Using Zorn’s Lemma: Baer’s Criterion

Zorn’s Lemma: If (\mathcal{P}, \leq) is a partially ordered set with the property that every totally ordered subset (often called a chain) has an upper bound, then there exists a maximal M\in \mathcal{P}. (that is, for N\in\mathcal{P}, we have M\not\leq N)

This “lemma” is a basic tool for dealing with large collections in a systematic fashion. (if you are interested in the background behind it, see the Wikipedia articles on Zorn’s Lemma and the axiom of choice) Here I’ll demonstrate a fun application in module theory, but there are hundreds of other examples throughout algebra and set theory.

To understand this example, you should be vaguely familiar with modules and cyclic modules.

If R is a commutative ring, it turns out that there exist R-modules Q with a fascinating property: If f: A \to Q is an R-module homomorphism and A\subset B, where A and B are R-modules, then there exists an R-module homomorphism F: B\to Q extending f. Such modules Q are called injective modules.

This is a surprising property, because in general we cannot make functions “bigger”. For example, if R=\mathbb{Z}, we have a map 2\mathbb{Z} \to \mathbb{Z}/2\mathbb{Z} taking 2a\mapsto \overline{a}, but in any map from \mathbb{Z}\to \mathbb{Z}/2\mathbb{Z} we have 2\mapsto \overline{0}.

It is not at all obvious at first that injective modules exist, (other than the zero module) let alone what they look like. Nevertheless, there is one very simple example: the \mathbb{Z}-module \mathbb{Q}. Intuitively, we can imagine dividing by elements of \mathbb{Z} to extend the domain of some function, but even in this case there is some difficulty in proving that \mathbb{Q} is injective. We need, as they say, to “develop some theory”.

Theorem: (Baer’s Criterion) Suppose that Q has the property that if I is any ideal of R and f:I\to Q is an R-module homomorphism, there exists an R-module homomorphism F:R\to Q extending f. Then Q is an injective R-module.

This is a powerful theorem because it is much easier to describe the ideals of a ring than it is to describe every possible homomorphism from every possible R-module. For R=\mathbb{Z} and Q=\mathbb{Q}, we see quite quickly that any map n\mathbb{Z} \to \mathbb{Q} can be extended to \mathbb{Z}\to \mathbb{Q}, so this theorem shows that \mathbb{Q} is injective.

But how to prove Baer’s Criterion? It helps to start by considering the simplest cases.

For a basic example, suppose that Ra \subset Rb is an inclusion of cyclic modules, (that is, modules generated by a single element) and f: Ra\to Q is a homomorphism. We will show that we can extend to b. Let I=\{r\in R | rb \in Ra\}. Then the map g: I\to Q defined by g(r) = f(rb) is a homomorphism, so we can extend to G:R\to Q, by the hypothesis. Define F(rb) = G(r). Some basic manipulation of the definitions show that F is a well-defined homomorphism that extends f.

Unfortunately, modules are not, in general, cyclic. But if A\subset A + Ra' for some a'\not\in A, and we have f:A\to Q, we can perform a nearly identical construction. Let I=\{r\in R | rb \in A\}, and extend g(r)=f(rb) to G:R\to Q. Then we can define F(a + ra') = f(a) + G(r). Again, it is easy to show that F is a well-defined homomorphism that extends f. (this is actually an elementary kind of pushout)

This shows that we can always increase the domain of f by adding a new element. In particular, this proves Baer’s criterion for finitely generated modules. But how can we extend to argument to the infinitely generated case? (for example, the \mathbb{Z}-module \mathbb{Q})

This is where Zorn’s Lemma comes in. Say that A\subset B and f: A\to Q is a homomorphism. We can form the partially ordered set of extensions of f. (where (C,g)\leq (D,h) if C\subset D and h extends g) Then if \{(A_i, f_i) | i\in I\} is a totally ordered set, we can form \cup f_i : \cup A_i \to Q, so this verifies the chain condition. Hence there exists a maximal f' : A' \to Q, by Zorn’s Lemma.

But if we have some b\in B-A', we can extend f' to A' + Rb, contradicting the maximality of f'. So we must have A'=B, and we are done.

Notice that Zorn’s lemma takes the place of a wishful induction. We want to say, “We keep adding elements to the domain until we are done,” but induction alone is not powerful enough to make this rigorous.


2 Responses to Using Zorn’s Lemma: Baer’s Criterion

  1. Tim Campion says:

    Thanks for posting this; I found it very helpful. I think something funny is going on in the baby case of cyclic rings, but the essential part was easy to follow. One typo: you sometimes say a’, sometimes b for what I believe is the same module generator.

  2. lydianrain says:

    Thanks for mentioning it, I did find a couple of typos. Hopefully everything’s fixed now.

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