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*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).

## Motivation: The Rank

The most natural analogue of a vector space in the category of modules isn't any old module, which might be quite bizarre and complicated, but rather the free module; that is, a module with a basis. In this setting it makes perfect sense to try and define the rank of a free $ R$-module $ M$ as the cardinality of a chosen basis. Of course, one must show that the rank is independent of the choice of basis. However, in many cases, this is **not true**! As an extreme case, consider the endomorphism ring $ R=\mathrm{End}(V)$ where $ V$ is a vector space of countably infinite dimension. Then as $ R$ modules, there is an isomorphism $ R^n \cong R^m$ (as $ R$-modules!) for any positive integers $ n,m$ (good exercise!).

Well, that was a noncommutative example and indeed, for a commutative ring $ R$, any two bases of a free $ R$-module $ M$ do in fact have the same cardinality. In fact, a well-defined rank for free modules happens whenever there is a ring homomorphism $ R\to F$ where $ F$ is a division ring. For instance, in the commutative case, dropping to the vector space $ M\otimes_R(R/m)$ where $ m$ is a maximal ideal (axiom of choice!) shows that any two bases do in fact have the same cardinality. Rings with a well-defined rank for free modules certainly deserve a label! We will say such rings have the invariant basis property (IBP). Rings with the IBP are sometimes also called IBN (invariant basis number) rings. More formally:

Let $ R$ be a ring. We say that $ R$ has the invariant basis property if $ R^n\not\cong R^m$ for any natural numbers $ m,n$. In this case, free $ R$-modules have a well-defined rank (even for infinite direct sums, though this is not completely trivial!).

## Rank in the Projective Case

Of course, one might want to extend the concept of rank to other kinds of modules besides free modules. One way to do this is to note first that a finitely generated projective module over a local ring is free. This is a typical result that is proved in an introductory algebra course.

So if $ R$ is commutative, then $ R$ has the IBP, and so does any localization. Now, if $ P$ is finitely generated and projective and $ p$ is a prime ideal, then $ P_p$ is a finitely generated projective module because $ P_p\cong P\otimes R_p$ and is finitely generated over the local ring $ R_p$. Thus $ P_p$ is free and we get the rank at the prime $ p$. This gives us a function $ \mathrm{rank}(P):\mathrm{Spec}(R)\to \mathbb{Z}$, which is actually continuous if $ \mathbb{Z}$ is given the discrete topology. Hence if $ \mathrm{Spec}(R)$ is connected then this rank function is constant and we have a well-defined number for the rank. And even if $ \mathrm{rank}(P)$ is not constant, we still get this handy rank function that can be used for all sorts of mischief.

## Can We Drop Finitely Generated?

Well, can we, please? What about those poor projectives that are not finitely generated? Well, I admit the concept of rank might be less useful in the infinitely generated case. Nevertheless, we can still ask:

**Are projective modules over local rings free?**

The answer is **yes**, and the proof was given by Irving Kaplansky in his paper,

Irving Kaplansky, "Projective Modules". The Annals of Mathematics, Second Series, Vol. 68, No. 2 (Sep., 1958), pp. 372-377

This is in fact the topic of this entire post! I will sketch the proof, and the reader is urged to look at Kaplansy's paper for more details and other things related to projective modules.

## The Proof: The Initial Steps

Let us state the result we are trying to prove:

**Theorem 0.**A projective module over a local ring is free.

The main theorem of Kaplansky's paper is as follows:

**Theorem 1.**Suppose $ R$ is a ring and $ M$ and $ R$-module. If $ M$ can be written as a direct sum of countably generated $ R$-modules, and $ P$ is a direct summand of $ M$, then $ P$ can also be written as a direct sum of countably generated $ R$-modules.

The proof is a basic transfinite argument, which I will avoid here, although the reader is encouraged to go through it. I should say though that the hard work is not really contained in this theorem. In fact this theorem is a nice technical tool that can be used for various other purposes.

Anyways, we see from Theorem 1 that any projective module is a direct sum of countably generated projectives. Indeed, if $ P$ is projective, then $ P$ is the direct summand of a free module, and any free module is by definition a direct sum of countably (i.e. one-element) generated modules. But then by Theorem 1, $ P$ can be written as a direct sum of countably generated modules. Since any direct summand of a projective is again projective, this actually gives $ P$ as a direct sum of countably generated *projective* modules.

This reduces the problem, because to prove Theorem 0, we just need to show that any countable projective module over a local ring is free. Actually, the reduction from arbitrary cardinality to the countable case makes the problem much more tractable, and we do this by using Kaplansky's Lemma 1, which can be formulated for our purposes as follows:

**Lemma 1.**Let $ R$ be a ring and $ M$ a countably generated $ R$-module. Suppose that any direct summand $ N$ of $ M$ has the following property: for any element of $ x\in N$, there exists a free direct summand $ F$ of $ N$ such that $ x\in F$. Then $M$ is free.

*Proof*. The proof, although not given in the paper, is not difficult: suppose $ \{x_i\}$ is a countable generating set of $ M$ with each $ x_i$ nonzero. Choose for $ x_1$ a free direct summand $ F_1$ of $ M$ such that $ x_1\in F_1$ and write $ M = F_1\oplus M_1$. Project the remaining generators onto $ M_1$, discard zeros, and repeat the process. Then we get $ M \cong \oplus F_i$.

Note also that the word "free" can be replaced by "finitely generated".

Finally, we just need to prove that projective modules over local rings satisfy the hypotheses of Lemma 1, which is the content of Lemma 2 of Kaplansky. In fact this works for any projective, though Lemma 1 only works in the countable case.

## The Proof: Final Step

So let us prove this: suppose $ P$ is projective over a local ring $ R$ and let $ x\in P$. We need to prove that there exists a free summand $ S$ of $ P$ such that $ x\in S$. Write $ F = P\oplus Q$ and choose a basis $ \{u_i\}$ of $ F$ so that the number of generators required to express $ x$ is minimal, and write $ x = a_1 u_1 + \dots + a_n u_n$ where $ a_i\in R$. Because of this minimality, it is not hard to show:

**Exercise.** In the expression for $ x$, for each $ i$, $ a_i$ cannot be written as a right $ R$-linear combination of the other $ a_j$, $ j\not= i$.

Now write

Then we can write $ x = a_1 u_1 + \dots+ a_n u_n = a_1 y_1 + \dots + a_n y_n$. We claim that the $ y_1,\dots,y_n$ together with $ \{ u_i\}$ for $ i\not\in \{1,\dots,n\}$ is in fact another basis of $ F$. If we can show this then we are done, because the submodule $ S = Ry_1 + \dots + Ry_n$ would in fact be the free summand of $ F$ (and hence of $ P$!) that we are looking for.

So how can we do this? Notice that we have not yet used the local ring assumption. In fact we are going to use it now! Write:

where the $ t_i$ is an $ R$-linear combination of the remaining basis elements not in $ \{u_1,\dots,u_n\}$. So in order to justify our claim about $ y_1,\dots,y_n$ together with the other $ u_j$ being a basis, **we just have to show that the matrix $ C = (c_{ij})$ is an invertible matrix**.

By putting the sum for $ y_i$ into the equation $ a_1u_1 + \dots+ a_nu_n = a_1y_1 + \dots + a_ny_n$, we see by comparing basis coefficients that for each $ j$,

Now we claim that $ c_{ii}$ is a unit for every $ i=1,\dots,n$. Indeed, by rearranging we get $ (1- c_{ii})$ cannot be a unit because otherwise $ a_i$ would be a right-linear combination of the others, which violates the exercise, so since $ R$ is a local ring $ c_{ii}$ is a unit. Similarly $ c_{ij}$ cannot be a unit for the same reason! Now we have a matrix with units on the diagonal and nonunits off the diagonal.

Thus we have reduced the problem to showing that such a matrix over a local ring is actually invertible. One might be tempted to think this trivial and do something with determinants. But remember that our ring isn't necessarily commutative, and determinants (i.e. their generalizations) in such a setting are fearsome and tricky entities indeed.

The proof of this was not included in the paper, but here is one proof: observe that such a matrix can be written as $ D + OD$ where $ D$ is the diagonal matrix of units and $ OD$ is the matrix of off-diagonal entries. Now an elementary result from noncommutative ring theory says that $ \mathrm{rad}M_n(R) = M_n(\mathrm{rad}R)$ which shows that $ OD$ is actually in the Jacobson radical of $ M_n(R)$, and $ D$ is clearly a unit, which shows the result.

## 2 Comments

Do you know an example of a module M (over a ring R) which is not free but M^n is free for some positive integer n? Or do you know where I can search it? Thank in advance.

Bien

Bien,

There are lots of examples provided by ideals in (commutative) rings of integers of algebraic number fields – because they are Dedekind Domains. For example, take R = Z[ \sqrt {-5}] = the ring of integers in Q(\sqrt {-5}). The class number of R is 2, so take any nonprincipal ideal for your module M (for example, the ideal M = (2, 1 + \sqrt{-5}) generated by 2 and 1 + \sqrt{-5}. Since M is nonprincipal (easy to check), it is not free. But M \oplus M is isomorphic to R \oplus M^2 (where M^2 is the square of the ideal), and since M^2 is principal ( = (2), in fact), M \oplus M is free of rank 2. For further information, check any reference that describes the Fundamental Theorem for Finitely Generated Modules over Dedekind Domains.