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Markus Maximilian Rudolf Tripp 2026-09-24 17:09:50 +02:00
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@ -129,7 +129,7 @@ In the case of a numerical monoid $\SGR$, the corresponding toric algebra $R$ sa
%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 $x^{n_1}\in{}^+R$. %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. 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. 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.
\subsection{Graded Krull–Schmidt theorem} \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[§ 2]{Shibuta2017}.