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\title{Noncommutative Resolutions of Monomial Curves}
\author{Markus M. R. Tripp}
\date{}

\begin{document}
\maketitle

\begin{abstract}
\end{abstract}

\section{Monomial curves}

\section{Frobenius maps on affine semigroups}
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 the associated semigroup algebra. Fix a positive integer $q$, and write $q^{-1}\SGR \coloneqq \{ a/q \mid a \in \SGR \} \subseteq \QQ^d$ for the affine semigroup $\SGR$ refined by $q$. Consider the semigroup algebra
\[ R^{1/q} \coloneqq \kk[q^{-1}\SGR] = \spn_\kk\{ x^{a/q} \mid a \in \SGR \} \]
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 characteristic $p$, where $R^{1/q}$ is the ring of $q$-th roots of elements in $R = \kk[\SGR]$. Further, for $r = \sum_{a \in \QQ^d} k_a x^{a}$, we set $\qpower{r} \coloneqq \sum_{a \in \QQ^d} k_a x^{aq}$, and write $\qpower{F}$ for the resulting \emph{$q$-th pseudo-Frobenius endomorphism} on $R$, which maps $r$ to $\qpower{r}$.

\begin{remark}
% 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 $F^e$.
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, in which case $\kk \subseteq R$ is fixed pointwise by $F^e$.
\end{remark}

\begin{definition}
Let $M$ be an $R$-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~$r\cdot m \coloneqq \qpower{r}m$  
% EXPLICIT DECLARATION % \[ \Bigl( \sum_{a \in S} k_a x^a \Bigr) \cdot_q \; m \coloneqq \bigl( \sum_{a \in S} k_a x^{aq} \bigr) m \]
for all $r \in R$ and $m \in M$.
\end{definition}

\begin{lemma}\label{lemma:cannonc-isom}
There is a cannonical $R$-module isomorphism 
\[ R^{1/q} \xrightarrow{\;\sim\;} \qpower{R}, \]
defined by $r = \sum_{a \in \SGR} k_a x^{a/q} \mapsto \qpower{r} = \sum_{a \in \SGR} k_a x^a$.
\end{lemma}

\begin{proof}
By abuse of notation, we also write $\qpower{F}$ for the map $R^{1/q} \to \qpower{R}$ above; it is clearly additive and bijective. To verify compatibility with the respective $R$-module structures, write $r = \sum_{a \in \SGR} k_a x^a \in R$ and $f = \sum_{a \in \SGR} \ell_a x^{a/q} \in R^{1/q}$. Then
\[ \qpower{F}(rf) = \qpower{F}\Bigl(\sum_{a, b \in \SGR} k_a\ell_b x^{a + b/q}\Bigr) = \sum_{a, b \in \SGR} k_a\ell_b x^{aq + b}, \]
which is precisely $r\cdot\qpower{F}(f)$ in $\qpower{R}$.
\end{proof}

\begin{definition}
We say that $R$ is \textbf{$q$-split} if the Frobenius map $\qpower{F} \colon R \to \qpower{R}$ splits as a map of $R$-modules. Moreover, we call $R$ \textbf{$q$-weakly normal} if, for every $r \in \kk[G(\SGR)]$\todo{$\Frac(R)$?}, $\qpower{r} \in R$ implies $r \in R$.
\end{definition}

\begin{remark}
In semigroup-thereotic terms, $R = \kk[\SGR]$ is $q$-weakly normal if and only if, for every $a \in G(\SGR)$, $qa \in S$ implies $a \in S$. In particular, $\SGR$ (or equivalently, $R$) is normal if and only if $R$ is $q$-weakly normal for all $q \in \NN$.
\end{remark}

In positive characteristic $p$, the Frobenius endomorphism yields $\qpower[p]{r} \in R$ if and only if $r^p \in R$, from which we obtain the following:

\begin{corollary}
If $\kk$ has characteristic $p >0$, then $R$ is $p$-weakly normal 
% MORE INFORMATION % (or, equivalently, $p^e$-weakly normal for all $e \in \NN$)
if and only if $R$ is weakly normal (i.e., for any $r \in \Frac(R)$, $r^p$ implies $r \in R$). In particular, $R$ is seminormal in this case.
\end{corollary}

\begin{proposition}\textnormal{\cite{Hochster1976}, \cite[Prop. 5.3]{Yasuda2009}}
$R$ is $q$-split if and only if it is $q$-weakly normal.
\end{proposition}

\begin{proof}
% FAKE PROOF % Suppose $R$ is $q$-split, with $\sigma \colon \qpower{R} \to R$ an $R$-module homomorphism satisfying $\sigma \circ \qpower{F} = \id_R$; in particular, $\sigma(1) = 1$. Now let $r \in \Frac(R)$ be such that $\qpower{r} \in R$. Then $r  = r\sigma(1) = \sigma(\qpower{r}) \in R$. 

Suppose $R$ is $q$-split, with $\sigma \colon \qpower{R} \to R$ an $R$-module homomorphism satisfying $\sigma \circ \qpower{F} = \id_R$. Let $r \in \kk[G(\SGR)]$ be such that $\qpower{r} \in R$, and choose $s \in R^\bullet$ with $sr \in R$. Then
\[ sr = \sigma\bigl( \qpower{F}(sr) \bigr) = \sigma(\qpower{s}\qpower{r}) = s\sigma(\qpower{r}), \]
and cancelling $s$ yields $r = \sigma(\qpower{r}) \in R$.

Conversely, suppose that $R$ is $q$-weakly normal. Let $\sigma \colon \qpower{R} \to R$ be the map that sends $f \in \qpower{R}$ to the unique $g \in R$ with $f = \qpower{g}$, if such a $g$ exists, and to $0$ otherwise. Clearly $\sigma \circ \qpower{F} = \id_R$, so it remains to show that $\sigma$ is a homomorphism of $R$-modules. As $\sigma$ is additive, it suffices to check that $\sigma(x^{aq+b}) = x^a \sigma(x^b)$ for $a$, $b \in \SGR$, which is straightforward using the assumption that $R$ is $q$-weakly normal. % MORE EXPLANATION % Suppose first that $b = qc$ for some $c \in \SGR$. Then $\sigma(x^{aq+b}) = \sigma(x^{q(a+c)}) = x^{a+c} = x^a \sigma(x^b)$. Now suppose there exists no $c \in \SGR$ with $b = qc$, so $\sigma(x^b) = 0$. If $aq+b = qc$ for some $c \in \SGR$, then $\qpower{x^{c-a}} = x^b \in R$, so $x^{c-a} \in R$ by assumption, giving $c-a \in \SGR$ and $b = q(c-a)$, a contradiction. Hence $\sigma(x^{aq+b}) = 0 = x^a\sigma(x^b)$.
\end{proof}

\begin{corollary}\textnormal{\cite[Cor. 6.3]{Bruns2006}}
$R$ is normal if and only if $R$ is $q$-split for all $q \in \NN$.
\end{corollary}

\begin{lemma}\textnormal{\cite[Lemma 5.0.3]{Blickle2013}}\label{lemma:q-split}
Let $q = p_1^{e_1} \cdots p_k^{e_k}$, where the $p_i$ are distinct primes. Then $R$ is $q$-split if and only if $R$ is $(p_1^{\ell_1}\cdots p_k^{\ell_k})$-split for all $(\ell_1, \dots, \ell_k) \in \NN_0^k$.
\end{lemma}

\begin{proof}
Assume that $R$ is $q$-split, i.e., there exists a $R$-module homomorphism $\sigma \colon \qpower{R} \to R$ such that $\sigma \circ \qpower{F} = \id_R$. We first claim that, for each $i \in [1,k]$, $R$ is $p_i^{\ell_i}$ split for every $\ell_i \in \NN_0$. To this end, fix an $i \in [1,k]$ and consider the $R$-module homomorphism $f_i \colon \qpower[p_i]{R} \to \qpower{R}$, defined by $r \mapsto \qpower[q/p_i]{r}$. Setting $\sigma_i \coloneqq \sigma \circ f_i$, we obtain that 
\[ \sigma_i(\qpower[p_i]{F}(r)) = \sigma(f_i(\qpower[p_i]{r})) = \sigma(\qpower{r}) = r \] 
for all $r \in R$, so $R$ is $p_i$-split. Now, an induction on $\ell_i$ shows that $\sigma_i^{\ell_i} \circ \qpower[p_i^{\ell_i}]{F} = \id_R$ for all $\ell_i \in \NN_0$, where $\sigma_i^{\ell_i}$ denotes the $\ell_i$-fold composition of $\sigma_i$. Moreover, $\sigma_i^{\ell_i}$, regarded as a map $\qpower[p_i^{\ell_i}]{R} \to R$, is readily verified to be $R$-linear, providing the claim.

Finally, fix $(\ell_1, \dots, \ell_k) \in \NN_0^k$, and set $q^\prime \coloneqq p_1^{\ell_1} \cdots p_k^{\ell_k}$. By the first part, $R$ is $p_i^{\ell_i}$-split for each $i \in [0,k]$. Hence, there exist $R$-module homomorphisms $\sigma_i^\prime \colon \qpower[p_i^{\ell_i}]{R} \to R$ such that $\sigma_i^\prime \circ \qpower[p_i^{\ell_i}]{F} = \id_R$. Setting $\sigma^\prime \coloneqq \sigma_1^\prime \circ \cdots \circ \sigma_k^\prime$, a direct verification shows that $\sigma^\prime$ gives the required splitting of $\qpower[q^\prime]{F}$.
\end{proof}

\section{$R$-module decomposition of $R^{1/q}$}

We now specialize to the case where $\SGR = \langle a_1, \dots, a_n \rangle$ %= \{ c_1a_1 + \cdots + c_na_n \mid c_i \in \NN_0\} \subseteq \NN_0 \}
is a numerical semigroup. Following Shibuta~\cite{Shibuta2017}, we sketch how $R^{1/q} = \kk[x^{a_1/q}, \dots, x^{a_n/q}]$ decomposes as an $R$-module. Observe that
\[
\Gamma_q \coloneqq \left\{ \bigl( x^{a_1/q} \bigr)^{e_1} \cdots \bigl( x^{a_n/q} \bigr)^{e_n} \mid e_i \in [0,q-1] \right\}
\]
is a homogeneous generating set for $R^{1/q}$. We partition $\Gamma_q$ by degree modulo $\ZZ$: writing each degree as
$a/q$ with $a \in \NN_0$, this amounts to grouping elements by the residue of $a$ modulo $q$. This yields
\[
\Gamma_q = \Gamma_{q,0} \sqcup \Gamma_{q,1} \sqcup \cdots \sqcup \Gamma_{q,q-1},
\]
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 $\kk[x^q \mid q \in \QQ_{\geq 0}]$
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} \},
\end{equation}
and, by construction, summing over $r$ gives the $R$-module decomposition
\[
R^{1/q} = \bigoplus_{r \in [0,q-1]} I_{q,r},
\]
where each summand is indecomposable (see Lemma~\ref{lemma:rank-1}).

% REDUNDANT % \begin{lemma}
%Each $R$-module $I_{q,r}$ has rank $1$ and is therefore indecomposable.
%\end{lemma}

%\begin{proof}
%For simplicity, let $Q \coloneqq \Frac(R)$. The localization $V_{q,r} \coloneqq (R\setminus \{ 0 \})^{-1} I_{q,r}$ is isomorphic to $Q \otimes_R I_{q,r}$ as a $Q$-vector space. For a fixed element $x^{a/q} \in I_{q,r} \subseteq V_{q,r}$, we claim that $V_{q,r}$ is $1$-dimensional over $Q$, generated by $x^{a/q}$.

%To see this, take any $x^{b/q} \in I_{q,r}$. Since $b \equiv a \pmod{q}$, we have $b-a = cq$ for some $c \in \ZZ$. Thus, $x^{b/q} = x^c x^{a/q} \in \spn_Q\{ x^{a/q} \}$. The claim follows because any element of $V_{q,r}$ is a $Q$-linear combination of homogeneous elements of $I_{q,r}$.
%\todo{note $\Frac{R} = \kk(x)$} 
%\end{proof}

Moreover, writing $m_{q,r}/q$ for the minimal degree occurring among
elements of $\Gamma_{q,r}$, we have
\[
I_{q,r} = x^{m_{q,r}/q} \cdot \langle x^\alpha \mid \alpha \in \NN_0, \, m_{q,r}/q + \alpha \in \Gamma_{q,r} \rangle_R,
\]
from which it follows that every indecomposable summand of $R^{1/q}$ is isomorphic to $I(A) \coloneqq \langle x^\alpha \mid \alpha \in A \cup \{ 0 \} \rangle_R$ for some $A \subseteq \gapset$\todo{name?}. We may always choose $A$ to be minimal with respect to the order $\preceq_\SGR$ induced by $\SGR$, where $\alpha \preceq_\SGR \beta$ means $\beta - \alpha \in \SGR$. Write
\[ \Min(\SGR) \coloneqq \{ A \in \gapset \mid A \text{ is minimal with respect to } \preceq_\SGR \}. \]
In particular, as $q$ ranges over $\NN$, only finitely many indecomposable summands of $R^{1/q}$ occur, up to isomorphism.

\begin{lemma}\label{lemma:rank-1}
For each $A \in \Min(\SGR)$, the $R$-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 homogeneous $R$-submodules of $\bar{R} = \kk[x]$.
\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, it is clear that $I(A)$, $A \in \Min(\SGR)$, are homogeneous fractional ideals of $R$, so the map is well-defined. To show injectivity, suppose we are given an $R$-module isomorphism $\varphi \colon I(A) \to I(B)$ for some $A$, $B \in \Min(\SGR)$. This induces an isomorphism of $Q$-vector spaces $\varphi_{(0)} \colon I(A)_{(0)} \to I(B)_{(0)}$, where both sides equal~$Q$.
%and $\varphi_{(0)}(f/1) = \varphi(f)$ for all $f \in I(A)$. 
In particular, $\varphi_{(0)}$ (and therefore $\varphi$) is given by multiplication with some $g_0 \in Q$, so $\varphi(f) = g_0 f$ for all $f \in I(A)$. Taking $f=1$ yields $g_0 = \varphi(1) \in I(B)$. Since $\varphi$ is surjective and $1 \in I(B)$, a degree argument yields $g_0 \in \kk\setminus\{ 0 \}$. Consequently, 
\[ I(B) = \varphi(I(A)) = g_0 \cdot I(A) = I(A), \]
and because $A$ and $B$ are both minimal with respect to $\preceq_\SGR$, it follows that~$A = B$. Now, let $I \subseteq \bar{R}$ be an arbitrary homogeneous $R$-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 $\preceq_\SGR$.
\end{proof}

\begin{remark}
We now present another perspective on the $R$-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 $R$-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
$I(A)_\mathfrak{m}$ 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 $\qpower{\bar{R}} = \bar{R}^{\oplus q}$ as $R$-module.
\end{lemma}

\begin{proof}
First, note that 
\[ \bar{R}^{1/q} = \kk[x^{1/q}] = \bigoplus_{r \in [0,q-1]} x^{r/q} \cdot \kk[x] \cong \bar{R}^{\oplus q} \]
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.
\end{proof}

\begin{example}\label{ex:large-powers}
Let $\SGR = \langle 3,4 \rangle \subseteq \NN_0$ and $R = \kk[\SGR] = \kk[x^3,x^4]$.

\begin{center}
\begin{figure}[htbp]
\begin{tikzpicture}[x=0.8cm]
    \draw[->] (0,0) -- (9.5,0);
    
    % labels
    \foreach \x in {0,1,2,5,6,7,8,9}
        \draw (\x,0.08) -- (\x,-0.08)
              node[below=3pt] {$\x$};
    
    % generators
    \foreach \x in {3,4}
        \draw[red] (\x,0.08) -- (\x,-0.08)
              node[below=3pt,red] {$\x$};

    % black dots for semigroups elements
    \foreach \x in {0,6,7,8,9}
        \fill (\x,0) circle (2.5pt);
    
    % red dots for generators
    \foreach \x in {3,4}
        \fill[red] (\x,0) circle (2.5pt);

    % bracket for frobenius number
    \draw[thick] (6,-0.18) -- (6,0.18);
    \draw[thick] (6,0.18) -- (6.15,0.18);
    \draw[thick] (6,-0.18) -- (6.15,-0.18);
\end{tikzpicture}
\caption{The numerical semigroup $\SGR = \langle 3, 4\rangle$}
\end{figure}
\end{center}

The set
\[ \{ x^{(3/q)e_1 + (4/q)e_2} : e_i \in [0,q-1] \} \]
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
\[ R^{1/2} = \langle 1, x^2 \rangle_R \oplus x^{3/2} \cdot \langle 1, x^2 \rangle_R \cong I(\{2 \})^{\oplus 2}  \] 
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
\[ R^{1/8} \cong \qpower[2]{(\qpower[4]{R})} = \qpower[2]{(\bar{R}^{\oplus 4})} = (\qpower[2]{\bar{R}})^{\oplus 4} \cong \bar{R}^{\oplus 8}, \]
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}.
\end{example}

% NOTE: next proposition holds when q \in \SGR

\begin{proposition}\label{prop:asy-behav}
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 $q$.
\end{proposition}

\begin{proof}
Fix $r \in [0,q-1]$. 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
\[ \supp(I_{q,r}) = \{ m_{q,r}/q + n \mid n \in \NN_0\}. \]
It now follows from~\eqref{eq:ind-summand-as-vector-space} that
\[ I_{q,r} = \spn_\kk \{ x^{m_{q,r}/q + n} \mid n \in \NN_0 \} = x^{m_{q,r}/q} \cdot \spn_\kk\{ x^n \mid n \in \NN_0 \} \cong k[x] . \]
\end{proof}

\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
\[ R^{1/5}\cong I(\{2\}) \oplus \bar{R}^{\oplus 4}. \]
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,
\[ \Min(\SGR) = \{ \emptyset,\{1\}, \{2\}, \{5\}, \{1,2\} \}, \]
so not every potential candidate $I(A)$ with $A \in \Min(\SGR)$ occurs as a direct summand.
\end{example}

\begin{corollary}
For each $q \geq 2$, $R$ is not $q$-split, meaning that $R$ does not occur as a direct summand of $R^{1/q}$. 
\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 $\bar{R}$, and therefore cannot contain $R$ as a 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 $e \in \NN$.
\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
\[ R^{1/(pq)} \cong \qpower[p]{(R^{1/q})} \cong \bigoplus_{i=1}^k \bigl(\qpower[p]{I(A_i)}\bigr)^{\oplus n_i}. \]
Thus, by assumption, the indecomposable summands occuring in the $R$-modules $\qpower[p]{I(A_i)}$, $i \in [1,k]$, 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 $e \in \NN$.
\end{proof}

\section{Global dimension of endomorphism algebras}

Throughout this section, let $\SGR$ be a numerical semigroup and $R = \kk[\SGR]$ the corresponding monomial curve.

Let $\Lambda$ be a commutative Noetherian ring and $M$ a finitely generated faithful $\Lambda$-module. The endomorphism algebra $\End_\Lambda(M)$ is called a \textbf{noncommutative resolution of singularities} of $\Lambda$ if it has finite global dimension.

\begin{theorem}
For $q \in \SGR$, the global dimension of $\End_R(R^{1/q})$ is equal to the Krull dimension of $R$. In particular, this is true for all sufficiently large $q$.
\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
\[ \gldim(\End_R(R^{1/q})) = \gldim(\bar{R}) = \gldim(\kk[x]) = 1 = \dim(R), \]
where the first equality follows from the Morita invariance of global dimension.
\end{proof}

Following the argument of Faber, Muller, and Smith~\cite[Theorem~6.13]{Faber2019} for the corresponding result on toric rings, we obtain the following:

\begin{theorem}
For a perfect field $\kk$ of prime characteristic, the ring $D_\kk(R)$ of differential operators on $R$ has finite global dimension.
\end{theorem}

\todo[inline]{for $\A \coloneqq \oplus_{A \in \Min(\SGR)} I(A)$, think about $\gldim(\End_R(\A))$ and/or $\gldim(\End_R(R^{1/q}))$ for small $q$}

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