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Revision of Bundle a family of objects into a single 'universal' object from Mon, 27/04/2009 - 18:29

Note iconIncomplete This article is incomplete. More examples needed.

Quick description

There are many situations in mathematics, especially in algebraic contexts, where a question about some infinite family of objects is equivalent to a question about a single 'universal' object of the same type. Often, it is easier to consider the question just for the universal object.

Prerequisites

Example 1

Note iconContributions wanted This section could use additional contributions. The Hasse principle seems a good example here, but I don't know enough number theory to do it justice. This example needs to be finished off.

Let f be a polynomial, and suppose we wish to determine if there are any integers x for which f(x)=0.

Certainly, if f(x) = 0 then f(x) \equiv 0 \mod n, for any positive integer n. In other words, the equation f(x)=0 has solutions in the ring \mathbb{Z}/n \mathbb{Z}. Consider the other direction: if we have solutions in \mathbb{Z}/n \mathbb{Z}, when do we have solutions in \mathbb{Z}? Ideally, we want to incorporate all the 'modulo n' cases into a single case. The way to do this is by forming an inverse limit, as explained below.

We have a family of rings \mathbb{Z}/n \mathbb{Z}, but this family has some structure to it: given positive integers m and n, there is a natural map from \mathbb{Z}/mn \mathbb{Z} to \mathbb{Z}/n \mathbb{Z} given by 'reduction modulo n'. Now consider sequences (x_n)_{n \in \mathbb{N}}, where x_n is an element of \mathbb{Z}/n \mathbb{Z}. Define addition and multiplication of sequences pointwise, and note that this forms a commutative ring, with identity given by the 'all 1' sequence. We say such a sequence is compatible if, for any positive integers m and n, we have x_{mn} \equiv x_n modulo n. Observe that the compatible sequences form a subring. This is the inverse limit of the rings \mathbb{Z}/n \mathbb{Z}, and is written \hat{\mathbb{Z}}.

It is clear that listing all the reductions modulo n of a given integer will define a compatible sequence, so \hat{\mathbb{Z}} contains \mathbb{Z} as a subring. Hence if want to rule out solutions in \mathbb{Z}, it suffices to rule out solutions in \hat{\mathbb{Z}}.

We could instead have only looked at solutions in the rings \mathbb{Z}/p^n \mathbb{Z}, where p is some prime. This time, the inverse limit is \mathbb{Z}_p, known as the p-adic integers, which is the ring of integers of the field \mathbb{Q}_p of p-adic numbers. Despite the appearance of the prime p here, this field actually has characteristic zero, and contains a copy of the rational numbers. It turns out that \hat{\mathbb{Z}} is isomorphic as a ring to the (unrestricted) Cartesian product of the rings \mathbb{Z}_p, as p ranges over all primes. So finding solutions in \hat{\mathbb{Z}} is equivalent to finding them separately in \mathbb{Z}_p for each prime p.

Example 2

(This is an example of a popular trick in group theory, 'try it for the free group'. This could perhaps be spun off to its own article.)

Let G be a group generated by d elements, and let H be a subgroup of index n. How many elements could it take to generate H?

At first sight, this seems an unreasonable question, as for d>1 the family of d-generated groups is rather diverse, and apparently has little structure in common. For instance, every finite simple group is generated by 2 elements. Moreover, G could be infinite; in this case, how do we even know that H is finitely generated? We don't want to be considering such a wide range of different examples.

However, we have a group that is in some sense universal for d-generator groups: this is the free group on d generators. Let's call our set of generators X = \{ x_1, \dots, x_d \}; we also need the symbols X^{-1} = \{ x^{-1}_1, \dots, x^{-1}_d \}. Then the free group F(X) on X consists of all equivalence classes of strings of symbols in X \cup X^{-1}, with the rule that we can 'cancel out' x_i and x^{-1}_i if we see them next to each other. Multiplication is given by concatenating the strings.

Now F(X) is universal in the following sense: given any function from the set X to a group G, this function extends to a homomorphism from F(X) to G. In particular, if the function takes X to a generating set for G, we obtain a surjective homomorphism \phi from F(X) to G. With a little more thought, it is clear how to reduce our original question to the special case of the free group: if H is a subgroup of G of index n, then its preimage \phi^{-1}(H) must have index n in F(X). Now if Y is a generating set for \phi^{-1}(H), then \phi(Y) generates H, and evidently |\phi(Y)| \le |Y|. So we can rephrase our question as follows:

Let G be a *free* group generated by d elements, and let H be a subgroup of index n. How many elements could it take to generate H?

The answer turns out to be surprisingly simple (although the proof is not obvious, and involves tricks beyond the scope of this article). The number of generators needed for H in this case is determined entirely by n and d, and is always precisely n(d-1) +1. This is known as the Schreier index formula. As an immediate consequence, we get the following answer to our original question:

Let G be a group generated by d elements, and let H be a subgroup of index n. Then the number of elements needed to generate H is at most n(d-1) +1. The free groups show that this bound cannot be improved.