Ozgur Esentepe - University of Leeds

DRP Turkiye - 5 Minute Talks - August 8, 2023

Slideas available at sntp.ca/ozgur

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.

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

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

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

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

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

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

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

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

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) $

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 $n$.

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.

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 $\mathrm{Ext}^n(V,N)$.

**$R$ 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 $M,N$.

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".

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.