Using Zorn’s Lemma: Baer’s Criterion

September 10, 2008

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(ra) 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.

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Universal Properties II: Tensor Products

August 31, 2008

Problem: Let I=(2,x)\subset \mathbb{Z}[x]. Show that 2\otimes x\neq x\otimes 2 in I\bigotimes_{\mathbb{Z}[x]} I.

If the notation is unfamiliar to you, read up on tensor products of modules.

What we are doing here is forming a \mathbb{Z}[x]-module comprised of sums of symbols of the form f\otimes g, where f,g\in I, subject to the condition that the map \Box \otimes \Box be R-bilinear, that is, linear in each variable and satisfying r(f\otimes g)=(rf)\otimes g = f\otimes (r g) for all r\in \mathbb{Z}[x].

It is possible to show that two different symbols represent the same element of I\bigotimes_{\mathbb{Z}[x]} I. A basic example is 2\otimes 4 = 2(2\otimes 2) = 4 \otimes 2. This illustrates an important thing about tensor products: in general, it is easy to prove that elements are equal and difficult to prove that they are not equal.

As in the previous article, the solution here is to approach this computation obliquely, through the universal property of tensor products, which we will describe briefly for commutative rings.

Suppose that R is a commutative ring and M and N are R-modules. Then the tensor product M\bigotimes_R N is universal in the following sense: let T be any R-module and \phi: M\times N \to T an R-bilinear map. Then there is a unique R-module homomorphism \Phi: M\bigotimes_R N \to T such that the following diagram commutes:

\begin{array}{rclc} M\bigotimes_R N \\ \iota \uparrow & \searrow \Phi \\ M\times N & \rightarrow & T \\ & \phi \end{array}

(again, apologies for the poor diagram) Here \iota: M\times N \to M\bigotimes_R N is simply the map \iota(m,n)=m\otimes n.

This suggests a general line of attack for our problem: to construct a bilinear map from I\times I to some \mathbb{Z}[x]-module that maps (2,x) and (x,2) to different elements.

One failure would be to define \phi: I\times I \to \mathbb{Z} by \phi(f, g)=f(0)g(0). This map is \mathbb{Z}[x]-bilinear, but it doesn’t distinguish between these two elements. A better choice is \phi(f,g)=f(0)g'(0), which sends (2,x) to 2, and (x,2) to {0}.

By the universal property above, it follows that there exists a homomorphism \Phi: I\bigotimes_{\mathbb{Z}[x]} I \to \mathbb{Z} such that \Phi(2\otimes x)=2 and \Phi(x\otimes 2)=0, and it follows that 2\otimes x\neq x\otimes 2.

A mistake that I used to make when working with tensor products was to try to define homomorphisms like this directly. In a proof you might write “Define \Phi by \Phi(\sum_i r_i (f_i\otimes g_i))= \sum_i r_i f(0)g'(0).” But this is incorrect; we do not know a priori that this is a real definition, that \Phi is a well-defined map. So understanding the universal property is really fundamental when working with tensor products.

Related Problem: Show that 2\otimes 2 + x\otimes x is not a simple tensor; that is, it cannot be written in the form f\otimes g for f,g\in I.

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Universal Properties I: Free Groups

August 22, 2008

Problem: Show that the group defined by the presentation \langle a, b | a^2=b^3=e\rangle is infinite.

If you are familiar with the definition of a group presentation, this problem should seem intuitively obvious to you.  But it is remarkably hard to come up with a good proof.

A first idea might be: show that the elements ab, abab, ababab, \ldots are distinct.  But to do this we need to prove that there is no sequence of basic operations on the constraints a^2=b^3=e that will allow us to conclude, for example, that abab = abababab.  It is certainly not immediately clear how to prove this, and a rigorous proof will probably involve several painful lemmas about words from the set \{a, b\} and when they can be equal to the identity.

There is another way  to proceed, which is more abstract and seems to be more standard in algebra, and that is to appeal to the universal property of free groups.  A free group F on a set S can be defined loosely as the group of words in elements of S, where the group operation is concatenation, taking care to add inverses and enforce relations like aa^{-1}=e.

But we can also refer to free groups by the following universal property: If H is any group and \phi: S \to H is any set map, then there exists a unique group homomorphism \Phi such that the following diagram commutes:

\begin{array}{rclc} F \\ \iota \uparrow & \searrow \Phi \\ S & \rightarrow & H \\ & \phi \end{array}

(let me know if you know how to make less ugly LaTeX diagrams in WordPress) Here \iota:S \to F is simply the natural inclusion map, and by commutes we just mean that \Phi \circ \iota = \phi.

How can we use this? Well, our original group G=\langle a, b | a^2=b^3=e\rangle can (and should!) be defined as a quotient F / N, where F is the free group on S=\{a,b\} and N is the normal closure of \langle a^2, b^3\rangle. It would help if we could construct a surjective homomorphism from F/N onto some infinite group H, (you could call this a realization of the group) and indeed, the universal property allows us to do this.

One such realization is a \to \alpha = \overline{\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}}, b \to \beta = \overline{\begin{pmatrix} 0 & -1 \\ 1 & 1 \end{pmatrix}}, in PSL_2(\mathbb{C}). (this is the projective special linear group) These matrices are chosen so that their product \alpha \beta = \overline{\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}} has infinite order.

What we would like to say is that there exists a homomorphism from G to PSL_2(\mathbb{C}) that maps a and b as above, but we do not know a priori that simply declaring this map will produce a well-defined homomorphism. Here is where the universal property comes in. We can clearly define the map \phi : S \to PSL_2(\mathbb{C}) by \phi(a) = \alpha, \phi(b) = \beta, and apply the universal property to produce a map \Phi : F \to H that maps a to \alpha and b to \beta.

If we can show that N \subset ker\ \Phi, then \Phi will naturally induce a map \overline{\Phi} : F/N \to H. But this is easy to verify: \alpha^2=\beta^3=e, (remember that a^2 and b^3 generate N) so we have now produced a map \overline{\Phi}: G \to H such that \overline{\Phi}(ab)=\alpha\beta has infinite order in H. It follows that ab has infinite order and we are done.

It is important to realize that we can do exactly this sort of realization whenever we can identify a group with generators that satisfy relations consistent with the relations in N. Furthermore, this approach is generalizable to many other situations where universal properties are involved, (tensor products, for example) and helps us avoid the (sometimes messy) definitions of the objects involved.

On an intuitive level, all that has been said here is: whenever a group G is given by a presentation and a group H has generators consistent with that presentation, H is a homomorphic image (hence a quotient) of G. The use of this fact is a key proof technique for problems like the one above, in which we essentially need to prove that a certain structure is not too degenerate.

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