Özgür Esentepe - University of Graz
Triangulated categories in algebra and geometry
Prague - September 2024
A: We have a decomposition \[ \mathsf{D}_{\mathrm{sg}}(R) = \mathsf{smd}(\mathsf{C}(x_1) * \ldots * \mathsf{C}(x_n) ) \]
This is joint work with Ryo Takahashi.
B: We have an upper bound \[ \dim \mathsf{D}_{\mathrm{sg}}(R) \leq \sum_{i=1}^n \dim \mathsf{D}_{\mathrm{sg}}(R/x_i R) + n - 1 \]assuming that no $x_i$ is a unit or a zerodivisor.
Until further notice, let $R$ be a commutative Noetherian and local with maximal ideal $\m$.
And assume all modules are finitely generated.
A ring element $x \in \m$ is a nonzerodivisor on a module $M$ if $xm = 0$ implies $m =0$.
A sequence of elements $x_1, \ldots, x_n \in \m$ is a regular sequence on a module $M$ if $x_1$ is a nonzerodivisor on $M$ and $x_{i+1}$ is a nonzerodivisor on $M/(x_1,\ldots, x_i)M$ for any $i = 1, \ldots, n-1$.
The depth of a module $M$ is the maximum length of a regular sequence on $M$.
The depth of a nonzero module is always bounded above by the Krull dimension of the ring.
We say that $R$ is Cohen-Macaulay if the depth of the regular module equals the Krull dimension.
We say that a module $M$ over a Cohen-Macaulay ring is (maximal) Cohen-Macaulay if the depth of $M$ equals the Krull dimension of $R$.
Let $0 \to M \to N \to L \to 0$ be a short exact sequence of $R$-modules.
Consider a short exact sequence \[ 0 \to \Omega M \to P \to M \to 0 \] where $P$ is free defining the syzygy $\Omega M$. Then, we have \[ \depth(\Omega M) \geq \min\{\depth (R), \depth(M) + 1\}. \]
Consider a short exact sequence \[ 0 \to M \xrightarrow{x} M \to M/xM \to 0 \quad . \] Then, we have \[ \depth(M/xM) \geq \min\{\depth (M), \depth(M) - 1\}. \]
If $M \in \MCM(R)$ and $x$ is a nonzerodivisor in $R$, then $\Omega(M/xM) \in \MCM(R)$.
Question: Can we understand the map \[ M \mapsto \Omega(M/xM) \] on $\MCM(R)$?
Let $S = \mathbb{C}[[x_0, \ldots, x_d]]$ and $f$ be a nonzero nonunit element in $S$.
Knörrer: Put $R=S[y]/(f+y^2)$. Then, for any $M \in \MCM(R)$, we have \[\Omega(M/yM) \cong M \oplus \Omega M .\]
Herzog-Popescu: Put $R=S[y]/(f+y^k)$. Then, for any $M \in \MCM(R)$, we have \[\Omega(M/y^{k-1}M) \cong M \oplus \Omega M .\]
More generally: Let $R$ be a commutative Noetherian ring and $M$ be an $R$-module. Assume $x$ is a nonzerodivisor on $M$. Then, we have \[ \Omega(M/xM) \cong M \oplus \Omega M \] if and only if $x\Ext_R^1(M, \Omega M) = 0$.
Fun fact: We always have an equality \[\sann_R(M):= \ann_R\sEnd_R(M) = \ann_R\Ext_R^1(M, \Omega M) \]
Let $M \in \MCM(R)$ and $x$ be a nonzerodivisor. Then, we have \[ \Omega (M/xM) \cong M \oplus \Omega M \iff x \in \sann_R(M) \]
Definition: \[ \C(x): = \{ M \in \MCM(R) \colon x \in \sann_R(M) \} \]
We will change this definition later.
\[ \C(x): = \{ M \in \MCM(R) \colon x \in \sann_R(M) \} \] It is closed under finite direct sums, direct summands, taking syzygies and it contains projectives.
$C(1) = \mathrm{proj}(R)$.
Let $S = \mathbb{C}[x_1, \ldots, x_n]$ and $M$ be a finitely generated $R$-module. Then,
the $n$th syzygy module of $M$ is free.
the projective dimension of $M$ is at most $n$.
$M$ is quasi-isomorphic to a perfect complex of length at most $n$.
Question: Over which rings any finitely generated module is quasi-isomorphic to a perfect complex?
Question: Over which rings any bounded complex of finitely generated modules is quasi-isomorphic to a perfect complex?
Auslander-Buchsbaum-Serre and more: Regular rings.
Singularity Category: \[ \Dsing(R) := \frac{\mathsf{D}^{b}(R)}{\mathrm{perf}(R)} \quad .\]
Let $R$ be a Gorenstein ring. Then,
$\MCM(R)$ $\hookrightarrow \mathrm{mod}(R)$ $\hookrightarrow \D^b(R)$ $\phantom{\MCM(R) \hookrightarrow \mathrm{mod}(R) \hookrightarrow} \downarrow$ $\phantom{\MCM(R) \hookrightarrow \mathrm{mod}(R) \hookrightarrow} \Dsing(R)$gives an equivalence \[\underline{\MCM}(R) \cong \Dsing(R) \] of triangulated categories.
Now, for any commutative Noetherian ring, we define $\C(x)$ as \[\{ X \in \Dsing(R) \colon x\cdot \End_{\Dsing(R)}(X) = 0\} \]
Note: $X$ belongs to $\C(x)$ if and only if multiplication by $x$ on $X$ factors through a perfect complex.
$\C(x)$ is closed under finite direct sums, direct summands, shifts and it contains perfect complexes.
$C(1) = \mathrm{perf}(R)$.
Let $R = \mathbb{C}[[x_0, \ldots, x_d]]/(f)$ be a hypersurface ring. Then, any MCM module is given by a matrix factorisation $(A,B)$ of $f$. Moreover, we have \[\underline{\MCM}(R) \cong \mathrm{HMF}(f) \] By the product rule, we have \[\frac{\partial}{\partial x_i} f = \frac{\partial}{\partial x_i} A \; B + A \frac{\partial}{\partial x_i} B \] In particular, multiplication by partial derivatives of $f$ is nullhomotopic. Hence, \[ \C(\frac{\partial}{\partial x_i} f) = \Dsing (R). \]
Let $\A$ be an additive category and $\B$ be a subcategory. We denote by $\smd(\B)$ the subcategory of $\A$ consisting of direct summands of objects in $\B$.
Let $\T$ be an additive category and $\X, \Y$ be subcategories. We denote by $\X * \Y$ the subcategory of $\T$ consisting of objects $E \in \T$ such that there exists an exact triangle $X \to E \to Y \to X[1]$ with $X \in \X$ and $Y \in \Y$.
We have that
$*$ is associative. (8)
\begin{align*}\smd(\smd (\X) * \Y) &= \smd(\X * \Y) \\ &= \smd(\X * \smd(\Y)) \quad {\color{rgb(230,182,28)}{(\mathrm{BvdB})}}\end{align*}
Proposition: Let $x_1, \ldots, x_n \in R$. Then, \[\C(x_1 \cdots x_n) = \smd(\C(x_1) * \cdots * \C(x_n)) \]
Corollary: Let $x_1, \ldots, x_n \in R$ be such that $x_1 \cdots x_n \in \ann \Dsing(R)$. Then, \[\Dsing(R) = \smd(\C(x_1) * \cdots * \C(x_n)) \]
The key observation which is used to prove the Proposition in the generality of all commutative Noetherian rings is the following:
\[\C(x) = \smd\{\K(x) \otimes X \colon X \in \Dsing(R) \} \]
For one direction, we recall that multiplication by $x$ on $\K(x)$ is nullhomotopic. For the other direction, we recall that we have a triangle $R \xrightarrow{x} R \to \K(x) \to R[1]$ in $\D^b(R)$. This gives us a triangle \[Y \xrightarrow{x} Y \to \K(x) \otimes Y \to Y[1] \] yielding an isomorphism $\K(x) \otimes Y \cong Y \oplus Y[1]$ when $Y \in \C(x)$.
Let $\T$ be a triangulated category and $\X$ be a subcategory.
Denote by $\langle \X \rangle$ the smallest subcategory that contains $\X$ and is closed under finite direct sums, direct summands and shifts.
We put $\langle \X \rangle_0 = 0$ and define $\langle \X \rangle_n = \langle \langle \X \rangle_{n-1} * \langle \X \rangle \rangle$ for $n \geq 1$. Note that $\langle \X \rangle_1 = \langle \X\rangle$.
We define the Rouquier dimension as \[\dim \T = \inf\{ n \geq 0 \colon \langle G \rangle_{n+1} = \T \text{ for some } G \in \T \} \quad . \]
Let $x_1, \ldots, x_n$ be elements of a commutative Noetherian ring $R$ such that the product $x_1 \cdot \cdot \cdot x_n$ belongs to $\mathrm{ann} \mathsf{D}_{\mathrm{sg}}(R) $. We have \[ \dim \mathsf{D}_{\mathrm{sg}}(R) \leq \sum_{i=1}^n \dim \mathsf{D}_{\mathrm{sg}}(R/x_i R) + n - 1 \]assuming that no $x_i$ is a unit or a zerodivisor.
Assume that \[\dim (\Dsing (R/x_i R)) = d_i < \infty \]
Pick $G_i \in \Dsing(R/x_i R)$ with \[\langle G \rangle_{d_i +1} = \Dsing(R/x_i R)\]
The surjection $R \to R/x_i R$ induces \[\Dsing(R/x_i R) \to \Dsing(R) \]
When $x$ is a nonzerodivisor, \begin{align*} \C(x) &= \{ \K(x) \otimes X \colon \Dsing(R) \} \\ &= \{R/x_i R \otimes^{\mathbf{L}} X \colon X \in \Dsing R \} \end{align*}
We have an inclusion \[\C(x_i) \subseteq \langle G_i \rangle_{d_i+1}^{\Dsing(R)} \]
Let $R = \mathbb{C}[[x,y]]/(x^a + y^b)$ with $2 \leq a \leq b$. Then, \[x^{a-1} \in \ann_R \Dsing(R)\] So, we have \begin{align*}\dim \Dsing(R) &\leq (a-1) \dim \Dsing(\mathbb{C}[y]/y^b) + a-1 - 1 \\& =a-2 \end{align*}
Recall the theorems of Knörrer and Herzog-Popescu?
Dugas and Leuschke: Some extensions of theorems of Knörrer and Herzog-Popescu (2021)
Our paper could have been called Some extensions of theorems of Dugas-Leuschke
Instead of nonzerodivisors, one can also use regular sequences.
One way to go about this is to use induction.
But we have another theorem which gives better results.
Example: $R = \mathbb{C}[[x,y,z,w]]/(f)$ with $f = x^3 + y^3 + xyz +w^2$.
We have \[x^3 = \frac{1}{3} x(3x^2+yz) - \frac{1}{3}z(xy) \text{ and } y^3 = \frac{1}{3} y(3y^2+xz) - \frac{1}{3}z(xy)\]
So, \[ \dim \Dsing(R) \leq 3 \cdot 3 \left( \frac{1}{3} + \frac{1}{3} \right)\cdot(1+1) - 1 = 11 \]
The ideal $\ann_R \Dsing(R)$ defines the singular locus of $R$ given that $\dim \Dsing(R)$ is finite (Liu 2023).
If $R$ is an isolated singularity, then $\ann_R \Dsing(R)$ is $\m$-primary.