ncr-of-monomial-curves/ncr-of-monomial-curves.tex

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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}
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}

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

We now specialize to the case where $\SGR = \langle a_1, \dots, a_n \rangle \subseteq \NN_0$ 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 $\bar{R} = \kk[x]$\todo{this also holds for $\kk[x,x^{-1}]$, interesting?}.
\end{lemma}

\begin{proof}
For simplicity, set $Q \coloneqq \Frac(R) = \kk(x)$\todo{note $\Frac{R} = \kk(x)$}  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}
It holds that $\qpower{\bar{R}} = \bar{R}^{\otimes q}$ as  $R$-module.
\end{lemma}

\begin{proof}
Note that $\bar{R}^{1/q} = \kk[x^{1/q}] = \bigoplus_{r \in [0,q]} x^{r/q} \cdot \kk[x]$
\end{proof}

%\begin{example}
%\end{example}

\begin{proposition}
For sufficiently large $q$, all indecomposable summands of $R^{1/q}$ are isomorphic to the normalization $\bar{R} = \kk[x]$ of $R$. That is, $R^{1/q} \cong \bar{R}^{\oplus q}$.
\end{proposition}

\begin{proof}
Let $q$ be such that $\ZZ_{\geq q} \subseteq S$, and fix $r \in [0,q-1]$. For simplicity, set 
\[ \supp(I_{q,r}) \coloneqq \{ a/q \mid a \in S, \, a \equiv r \pmod{q} \} \]
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 $nq \in S$, and hence $m_{q,r} + nq \in S$. 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{proposition}
% R is not q-split 
%\end{proposition}

%\begin{proposition} 
% if two consecutive powers have same indecomposable summands, then so do all larger powers
%\end{proposition}

\section{Global dimension of endomorphism algebras}

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

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