Examples: Projective Modules that are Not Free

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.

Noetherian, Artinian, but not Semisimple

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?

Solution: Kaplansky’s Commutative Rings 4.3.2

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 $\chi(R/I) = 0$.

The Nonzero K-Theory of Finite Rings is Finite

Posted by Jason Polak on 03. May 2014 · Write a comment · Categories: modules · Tags:

Let $R$ be a finite ring. The example we'll have in mind at the end is the ring of $2\times 2$ matrices over a finite field, and subrings. A. Kuku proved that $K_i(R)$ for $i\geq 1$ are finite abelian groups. Here, $K_i(R)$ denotes Quillen's $i$th $K$-group of the ring $R$. In this post we will look at an example, slightly less simple than $K_1$ of finite fields, showing that these groups can be arbitrarily large. Before we do this, let us briefly go over why this is true

But even before this, can you think of an example showing why this is false for $i=0$?
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A Case of No Positive Finite Projective Dimension

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 »

Wild Spectral Sequences Ep. 4: Schanuel's Lemma

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 »

Wild Spectral Sequences Ep. 2: Five, Isomorphism!

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

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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Projectives and the Devious Determinant

The determinant is certainly a fascinating beast. But what is the determinant? Is it a really just a number or a function on matrices? In this post I hope to convince you that the answer is 'no'. In fact, we will see that the determinant, suitably modified, can be used to classify certain types of projective modules over nice rings.

Determinants of Matrices

Let $R$ be a commutative ring and $n$ be a natural number. Just as in the case of vector spaces, an $R$-module map $f:R^n\to R^n$ can be given by an $n\times n$ matrix with coefficients in $R$. Moreover, we can compute the determinant of this matrix just as in linear algebra. In fact, various notions of "determinants" also exist when $R$ is not commutative, but we will stick with the commutative case.
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Projective Modules over Local Rings are Free

Posted by Jason Polak on 06. May 2012 · 2 comments · Categories: commutative-algebra, modules · Tags: , , , ,

Conventions and definitions: Rings are unital and not necessarily commutative. Modules over rings are left modules. A local ring is a ring in which the set of nonunits form an ideal. A module is called projective if it is a direct summand of a free module.

Today I shall share with you the wonderful result that any projective module over a local ring is free. We shall follow Kaplansky (reference given below), who first proved this result.

Now modules are in fact my favourite mathematical objects. They are like vector spaces, except that they are interesting. Of course, this "interesting" can be irksome if one has to solve a problem and these interesting properties throw a wrench in the works. However, by themselves modules are certainly curious creatures worthy of intense and gruelling analysis!

Of course, when the idea of a module was first conceived, mathematicians attempted to port all kinds of ideas from vector spaces into the world of modules. Some, like the direct sum construction, worked flawlessly. Other concepts such as rank, fortunately or unfortunately depending on your perspective, did not turn out so well (think about it: if everything worked well with modules then there'd be much less interesting math).
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Infinite Integer Product Not Free

Posted by Jason Polak on 01. February 2012 · Write a comment · Categories: commutative-algebra, modules · Tags: ,

Introduction

Assuming the axiom of choice, any vector space possesses the pleasant but prosaic property* that it is determined up to isomorphism by the cardinality of its basis.

For instance, consider $\prod_\omega \mathbb{Z}/2$ and $\oplus_{2^\omega} \mathbb{Z}/2$. Both are vector spaces over the finite field $\mathbb{Z}/2$ so to show that they are isomorphic, we need to show that their respective bases have the same cardinality. The vector space on the right is written as a direct summand and so we can see that its basis must have size $\mathfrak{c}$. On the other hand, the vector space $\prod_\omega \mathbb{Z}/2$ has cardinality $\mathfrak{c}$ over a finite field, so its basis must have the same cardinality as the space itself. Aren't vector spaces a walk in the park; a piece of cake; easy as pie (ok, enough metaphors?!)?

From $\mathbb{Z}/2$ to $\mathbb{Z}$ Modules

But what if we sent the above proof to a publisher who didn't yet have the "2" character or the "/" installed on her printing press? Then all hell would break loose because $\prod_\omega \mathbb{Z}$ and $\oplus_{2^\omega} \mathbb{Z}$ aren't vector spaces any more, and the previous paragraph would be rife with errors. But they certainly are abelian groups, and they have a bit more spice than those vector spaces. So are they isomorphic? They do have the same cardinality. Fortunately for us, Baer ("Abelian Groups without Elements of Finite Order", Duke Math J. 3 (1937), pp. 88-122) answered this question in the negative (fortunately, because otherwise abelian groups would be less exciting). In fact, this question is particularly interesting to me because I had wondered about it a few months ago, and now I have the answer, thanks to Faith's book for the references.
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