repair section sign

ncr-of-monomial-curves.tex

@@ -102,7 +102,7 @@ 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}. 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, $a_n \in \NN_0$. 
+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[\S~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}
@@ -131,7 +131,7 @@ Repeating this argument successively for the remaining gaps shows that $x^{n_1}$
 
 \subsection{Graded Krull--Schmidt theorem}  
 
-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}.
+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[\S~2]{Shibuta2017}.
 
 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
 \[ M \shcong M_1 \oplus \cdots \oplus M_k \]
@@ -157,7 +157,7 @@ Assume that $\kk$ is a field of prime characteristic $p$ and that \mbox{$q = p^e
 \begin{proof}
 First, note that for $r = \sum_{a \in \SGR} k_a x^a \in R$, we have $F^e(r) = \sum_{a \in \SGR} k_a^{q} x^{qa}$. Hence, $\qpower{F}$ and $F^e$ agree precisely when $F^e$ fixes $\kk\subseteq R$ pointwise.
 
-Suppose first that $\kk$ is finite of order $p^f$. By Lagrange's theorem, $k^{p^f} = k$ for every $k \in \kk$. If $f \mid e$, then repeated application of this identity gives $k^{p^e} = k$ for all $k \in \kk$. Now assume that $f \nmid e$. Since the multiplicative group of a finite field is cyclic~\cite[§ 9, Prop. 18]{Dummit2004}, there exists an element $k \in \kk \setminus \{ 0 \}$ of order $p^f -1$. We claim that $k^{p^e} \neq k$. Suppose, for contradiction, that $k^{p^e} = k$. Then $k^{p^e -1 } = 1$, and hence $p^f -1 \mid p^e -1$. It follows that 
+Suppose first that $\kk$ is finite of order $p^f$. By Lagrange's theorem, $k^{p^f} = k$ for every $k \in \kk$. If $f \mid e$, then repeated application of this identity gives $k^{p^e} = k$ for all $k \in \kk$. Now assume that $f \nmid e$. Since the multiplicative group of a finite field is cyclic~\cite[\S~9, Prop. 18]{Dummit2004}, there exists an element $k \in \kk \setminus \{ 0 \}$ of order $p^f -1$. We claim that $k^{p^e} \neq k$. Suppose, for contradiction, that $k^{p^e} = k$. Then $k^{p^e -1 } = 1$, and hence $p^f -1 \mid p^e -1$. It follows that 
 \[ p^f -1 = \gcd(p^e-1,p^f-1) = p^{\gcd(e,f)}-1, \]
 where the last equality is a standard identity; see, for example,~\cite[Example 1.14]{Mollin2008}. Therefore, $f = \gcd(e,f)$, and hence $f \mid e$, contradicting our assumption. Finally, if $\kk$ is infinite, the polynomial $x^q - x \in \kk[x]$ cannot vanish on all of $\kk$, so there exists some $k \in \kk$ such that $k^q \neq k$.
 \end{proof}
@@ -207,7 +207,7 @@ We say that $R$ is \textbf{$q$-split} if the pseudo-Frobenius map $\qpower{F} \c
 \end{definition}
 
 \begin{remark}
-%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}.
+%Recall that $R$ is $q$-split if and only if $R$ is a direct summand of $\qpower{R}$; see, for example,~\cite[\S 10, Prop. 26]{Dummit2004}.
 By~\cite[Prop. 2.3.1]{Nastasescu2004}, the map $\qpower{F}$ splits as an $R$-module homomorphism if and only if it splits in the category of $\QQ^d$-graded $R$-modules. In other words, whenever $\qpower{F}$ splits, the splitting map $\sigma$ may be chosen to be $\QQ^d$-graded. Hence, $R$ is $q$-split if and only if it is a graded direct summand of $\qpower{R}$, that is, if there exists a $\QQ^d$-graded $R$-module $N$ such that $\qpower{R}\grcong R \oplus N$.
 \end{remark}