## Annihilators of the singularity category

Ozgur Esentepe - University of Leeds
DRP Turkiye - 5 Minute Talks - August 8, 2023

Slideas available at sntp.ca/ozgur

### In primary school:

We learn numbers and we learn that if $xy = 0$, then either $x = 0$ or $y = 0$. Equivalently, two nonzero numbers can not multiply to zero.

### In secondary school:

We learn polynomials and we learn that if $pq = 0$, then either $p = 0$ or $q = 0$. Equivalently, two nonzero polynomials can not multiply to zero. This is how we solve polynomial equations like $(x-1)(x-3) = 0$. I am in secondary school so characteristic is zero.

### We get older:

We learn matrices and we learn that matrix multiplication is not as well behaved as well ordinary multiplication. Two nonzero matrices may multiply to zero \begin{align*} \begin{bmatrix}1 & 1 \\ 1 & 1 \end{bmatrix} \begin{bmatrix}1 & 1 \\ -1 & -1 \end{bmatrix} = \begin{bmatrix}0 &0 \\ 0 & 0 \end{bmatrix} \end{align*}

### My column space is in your null space.

Unpacking the definition of matrix multiplication, we see that \begin{align*} AB = 0 \iff \mathrm{col}(B) \subseteq \mathrm{null}(A) \end{align*}.

We may have $0 = \mathrm{col}(B) \subseteq \mathrm{null}(A)$.

We may have $0 \subseteq \mathrm{col}(B) = \mathrm{null}(A)$.

## The question is:

How different is the column space of $B$ than the null space of $A$?

We can measure this by the quotient \begin{align*} H = \mathrm{null(A)}/\mathrm{col(B)}. \end{align*}

### Chain complexes

\begin{align*} V_1 \xrightarrow{f_1} V_2 \xrightarrow{f_2} V_3 \end{align*} with the property that $f_2 \circ f_1 =0$.

Equivalently, $\mathrm{im}(f_1) \subseteq \mathrm{ker}(f_2)$.

Interested in $H = \mathrm{ker}(f_2)/\mathrm{im}(f_1)$.

### Make it longer.

\begin{align*} V_0 \xrightarrow{f_0} V_1 \xrightarrow{f_1} V_2 \xrightarrow{f_2} V_3 \xrightarrow{f_3} V_4 \end{align*}

with the property that $f_{i+1} \circ f_i =0$ for all $i = 0,1,2$.

### And longer.

\begin{align*} \ldots \to V_0 \xrightarrow{f_0} V_1 \xrightarrow{f_1} V_2 \xrightarrow{f_2} V_3 \xrightarrow{f_3} V_4 \to \ldots \end{align*}

with the property that $f_{i+1} \circ f_i =0$ for all $i \in \mathbb{Z}$.

### You can do this for every ring.

Let $R$ be a ring and $V_i$ be $R$-modules with $R$-linear maps $f_i$ in between them. \begin{align*} C \colon \ldots \to V_0 \xrightarrow{f_0} V_1 \xrightarrow{f_1} V_2 \xrightarrow{f_2} V_3 \xrightarrow{f_3} V_4 \to \ldots \end{align*}

with the property that $f_{i+1} \circ f_i =0$ for all $i \in \mathbb{Z}$.

Interested in $H_i(C) = \mathrm{ker}(f_{i+1})/\mathrm{im}(f_i)$

### Every vector space has a basis.

If $R$ is a vector space, then any finitely generated $R$-module $V$ (i.e. a vector space over $R$) has a basis. This basis gives us an isomorphism $V \cong R^n$ for some $n$.

### Free modules

If $R$ is some arbitrary(*) ring, then we say that a finitely generated module $V$ is free if it is isomorphic to $R^n$ for some $n$.

Not every module is free.

BUT

There is a "notion of isomorphism" which you can get after a "technical construction" which allows you to say that every module is "isomorphic in that sense" to a chain complex of free modules.

### Free resolutions

Given an $R$-module $V$, we can create a complex $C$ such that \begin{align*} H_n(C) \cong \begin{cases} V & \text{ if } n = 0 \\ 0 & \text{ else} \end{cases} \end{align*}

### Ext.

Let $V$ be a module and $F$ be a free resolution of $V$. $F \colon \ldots \to F_{-1} \xrightarrow{d_{-1}} F_0 \xrightarrow{d_0} F_1 \xrightarrow{d_1} F_2 \xrightarrow{d_2} F_3 \to \ldots$

That is, we have $\mathrm{ker}{d_{i+1}} = \mathrm{im}{d_i}$ everywhere except one place and that place we have homology isomorphic to $V$.

In fact, $F$ can be made to look like this: $F \colon \ldots \to F_{-2} \xrightarrow{d_{-2}} F_{-1} \xrightarrow{d_{-1}} F_0 \xrightarrow{d_0} 0$

Given another module $N$, now construct $\mathrm{Hom}(F,N)$:

$\mathrm{Hom}(F,N) \colon \ldots \leftarrow \mathrm{Hom}(F_{-2},N) \xleftarrow{\partial_{-2}} \mathrm{Hom}(F_{-1},N) \xleftarrow{\partial_{-1}} \mathrm{Hom}(F_0,N) \xleftarrow{\partial_0} 0$ where $\partial(f) = f \circ d$

This is still a complex! That is, we still have $\partial \circ \partial = 0.$

BUT while $F$ had almost no homologies, this new complex may have!

Almost no homologies: $F \colon \ldots \to F_{-2} \xrightarrow{d_{-2}} F_{-1} \xrightarrow{d_{-1}} F_0 \xrightarrow{d_0} 0$

May have homologies: $\mathrm{Hom}(F,N) \colon \ldots \leftarrow \mathrm{Hom}(F_{-2},N) \xleftarrow{\partial_{-2}} \mathrm{Hom}(F_{-1},N) \xleftarrow{\partial_{-1}} \mathrm{Hom}(F_0,N) \xleftarrow{\partial_0} 0$

Let's call the $n$th homology of this $\mathrm{Ext}^n(V,N)$.

### Examples

$R$ is a field: $\mathrm{Ext}^n(M,N) = 0$ for all $n > 0$.

$R= \mathbb{C}[x]$: $\mathrm{Ext}^n(M,N)=0$ for all $n >1$.

$R= \mathbb{C}[x,y]$: $\mathrm{Ext}^n(M,N)=0$ for all $n >2$.

$R= \mathbb{C}[x_1, \ldots, x_m]$: $\mathrm{Ext}^n(M,N)=0$ for all $n >m$.

These are for ALL modules $M,N$.

### Singularities

Is it always true that there exists a threshold $m$ such that for all $n>m$ we have $\mathrm{Ext}^n(M,N) = 0$?

The answer is yes if the "geometric object corresponding to your ring" is "smooth", the answer is no if it has "singularities".

### What did I do in my thesis?

I studied the ring elements which "see" all the "large enough" Exts as zero. That is, I studied ring elements which uniformly annihilate $\mathrm{Ext}^n(M,N)$ for large $n$ and for all $M,N$. Not so surprisingly, they form an ideal and the vanishing locus of this ideal is the singular locus of the corresponding geometric object. They can be useful tools to study different properties of your rings.