# More about Ext Calculations with Regular Sequences

Posted by Jason Polak on 21. March 2017 · Write a comment · Categories: commutative-algebra · Tags:

This post is a continuation of this previous one, though I repeat the main definitions for convenience.

Let $R$ be a commutative ring and $A$ and $R$-module. We say that $x_1,\dots,x_n\in R$ is a regular sequence on $A$ if $(x_1,\dots,x_n)A\not = A$ and $x_i$ is not a zero divisor on $A/(x_1,\dots,x_{i-1})A$ for all $i$. Last time, we looked at the following theorem:

Theorem. Let $A$ and $B$ be $R$-modules and $x_1,\dots,x_n$ a regular sequence on $A$. If $(x_1,\dots,x_n)B = 0$ then
$${\rm Ext}_R^n(B,A) \cong {\rm Hom}_R(B,A/(x_1,\dots,x_n)A)$$

When $R$ is a Noetherian ring, $I$ a proper ideal of $R$, and $A$ a finitely-generated $R$-module, this theorem for $B = R/I$ says that the least integer $n$ such that ${\rm Ext}_R^n(R/I,A)\not= 0$ is exactly the length of a maximal regular sequence in $I$ on $A$.

The Noetherian and finitely generated hypotheses are crucial. Why is this? It’s because you need to have control over zero divisors. In fact you can see this by looking at the case $n = 0$:

Theorem. Let $R$ be a Noetherian ring, $I$ a proper ideal of $R$, and $A$ a finitely-generated $R$-module. Then every element of $I$ is a zero divisor on $A$ if and only if ${\rm Hom}_R(R/I,A)\not= 0$.
Proof. Since $A$ is a finitely generated $R$-module, that every element of $I$ is a zero divisor on $A$ is equivalent to $I$ being contained in the annihilator of a single nonzero element $a\in A$, which is in turn equivalent to every element of $I$ being sent to zero under the homomorphism
$$R\to A\\ 1\mapsto a.$$
Such homomorphisms are the same as nonzero homomorphisms $R/I\to A$. QED.

Here we are using this crucial fact:

Cool Theorem. For a finitely generated module $A$ over a Noetherian ring $R$, the zero divisors $Z(A)$ of $A$ in $R$ are a union of prime ideals of $R$, each of which are ideals maximal with respect to the property of being in $Z(A)$. Furthermore, each such prime is the annihilator of a single nonzero element of $A$.

In general, primes that are equal to the annihilator of a single element of a module $M$ are called the associated primes of $M$, and of course the theory of associated primes and primary decomposition is much more vast than this simple ‘Cool Theorem’, as is evident from Eisenbud’s 30-page treatment of them in his book Commutative Algebra. In practice however, I only ever seem to need this simple version of the ‘Cool Theorem’.

# Regular Sequences and Ext Calculations

Posted by Jason Polak on 22. February 2017 · Write a comment · Categories: commutative-algebra, homological-algebra · Tags:

Let $R$ be a commutative ring and $A$ and $R$-module. We say that $x_1,\dots,x_n\in R$ is a regular sequence on $A$ if $(x_1,\dots,x_n)A\not = A$ and $x_i$ is not a zero divisor on $A/(x_1,\dots,x_{i-1})A$ for all $i$. Regular sequences are a central theme in commutative algebra. Here’s a particularly interesting theorem about them that allows you to figure out a whole bunch of Ext-groups:

Theorem. Let $A$ and $B$ be $R$-modules and $x_1,\dots,x_n$ a regular sequence on $A$. If $(x_1,\dots,x_n)B = 0$ then
$${\rm Ext}_R^n(B,A) \cong {\rm Hom}_R(B,A/(x_1,\dots,x_n)A)$$

This theorem tells us we can calculate the Ext-group ${\rm Ext}_R^n(B,A)$ simply by finding a regular sequence of length $n$, and calculating a group of homomorphisms. We get two cool things out of this theorem: first, a corollary of this theorem is that any two maximal regular sequences on $A$ have the same length if they are both contained in some ideal $I$ such that $IA\not= A$, and second, it enapsulates a whole range of Ext-calculations in an easy package.

For example, let’s say we wanted to calculate ${\rm Ext}_\Z^1(\Z/2,\Z)$. Well, $2\in\Z$ is a regular sequence, and so the above theorem tells us that this Ext-group is just ${\rm Hom}_\Z(\Z/2,\Z/2) \cong\Z/2$.

Another example: is ${\rm Ext}_{\Z[x]}^1(\Z,\Z[x])\cong\Z$.

Of course, the above theorem is really just a special case of a Koszul complex calculation. However, it can be derived without constructing the Koszul complex in general, and so offers an instructive and minimalist way of seeing that for Noetherian rings and finitely generated modules, the notion of length of a maximal regular sequence is well-defined.

# A Very Short Introduction to Regular Sequences

Take yourself away from this cold day in December and transport yourself to the world of commutative rings with identity. In this land there is a wonderful tool called the theory of regular sequences, which we will examine in this post. Our aim will be to get a quick idea of what regular sequences are, without going into too much tedious detail, with the hope that everyone reading this will think regular sequences are cool.

Now before I even define regular sequences, let us look at some examples of regular sequences:

1. In the ring $k[x,y,z]$, the sequence $x,y,z$.
2. In the ring $\mathbb{Z}[x]$, the sequence $2,x$.
3. In the ring $\mathbb{Z}$, the sequence $4$