If $R$ is a commutative ring and $M$ an $R$-module, a regular sequence on $M$ is a sequence $x_1,\dots,x_n\in R$ such that $(x_1,\dots,x_n)M \not=M$ and for each $i$, the element $x_{i+1}$ is not a zero divisor on the module $M/(x_1,\dots,x_{i})$. The term regular sequence in $R$ just refers to a regular sequence on $R$ as an $R$-module over itself. The length of any regular sequence is the number of elements in the sequence.

The projective dimension of an $R$-module $M$ is the infimum over all lengths of projective resolutions of $M$, and hence is either a nonnegative integer or infinity. We write ${\rm pd}_R(M)$ for the projective dimension of an $R$-module $M$. (Clearly, this definition also makes sense for noncommutative rings.)

In particular, any ideal of $R$ is an $R$-module by definition, so we can look at the projective dimension of ideals. For instance, if $x\in R$ is a nonzerodivisor (=an element that is not a zero divisor), then the ideal $Rx$ is a left $R$-module that is free because $x$ is not a zero divisor. The following is a generalisation of this remark:

Problem. If $I$ is an ideal in a commutative ring generated by a regular sequence of length $n > 0$ then ${\rm pd}_R(I) = n-1$.

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Let $R$ be a ring and $M$ an $R$-module with a finite free resolution (an “FFR module”). That is, there exists an exact sequence $0\to F_n\to F_{n-1}\to \cdots\to F_0\to M\to 0$ with each $F_i$ a finitely generated free $R$-module. If we denote by $r_i$ the rank of $F_i$, then the Euler characteristic of $M$ is defined to be $\chi(M) = r_0 – r_1 + \cdots + (-1)^nr_n$. One can easily prove that this is independent of the finite free resolution chosen and hence is a well-defined integer. In the post The Alternating Binomial Sum Vanishes we saw that it’s possible to prove that the alternating sum $\sum (-1)^i\binom{n}{i}$ vanishes by using some facts about Euler characteristics together with the Koszul complex.

Today’s problem shows that the Euler characteristic ins’t that interesting for ideals or homomorphic images of rings. First, here are some facts about the Euler characteristic, some of which might be useful for solving the problem:

  1. Any FFR module over a commutative ring has nonnegative Euler characteristic
  2. If an FFR module over a commutative ring has nonzero Euler characteristic, then the annihilator of $A$ is nil
  3. (Stallings) If $A$ is an FFR module with annihilator $I$ then $\chi(A)I = 0$.

We refer to these as Fact 1, Fact 2, and Fact 3.

Problem. If $I$ is a nonzero (!) ideal of a ring $R$ with a finite free resolution, show that $R/I$ also has a finite free resolution and that $\chi(I) = 1$ and $\chi(R/I) = 0$.

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Posted by Jason Polak on 23. August 2014 · Write a comment · Categories: homological-algebra

Suppose $k$ is a ring and $R$ is the ring of $2\times 2$ upper triangular matrices with entries in $k$. Via left matrix multiplication, $k^2$ viewed as column vectors becomes a left $R$-module. Now it turns out that $k^2$ is projective as a left $R$-module. Today we’ll see one methods to show this. Before we do this, let us briefly outline another solution to this problem that will make the reader appreciate the second method. The most straightforward approach to this problem is perhaps using the lifting criterion, or at least the following variation of it: define a surjection $R\to k^2$ by $1\mapsto [0,1]^t$. Now, one just has find a splitting of this surjection. Exercise: find a map that’s a splitting and prove carefully that it is actually an $R$-module homomorphism.

Now let’s see another way to do this. Instead of constructing a splitting, we could continue the surjection $R\to k^2$ defined by $1\mapsto [0,1]^t$ above into a projective resolution. Since maps out of free $R$-modules are easy to construct, if the resoluting projective resolution is small, then it will be easy to see by inspecting the resolution whether $k^2$ is projective (or so we hope).

To continue the projective resolution, we just find the kernel of $R\to k^2$ and map onto it with another free module. The reader may wish to write this process down before reading on.
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A commutative Noetherian local ring $ R$ with maximal ideal $ M$ is called a regular local ring if the Krull dimension of $ R$ is the same as the dimension of $ M/M^2$ as a $ R/M$-vector space.

In studying regular local rings one often uses the following lemma in inductive arguments: if $ R$ is an arbitrary commutative Noetherian local ring with maximal ideal $ M$, and $ M$ consists entirely of zero divisors, then the projective dimension $ \mathrm{pd}_R(A)$ is either zero or infinity for any finitely-generated $ R$-module $ A$. In other words, the only $ R$-modules of finite projective dimension are the projective (hence free) modules.

This is a neat little result that has a fun More »

It’s time for another installment of Wild Spectral Sequences! We shall start our investigations with a classic theorem useful in many applications of homological algebra called Schanuel’s lemma, named after Stephen Hoel Schanuel who first proved it.

Consider for a ring $ R$ the category of left $ R$-modules, and let $ A$ be any $ R$-module. Schanuel’s lemma states: if $ 0\to K_1\to P_1\to A\to 0$ and $ 0\to K_2\to P_2\to A\to 0$ are exact sequences of $ R$-modules with $ P$ projective, then $ K_1\oplus P_2\cong K_2\oplus P_1$.

We shall prove this using spectral sequences. I came up with this proof while trying to remember the “usual” proof of Schanuel’s lemma and I thought that this would be a good illustration of how spectral sequences can be used to eliminate the dearth of clarity in the dangerous world of diagram chasing.

Before I start, I’d like to review a pretty cool fact I which I think of as expanding the kernel, which is pretty useful More »

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$

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Here is a fact from linear algebra: if $ V$ is a finite dimensional vector space over a field $ k$ then $ V$ is naturally isomorphic to its double dual. Of course, it is also isomorphic to its dual as well, but it is the natural isomorphism that is more interesting.

Write $ V^* = \mathrm{Hom}_k(V,k)$ for the dual space of $ V$.

This is not true in general for infinite dimensional vector spaces. Before I actually talk about the subject matter, let us describe a counterexample:if we use the field $ k = \mathbb{Q}$, and $ I$ is a countable set, then $ \mathbb{Q}^{(I)}$ is also countable (here, the notation $ \mathbb{Q}^{(I)}$ refers to the direct sum of $ I$ copies of $ \mathbb{Q}$). The dual space of any vector space has cardinality at least as great as the original space, but $ \mathrm{Hom}_k(\mathbb{Q}^{(I)},\mathbb{Q})\cong\prod_{i\in I}\mathbb{Q}$, which is uncountable.

Anyways, the point is $ V\cong V^{**}$. Why is this? Of course, one can provide a short and direct proof: define $ \Phi:V\to V^{**}$ by $ \Phi(v)(f) = f(v)$. This map is injective and since $ V$ is finite dimensional, it is surjective.

Is this really satisfying though? Sometimes in mathematics, it is not completely satisfying just to prove something. Instead, one must also find the proper context for results. The proper context should be enlightening in a variety of situations. For instance, that we know $ V\cong V^{**}$ where $ V$ is a finite-dimensional vector space isn’t all that enlightening. We have a natural isomorphism, so we ought to ask: is there an equivalence of categories hiding somewhere? Indeed there is, and it’s called Morita theory!
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Posted by Jason Polak on 29. May 2013 · Write a comment · Categories: homological-algebra · Tags: ,

Let $ R$ be any associative ring with unit and $ A$ an $ R$-module. If $ P$ is a projective module and $ A\to P\to A = 1_A$, is $ A$ necessarily projective?

Last time in Wild Spectral Squences 2, we saw how to prove the five lemma using a spectral sequence. Today, we’ll see a very simple application of spectral sequences to the concept of cohomological dimension in group cohomology.

We will define the well-known concept of cohomological dimension of a group $ G$, and then show how the dimension of $ G$ relates to the dimension of $ G/N$ and $ N$ for a normal subgroup $ N$. We do this with a spectral sequence. Although this application will appear to be very simple, it might be a good exercise for those just learning about spectral sequences.

The Category

For the sake of conreteness, let us work in the category of $ G$-modules where $ G$ is a profinite group. The $ G$-modules are the $ \mathbb{Z}G$-modules $ A$ with a continuous action of $ G$ where $ A$ is given the discrete topology, but we could be working with any group $ G$ with suitable, minor modifications, or Lie algebras, etc.

Like in all homology theories, there is the notion of cohomological dimension: a profinite group $ G$ has cohomological dimension $ n\in \mathbb{N}$ if for every $ r > n$ and every torsion $ G$-module $ A$, the group $ H^r(G,A)$ is trivial. If no such $ n$ exists, we say that $ G$ has cohomological dimension $ \infty$. The usual arithmetic rules in working with $ \infty$ apply.
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Last time on Wild Spectral Sequences, we conquered the snake lemma using a spectral sequence argument. This time, we meet a new beast: the five lemma. The objective is the usual: prove the five lemma using spectral sequences.

Recall that the five lemma states that given a diagram

fivediag

in an abelian category, if the rows are exact and $ a,b,d,e$ are isomorphisms, then so is $ c$. Actually, the hypotheses are too strong. It suffices to have $ b,d$ isomorphisms, $ a$ an epimorphism and $ e$ a monomorphism. One can deduce this via J. Leicht’s “strong four lemma” (which we might try and prove via a spectral sequence too) or just by using the regular diagram-chasing proof of the five lemma.
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