Here are nine examples of projective modules that are not free, some of which are finitely generated.

Direct Products

Consider the ring $R= \Z/2\times\Z/2$ and the submodule $\Z/2\times \{0\}$. It is by construction a direct summand of $R$ but certainly not free. And it's finitely generated! Another example is the submodule $\Z/2\subset \Z/6$, though this is the same kind of thing because $\Z/6\cong\Z/2\times\Z/3$. This was the first example I ever saw of a nonfree projective module.

Infinite Direct Products

One can modify the above construction for infinite direct products of rings, too. For instance, $R = \prod_{i=1}^\infty \Z$ contains $\Z$ as a direct summand. Hence $\oplus_{i=1}^\infty\Z$ is a projective $R$ module, yet cannot be free since nonzero free modules are uncountable.

Ideals in Dedekind Domains

In a Dedekind domain $R$, take an ideal representing a nontrivial element in the class group. It will then be projective. As an example, the class number of $\Z[\sqrt{5}]$ is two, and the ideal $(2,1+\sqrt{5})$ represents the nontrivial element in the class group. It is not free since it is not principal, and it is finitely generated projective since it is invertible.

More generally, for any ring extension of commutative rings $R\subseteq S$, one may define invertible $R$-submodules of $S$ as it is done for Dedekind domains. Then any invertible $R$-submodule of $S$ will be finitely-generated and projective. For more details and a further example, see Lam's 'Lectures on Modules and Rings', Sections 2B-2C.

Rings of Continuous Functions

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Posted by Jason Polak on 19. December 2014 · Write a comment · Categories: homological-algebra, modules · Tags:

There are many ways to define the propery of semisimple for a ring $R$. My favourite is the "left global dimension zero approach": a ring $R$ is left semisimple if every left $R$-module is projective, which is just the same thing as saying that every left $R$-module is injective. In particular, ideals are direct summands, and an easy application of Zorn's lemma shows that $R$ can be written as a direct product of minimal left ideals, which is actually a finite sum because $R$ contains $1$.

An attack of Schur's lemma yields the famous Wedderburn-Artin theorem: a ring $R$ is semisimple if and only if it is the finite direct product of matrix rings over division rings.

Since $R$ can be written as a finite direct product of minimal left ideals, we see that $R$ must be Noetherian and Artinian. Is the converse true?

Of course not! Here is a minimal counterexample: $\Z/4$. This ring cannot be semisimple. Indeed if it were, by the Wedderburn-Artin theorem, it would be a direct product of fields since it is commutative. It is not a field so it is not $\F_4$, and the only other possibility is $\Z/2\times\Z/2$, which it is also not isomorphic to since $\Z/4$ is cyclic.

We don't have to appeal to the Wedderburn-Artin theorem however: the reduction map $\Z/4\to\Z/2$ makes $\Z/2$ into a $\Z/4$-module. If $\Z/4$ were semisimple, then $\Z/2$ would be a projective $\Z/4$-module, and hence at the very least as abelian groups, $\Z/2$ would be a direct summand of $\Z/4$, which is also nonsensical.

Can you think of a noncommutative example?

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?