small changes

ncr-of-monomial-curves.tex

@@ -82,9 +82,9 @@
 \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
+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 its semigroup algebra, which defines a toric variety. 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 $\qpower{r}$.
+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 prime characteristic $p$, where $R^{1/q}$ is the ring of $q$-th roots of elements in $R$. 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 (ring) endomorphism} of $R$, 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 $F^e$.
@@ -104,9 +104,9 @@ defined by $r = \sum_{a \in \SGR} k_a x^{a/q} \mapsto \qpower{r} = \sum_{a \in \
 \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
+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 $m = \sum_{a \in \SGR} l_a x^{a/q} \in R^{1/q}$. Then
-\[ \qpower{F}(rf) = \qpower{F}\Bigl(\sum_{a, b \in \SGR} k_a\ell_b x^{a + b/q}\Bigr) = \sum_{a, b \in \SGR} k_a\ell_b x^{aq + b}, \]
+\[ \qpower{F}(rm) = \qpower{F}\Bigl(\sum_{a, b \in \SGR} k_a l_b x^{a + b/q}\Bigr) = \sum_{a, b \in \SGR} k_a l_b x^{aq + b}, \]
-which is precisely $r\cdot\qpower{F}(f)$ in $\qpower{R}$.
+which is precisely $r\cdot\qpower{F}(m)$ in $\qpower{R}$.
 \end{proof}
 
 \begin{definition}
@@ -120,7 +120,7 @@ In semigroup-thereotic terms, $R = \kk[\SGR]$ is $q$-weakly normal if and only i
 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 
+If $\kk$ has prime characteristic $p$, 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}
@@ -132,7 +132,7 @@ $R$ is $q$-split if and only if it is $q$-weakly normal.
 \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
+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 $s \in R\setminus \{0 \}$ 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 $r = \sigma(\qpower{r}) \in R$.
 
@@ -150,7 +150,7 @@ Let $q = p_1^{e_1} \cdots p_k^{e_k}$, where the $p_i$ are distinct primes. Then
 \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.
+for all $r \in R$, so $R$ is $p_i$-split. Now, an induction on $\ell_i$ shows that, as maps of sets, $\sigma_i^{\ell_i} \circ \qpower[p_i^{\ell_i}]{F} = \id_R$ 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}
@@ -158,7 +158,7 @@ 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 $\SGR = \langle a_1, \dots, a_n \rangle$ %= \{ 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
+is a numerical semigroup and $R = \kk[\SGR]$ is the corresponding monomial curve. 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
 \[
 \Gamma_q \coloneqq \left\{ \bigl( x^{a_1/q} \bigr)^{e_1} \cdots \bigl( x^{a_n/q} \bigr)^{e_n} \mid e_i \in [0,q-1] \right\}
 \]
@@ -167,7 +167,7 @@ $a/q$ with $a \in \NN_0$, this amounts to grouping elements by the residue of $a
 \[
 \Gamma_q = \Gamma_{q,0} \sqcup \Gamma_{q,1} \sqcup \cdots \sqcup \Gamma_{q,q-1},
 \]
-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 $\kk[x^q \mid q \in \QQ_{\geq 0}]$
+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 $\kk[x^q \mid q \in \QQ]$
 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} \},
@@ -207,11 +207,11 @@ defines a bijection between $\Min(\SGR)$ and the set of isomorphism classes of h
 \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~$Q$.
+For the second part, it is clear that $I(A)$, $A \in \Min(\SGR)$, are homogeneous $R$-submodules of $\bar{R}$, 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 $\preceq_\SGR$.
+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}
@@ -282,7 +282,7 @@ The set
 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
-\[ R^{1/8} \cong \qpower[2]{(\qpower[4]{R})} = \qpower[2]{(\bar{R}^{\oplus 4})} = (\qpower[2]{\bar{R}})^{\oplus 4} \cong \bar{R}^{\oplus 8}, \]
+\[ R^{1/8} \cong \qpower[2]{(\qpower[4]{R})} \cong \qpower[2]{(\bar{R}^{\oplus 4})} = (\qpower[2]{\bar{R}})^{\oplus 4} \cong \bar{R}^{\oplus 8}, \]
 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}
 
@@ -302,6 +302,8 @@ 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}
 
+\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
 \[ R^{1/5}\cong I(\{2\}) \oplus \bar{R}^{\oplus 4}. \]
@@ -332,9 +334,9 @@ Thus, by assumption, the indecomposable summands occuring in the $R$-modules $\q
 
 \section{Global dimension of endomorphism algebras}
 
-Throughout this section, let $\SGR$ be a numerical semigroup and $R = \kk[\SGR]$ the corresponding monomial curve.
+Throughout this section, let $\SGR$ again 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.
+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.
 
 \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 $q$.
@@ -346,13 +348,17 @@ For $q \in \SGR$, Proposition~\ref{prop:asy-behav} gives $R^{1/q}\cong \bar 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:
+\todo[inline]{is there a notion of crepant for non-normal, non-Gorenstein rings?}
+
+\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 $q$}
+
+Following the argument of Faber, Muller, and Smith~\cite[Theorem~6.13]{Faber2019} for the corresponding result on normal toric varieties, 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 $q$}
+\todo[inline]{is there machinery to show equality for the global dimension of $D_\kk(R)$ and $\dim(R)$?}
 
 \printbibliography