \in \Min(\SGR)$ and a maximal ideal $\mathfrak{m}$ of $R$. Since $R$ is a domain of Krull dimension $1$, every maximal ideal has height $1$; in particular $\dim(R_\mathfrak{m}) = \hgt(\mathfrak{m}) = 1$. The localization
+By the Cohen--Seidenberg theorems, $\dim(R) = \dim(\overline{R}) = \dim(\kk[x]) = 1$. Fix $A \in \Min(\SGR)$ and a maximal ideal $\mathfrak{m}$ of $R$. Since $R$ is a domain of Krull dimension $1$, every maximal ideal has height $1$; in particular $\dim(R_\mathfrak{m}) = \hgt(\mathfrak{m}) = 1$. The localization
 $I(A)_\mathfrak{m}$ is a non-zero $R_\mathfrak{m}$-submodule of $\kk(x)$, hence torsion-free, so $1 \leq \depth(I(A)_\mathfrak{m}) \leq  \dim(R_\mathfrak{m}) = 1$. Hence, $I(A)$ is maximal Cohen--Macaulay.
 \end{proof}
 
 \todo{add direct sums are Cohen--Macaulay (Bruns/Herzog, 1993, Prop. 1.2.9)}
 
 \begin{lemma}
-For $q \in \NN$, it holds that $\qpower{\bar{R}} \shcong \bar{R}^{\oplus q}$.
+For $q \in \NN$, it holds that $\qpower{\overline{R}} \shcong \overline{R}^{\oplus q}$.
 \end{lemma}
 
 \begin{proof}
 First, note that 
-\[ \bar{R}^{1/q} = \kk[x^{1/q}] = \bigoplus_{r \in [0,q-1]} x^{r/q} \cdot \kk[x] \cong \bar{R}^{\oplus q} \]
+\[ \overline{R}^{1/q} = \kk[x^{1/q}] = \bigoplus_{r \in [0,q-1]} x^{r/q} \cdot \kk[x] \shcong \overline{R}^{\oplus q} \]
-as $R$-modules. By Lemma~\ref{lemma:cannonc-isom}, $\bar{R}^{1/q}$ 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.
+as $R$-modules. By Lemma~\ref{lemma:cannonc-isom}, $\overline{R}^{1/q}$ and $\qpower{\overline{R}}$ are isomorphic as $\QQ_{\geq 0}$-graded modules over $\overline{R}$, and hence, by restriction of scalars along $R \subseteq \overline{R}$, also as $\QQ_{\geq 0}$-graded modules over $R$.
 \end{proof}
 
 \begin{example}\label{ex:large-powers}
@@ -389,19 +410,19 @@ Let $\SGR = \langle 3,4 \rangle \subseteq \NN_0$ and $R = \kk[\SGR] = \kk[x^3,x^
 \end{center}
 
 The set
-\[ \{ x^{(3/q)e_1 + (4/q)e_2} : e_i \in \intval{0,q-1} \} \]
+\[ \bigl\{ x^{(3/q)e_1 + (4/q)e_2} : e_i \in \intval{0,q-1} \bigr\} \]
-generates $R^{1/q} = \kk[x^{3/q},x^{4/q}]$ as an $R$-module. For $q =2$, this 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
+generates $R^{1/q} = \kk[x^{3/q},x^{4/q}]$ as an $R$-module. For $q =2$, this homogeneous generating set is $\{ 1, x^{3/2}, x^2, x^{7/2} \}$, and grouping the numerators of the exponents by residue modulo $2$ yields the $\QQ_{\geq 0}$-graded $R$-module decomposition
-\[ R^{1/2} = \langle 1, x^2 \rangle_R \oplus x^{3/2} \cdot \langle 1, x^2 \rangle_R \cong I(\{2 \})^{\oplus 2}  \] 
+\[ R^{1/2} = \langle 1, x^2 \rangle_R \oplus x^{3/2} \cdot \langle 1, x^2 \rangle_R \shcong I(\{2 \})^{\oplus 2}  \] 
-For $q = 4$, a direct computation shows $R^{1/4} \cong \bar{R}^{\oplus 4}$. 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
+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
-\[ R^{1/8} \cong \qpower[2]{(\qpower[4]{R})} \cong \qpower[2]{(\bar{R}^{\oplus 4})} = (\qpower[2]{\bar{R}})^{\oplus 4} \cong \bar{R}^{\oplus 8}, \]
+\[ R^{1/8} \grcong \qpower[8]{R} = \qpower[2]{(\qpower[4]{R})} \grcong \qpower[2]{(R^{1/4})}. \]
-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}.
+Hence, by Lemma~\ref{lemma:qpower-under-shift}, it holds that $R^{1/8} \shcong (\qpower[2]{\overline{R}})^{\oplus 4}$. The preceding lemma then yields $R^{1/8} \shcong \overline{R}^{\oplus 8}$. By induction, we obtain that $R^{1/2^e} \shcong \overline{R}^{\oplus 2^e}$ for all $e \geq 2$. Consequently, there exist summands appearing in the $\QQ_{\geq 0}$-graded direct sum decomposition of $R^{1/2^e}$ for some $e \in \NN$ that do \emph{not} appear in the decomposition of $R^{1/2^\ell}$ for large $\ell$. This stands in contrast to the case where $\SGR$ is a normal affine semigroup and $R=\kk[\SGR]$ is a normal toric algebra~\cite{Bruns2005}.
 \end{example}
 
 % NOTE: next proposition holds when q \in \SGR
 
 \begin{proposition}\label{prop:asy-behav}
-It holds that $q \in \SGR$ if and only if every indecomposable summand of $R^{1/q}$ is isomorphic to the normalization $\bar{R} = \kk[x]$ of $R$, that is, $R^{1/q} \cong \bar{R}^{\oplus q}$. In particular, this holds for all sufficiently large $q$.
+It holds that $q \in \SGR$ if and only if every indecomposable summand of $R^{1/q}$ is isomorphic to the normalization $\overline{R} = \kk[x]$ of $R$, that is, $R^{1/q} \cong \overline{R}^{\oplus q}$. In particular, this holds for all sufficiently large $q$.
-%For $q \in \SGR$, every indecomposable summands of $R^{1/q}$ is isomorphic to the normalization $\bar{R} = \kk[x]$ 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 $\overline{R} = \kk[x]$ of $R$. That is, $R^{1/q} \cong \overline{R}^{\oplus q}$. In particular, this holds for all sufficiently large $q$.
 \end{proposition}
 
 \begin{proof}
@@ -420,8 +441,8 @@ It now follows from~\eqref{eq:ind-summand-as-vector-space} that
 
 \begin{example}
 Revisiting Example~\ref{ex:large-powers}, it remains to determine the decomposition of $R^{1/5}$ in order to obtain a complete description of the decompositions of $R^{1/q}$ for $q\in\NN$. A direct computation yields
-\[ R^{1/5}\cong I(\{2\}) \oplus \bar{R}^{\oplus 4}. \]
+\[ R^{1/5}\cong I(\{2\}) \oplus \overline{R}^{\oplus 4}. \]
-Since $\bar{R} = \kk[x]$ corresponds to $I(\{1,2\})$, the indecomposable summands that occur in $R^{1/q}$ for some $q \in \NN$ are precisely $I(\{2\})$ and $I(\{1,2\})$. On the other hand,
+Since $\overline{R} = \kk[x]$ corresponds to $I(\{1,2\})$, the indecomposable summands that occur in $R^{1/q}$ for some $q \in \NN$ are precisely $I(\{2\})$ and $I(\{1,2\})$. On the other hand,
 \[ \Min(\SGR) = \{ \emptyset,\{1\}, \{2\}, \{5\}, \{1,2\} \}, \]
 so not every potential candidate $I(A)$ with $A \in \Min(\SGR)$ occurs as a direct summand.
 \end{example}
@@ -433,7 +454,7 @@ For each $q \geq 2$, $R$ is not $q$-split, meaning that $R$ does not occur as a
 \todo{\emph{note:} this can also be easily proven directly!}
 
 \begin{proof}
-Fix $q \geq 2$, and let $p$ be a prime divisor of $q$. Suppose, for contradiction, that $R$ is $q$-split. By Lemma~\ref{lemma:q-split}, $R$ is then $p^e$-split for every $e \in \NN$. This contradicts Proposition~\ref{prop:asy-behav}, which shows that, for all sufficiently large $e$, $R^{1/p^e}$ decomposes as a direct sum of copies of $\bar{R}$, and therefore cannot contain $R$ as a direct summand.
+Fix $q \geq 2$, and let $p$ be a prime divisor of $q$. Suppose, for contradiction, that $R$ is $q$-split. By Lemma~\ref{lemma:q-split}, $R$ is then $p^e$-split for every $e \in \NN$. This contradicts Proposition~\ref{prop:asy-behav}, which shows that, for all sufficiently large $e$, $R^{1/p^e}$ decomposes as a direct sum of copies of $\overline{R}$, and therefore cannot contain $R$ as a direct summand.
 \end{proof}
 
 \begin{proposition} 
@@ -457,8 +478,8 @@ For $q \in \SGR$, the global dimension of $\End_R(R^{1/q})$ is equal to the Krul
 \end{theorem}
 
 \begin{proof}
-For $q \in \SGR$, Proposition~\ref{prop:asy-behav} gives $R^{1/q}\cong \bar R^{\oplus q}$. By~\cite[Cor. 2.3]{Leuschke2012}, multiplicities of direct summands do not affect the endomorphism ring up to Morita equivalence; hence $\End_R(R^{1/q})$ is Morita equivalent to $\End_R(\bar{R})$. As in the proof of Lemma~\ref{lemma:rank-1}, since $\bar{R}$ has rank $1$, any $R$-module endomorphism of $\bar{R}$ is induced by multiplication with an element of $\bar{R}$, and thus $\End_R(\bar{R}) \cong \bar{R}$. Consequently, we obtain
+For $q \in \SGR$, Proposition~\ref{prop:asy-behav} gives $R^{1/q}\cong \overline R^{\oplus q}$. By~\cite[Cor. 2.3]{Leuschke2012}, multiplicities of direct summands do not affect the endomorphism ring up to Morita equivalence; hence $\End_R(R^{1/q})$ is Morita equivalent to $\End_R(\overline{R})$. As in the proof of Lemma~\ref{lemma:rank-1}, since $\overline{R}$ has rank $1$, any $R$-module endomorphism of $\overline{R}$ is induced by multiplication with an element of $\overline{R}$, and thus $\End_R(\overline{R}) \cong \overline{R}$. Consequently, we obtain
-\[ \gldim(\End_R(R^{1/q})) = \gldim(\bar{R}) = \gldim(\kk[x]) = 1 = \dim(R), \]
+\[ \gldim(\End_R(R^{1/q})) = \gldim(\overline{R}) = \gldim(\kk[x]) = 1 = \dim(R), \]
 where the first equality follows from the Morita invariance of global dimension.
 \end{proof}
 

references.bib

@@ -176,3 +176,35 @@ year      = {2011},
 isbn      = {978-0-8218-4819-2},
 doi       = {10.1090/gsm/124}
 }
+
+
+@article{Geroldinger2025,
+title = {On algebraic properties of power monoids of numerical monoids},
+volume = {265},
+issn = {0021-2172, 1565-8511},
+url = {https://link.springer.com/10.1007/s11856-024-2683-0},
+doi = {10.1007/s11856-024-2683-0},
+pages = {867--900},
+number = {2},
+journaltitle = {Israel Journal of Mathematics},
+shortjournal = {Isr. J. Math.},
+author = {Bienvenu, Pierre-Yves and Geroldinger, Alfred},
+urldate = {2026-09-16},
+date = {2025-02},
+}
+
+@book{Rosales2009,
+location = {New York, {NY}},
+title = {Numerical Semigroups},
+volume = {20},
+rights = {https://www.springernature.com/gp/researchers/text-and-data-mining},
+isbn = {978-1-4419-0159-0 978-1-4419-0160-6},
+url = {https://link.springer.com/10.1007/978-1-4419-0160-6},
+doi = {10.1007/978-1-4419-0160-6},
+series = {Developments in Mathematics},
+publisher = {Springer New York},
+author = {Rosales, J.C. and García-Sánchez, P. A.},
+urldate = {2026-09-16},
+date = {2009},
+langid = {english},
+}