add tex pipeline

.forgejo/workflows/build-document.yml

@@ -0,0 +1,57 @@
+name: Build TeX document
+
+on:
+  push:
+    branches:
+      - main
+  workflow_dispatch:
+
+jobs:
+  build-and-publish:
+    runs-on: ubuntu-latest
+
+    steps:
+      - name: Checkout repository
+        uses: https://github.com/ischanx/checkout@06c611454073903c9d44641d7e085b21115d3dcf
+
+      - name: Install Tectonic
+        shell: bash
+        env:
+          TECTONIC_VERSION: 0.15.0
+          TECTONIC_SHA256: dfb82876f2986862996e564fa507a9e576e0c1e3bee63c2c1bd677c2543e6407
+        run: |
+          set -euo pipefail
+          archive="tectonic-${TECTONIC_VERSION}-x86_64-unknown-linux-musl.tar.gz"
+          url="https://github.com/tectonic-typesetting/tectonic/releases/download/tectonic%40${TECTONIC_VERSION}/${archive}"
+          curl --fail --location --silent --show-error "$url" --output "$archive"
+          echo "${TECTONIC_SHA256}  ${archive}" | sha256sum --check --strict
+          tar -xzf "$archive"
+          install -m 0755 tectonic "$RUNNER_TEMP/tectonic"
+          echo "$RUNNER_TEMP" >> "$GITHUB_PATH"
+
+      - name: Compile document
+        shell: bash
+        run: |
+          set -euo pipefail
+          mkdir -p dist/build dist/release
+          tectonic --keep-logs --outdir dist/build document.tex
+          cp dist/build/document.pdf dist/release/document.pdf
+
+      - name: Retain workflow artifact
+        uses: https://code.forgejo.org/actions/upload-artifact@c6a3b2bd78b3985e4b2f15397fec357f0fd808de
+        with:
+          name: ncr-of-monomial-curves
+          path: dist/release/ncr-of-monomial-curves.pdf
+          if-no-files-found: error
+
+      - name: Update stable release
+        uses: https://code.forgejo.org/actions/forgejo-release@98265452477dafb3f0f27ba9c462c90b18cb44fd
+        with:
+          direction: upload
+          url: https://imsc.uni-graz.at/git/
+          token: ${{ forge.token }}
+          tag: latest
+          title: Latest compiled document
+          release-dir: dist/release
+          release-notes: Automatically built from the latest commit on the main branch.
+          override: true

.gitlab-ci.yml

@@ -1,9 +0,0 @@
-compile_latex:
-  image: texlive/texlive:latest
-
-  script:
-    - latexmk -pdf -interaction=nonstopmode -halt-on-error ncr-of-monomial-curves.tex
-
-  artifacts:
-    paths:
-      - ncr-of-monomial-curves.pdf

ncr-of-monomial-curves.tex

@@ -264,7 +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 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
+is a numerical monoid and $R = \kk[\SGR]$ is the corresponding one-dimensional toric algebra. To exclude the trivial case, we assume that $\SGR$ is proper, that is, $\SGR\subsetneq\NN_0$. 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
 \[
 \Gamma_q \coloneqq \left\{ \bigl( x^{a_1/q} \bigr)^{e_1} \cdots \bigl( x^{a_n/q} \bigr)^{e_n} \mid e_i \in \intval{0,q-1} \right\}
 \]
@@ -296,7 +296,7 @@ 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 $\overline{R}=\kk[x]$ given by
+Let $A\subseteq \NN$. The \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}
 
@@ -330,7 +330,7 @@ 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 $\overline{R} = \kk[x]$.
+defines a bijection between $\Min(\SGR)$ and the set of $R$-module isomorphism classes of $\NN_0$-graded $R$-submodules of $\overline{R} = \kk[x]$.
 \end{lemma}
 
 \begin{proof}
@@ -399,7 +399,7 @@ Let $\SGR = \langle 3,4 \rangle \subseteq \NN_0$ and $R = \kk[\SGR] = \kk[x^3,x^
     \draw[thick] (6,0.18) -- (6.15,0.18);
     \draw[thick] (6,-0.18) -- (6.15,-0.18);
 \end{tikzpicture}
-\caption{The numerical semigroup $\SGR = \langle 3, 4\rangle$}
+\caption{The numerical monoid $\SGR = \langle 3, 4\rangle$}
 \end{figure}
 \end{center}
 
@@ -422,48 +422,48 @@ It holds that $q \in \SGR$ if and only if every $\QQ^d$-graded indecomposable su
 \begin{proof}
 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 $\supp(I_{q,r}) \subseteq \{ \frac{m_{q,r}}{q} + \alpha \mid \alpha \in \NN_0\}$, 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
+and observe that $\supp(I_{q,r}) \subseteq \bigl\{ \frac{m_{q,r}}{q} + \alpha \mid \alpha \in \NN_0\bigr\}$, 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
-\[ \supp(I_{q,r}) = \{ \frac{m_{q,r}}{q} + \alpha \mid \alpha \in \NN_0\}. \]
+\[ \supp(I_{q,r}) = \Bigl\{ \frac{m_{q,r}}{q} + \alpha \mid \alpha \in \NN_0\Bigr\}. \]
 It now follows from~\eqref{eq:ind-summand-as-vector-space} that
-\[ I_{q,r} = \spn_\kk \{ x^{m_{q,r}/q + \alpha} \mid \alpha \in \NN_0 \} = x^{\frac{m_{q,r}}{q}} \cdot \spn_\kk\{ x^\alpha \mid \alpha \in \NN_0 \} \cong k[x] . \]
+\[ I_{q,r} = \spn_\kk \bigl\{ x^{\frac{m_{q,r}}{q} + \alpha} \, \big| \, \alpha \in \NN_0 \bigr\} = x^{\frac{m_{q,r}}{q}} \cdot \spn_\kk\{ x^\alpha \mid \alpha \in \NN_0 \} \cong k[x] . \]
-
 
+Conversely, suppose that every indecomposable $\QQ^d$-graded summand of $R^{1/q}$ is, up to a degree shift, isomorphic to $\overline{R}$. In particular, $I_{q,0} \shcong \overline{R}$. Since both are gap modules, Lemma~\ref{lemma:rank-1} yields $I_{q,0}=\overline{R}=\kk[x]$. The description of $I_{q,0}$ then implies that $q \in \SGR$.
 \end{proof}
 
 \todo[inline]{think about what can be said about decompositions of $R^{1/q}$ for $q \notin \SGR$}
 
 \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
+Revisiting Example~\ref{ex:large-powers}, it remains to determine the $\QQ_{\geq 0}$-graded 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
-\[ R^{1/5}\cong I(\{2\}) \oplus \overline{R}^{\oplus 4}. \]
+\[ R^{1/5} \shcong I(\{2\}) \oplus \overline{R}^{\oplus 4}. \]
-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,
+Since $\overline{R} = \kk[x]$ corresponds to $I(\{1,2\})$, the indecomposable $\QQ_{\geq 0}$-graded summands occurring in $R^{1/q}$ for some $q\in\NN$ are, up to degree shift, precisely $I(\{2\})$ and $I(\{1,2\})$. On the other hand,
-\[ \Min(\SGR) = \{ \emptyset,\{1\}, \{2\}, \{5\}, \{1,2\} \}, \]
+\[ \Min(\SGR)=\{\emptyset,\{1\},\{2\},\{5\},\{1,2\}\}, \]
-so not every potential candidate $I(A)$ with $A \in \Min(\SGR)$ occurs as a direct summand.
+so not every potential summand $I(A)$ with $A \in \Min(\SGR)$ actually occurs.
 \end{example}
 
 \begin{corollary}
-For each $q \geq 2$, $R$ is not $q$-split, meaning that $R$ does not occur as a direct summand of $R^{1/q}$. 
+Let $\SGR \subsetneq \NN_0$. Then, for every $q \in \NN_{\geq 2}$, the toric algebra $R$ is not $q$-split, or equivalently, $R$ does not occur as a $\QQ_{\geq 0}$-graded direct summand of $R^{1/q}$.
 \end{corollary}
 
 \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 $\overline{R}$, and therefore cannot contain $R$ as a direct summand.
+Fix $q \in \NN_{\geq 2}$, 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 shifted copies of $\overline{R}$, and therefore cannot contain $R$ as a $\QQ_{\geq 0}$-direct summand.
 \end{proof}
 
 \begin{proposition} 
-Let $p$ be a prime divisor of $q$. If $R^{1/q}$ and $R^{1/(pq)}$ have the same indecomposable summands up to isomorphism, then the same holds for $R^{1/(p^e q)}$ for every $e \in \NN$.
+Let $p \in \NN$. If the indecomposable $\QQ_{\geq 0}$-graded summands of $R^{1/q}$ and $R^{\frac{1}{pq}}$ coincide up to degree shift, the the same is holds for $R^{1/q}$ and $R^{\frac{1}{p^e q}}$ for every $e \in \NN$.
 \end{proposition}
 
 \begin{proof}
-Decomposing $R^{1/q} \cong \bigoplus_{i=1}^k I(A_i)^{\oplus n_i}$ for pairwise distinct $A_i \in \Min(\SGR)$, it holds that
+Decompose $R^{1/q} \shcong \bigoplus_{i \in \intval{1,k}} I(A_i)^{\oplus m_i}$, where $A_1$, \dots, $A_k \in \Min(\SGR)$ are pairwise distinct and $m_i \in \NN$. By Lemma~\ref{lemma:qpower-under-shift}, it follows that
-\[ R^{1/(pq)} \cong \qpower[p]{(R^{1/q})} \cong \bigoplus_{i=1}^k \bigl(\qpower[p]{I(A_i)}\bigr)^{\oplus n_i}. \]
+\[  R^{\frac{1}{pq}} \grcong \qpower[p]{(R^{1/q})} \shcong \bigoplus_{i \in \intval{1,k}} \left(\qpower[p]{I(A_i)}\right)^{\oplus m_i}. \]
-Thus, by assumption, the indecomposable summands occuring in the $R$-modules $\qpower[p]{I(A_i)}$, $i \in \intval{1,k}$, are, up to isomorphism, precisely $I(A_1)$, \dots $I(A_k)$. Repeated application of this observation shows that $R^{1/(p^e q)}$ has the same indecomposable summands, up to isomorphism, for every $e \in \NN$.
+Thus, by assumption, the indecomposable $\QQ_{\geq 0}$-graded direct summands occurring in $\qpower[p]{I(A_i)}$, for $i \in \intval{1,k}$, are, up to degree shift, precisely $I(A_1)$, \dots, $I(A_k)$. Iterating this argument shows that $R^{\frac{1}{p^e q}}$ has the same indecomposable $\QQ_{\geq 0}$-graded direct summands, up to degree shift, for every $e \in \NN$. 
 \end{proof}
 
 \section{Global dimension of endomorphism algebras}
 
-Throughout this section, let $\SGR$ again be a numerical semigroup and $R = \kk[\SGR]$ the corresponding monomial curve.
+Throughout this section, let $\SGR$ again be a numerical monoid and $R = \kk[\SGR]$ the corresponding toric algebra.
 
 For a commutative Noetherian ring $\Lambda$ and a finitely generated faithful $\Lambda$-module $M$, the endomorphism algebra $\End_\Lambda(M)$ is called a \textbf{noncommutative resolution of singularities} of $\Lambda$ if it has finite global dimension.
 
@@ -472,7 +472,7 @@ 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 $R^{1/q}\cong \overline R^{\oplus q}$. 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
+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
 \[ \gldim(\End_R(R^{1/q})) = \gldim(\overline{R}) = \gldim(\kk[x]) = 1 = \dim(R), \]
 where the first equality follows from the Morita invariance of global dimension.
 \end{proof}