Tricki

## Use finite fields

### Quick description

If you have an algebraic problem about the complex numbers, it might be possible to solve it by solving a similar, but easier, problem about finite fields.

### Prerequisites

Basic notions of commutative and linear algebra (more sophisticated ideas are required to understand the proofs of the results quoted, but this is not needed to apply the trick).

This trick depends on the following proposition.

Proposition 1 (Reduction mod primes) Suppose that is a finitely-generated subring of . Then the quotient of by a maximal ideal is always a finite field, and .

In fact for the first example below we only require the fact that has some maximal ideal such that the quotient is a finite field. The proof of this (and a fortiori of the first statement in the proposition) is not particularly easy to find in the literature. Some links are given below, and perhaps the standard reference would be Bourbaki's Algebre Commutative, Ch. V, Sec 3, no. 4, Corollaire I, p.64.

### Example 1: Ax-Grothendieck theorem

The discussion here is cribbed from this article by Serre and a blog by Tao that followed from it, together with comments on that blog.

Theorem 2 Suppose that is an injective polynomial map. Then is also surjective.

The proof is by contradiction. If is injective but not surjective then

• vanishes whenever does and

• there is some such that vanishes whenever does (i.e. never).

The rather odd way of expressing these statements is so that we may apply Hilbert's Nullstellensatz. The two statements imply, in turn, that

• some power lies in the ideal generated by and

• some power lies in the ideal generated by .

More concretely,

• There is a polynomial such that and

• There is a polynomial such that .

Now let be the subring of generated by the (finite list of) coefficients of and . Let be a maximal ideal of , and consider the reduction map . By Proposition 1 the image is a finite field . Let us abuse notation by omitting explicit mention of the reduction map in the next few lines. We still have the polynomial relations

over , which means that the polynomial , regarded as a map from to , is injective but not surjective. This is manifestly nonsense on cardinality grounds.

### Example 2: Jordan-Chevalley decomposition in algebraic subgroups of

Suppose that is an invertible matrix, where is any field (actually, it needs to be a perfect field, but the fields we mention below are all perfect). The Jordan–Chevalley Decomposition of is a factorization into commuting semisimple and unipotent parts and . Semisimple is the same thing as diagonalizable over the algebraic closure , and a matrix is unipotent if and only if , or equivalently if may be conjugated to an upper-triangular matrix with ones on the diagonal. It is a well-known theorem of linear algebra that the Jordan-Chevalley decomposition exists and is unique.

Theorem 3 Let be an algebraic (Zariski-closed) subgroup, that is to say a subgroup which is also the zero set of a collection of polynomials in variables (the entries of the matrices). Then is closed under taking components of the Jordan-Chevalley decomposition; that is, if then the semisimple and unipotent parts and also lie in .

We will deduce this from the following finite field version of the result.

Theorem 4 Let be a group, where is some finite field. Then is closed under taking components of the Jordan normal form; that is, if then the semisimple parts and also lie in .

Let us give the deduction of Theorem 3, as this is the main point of this discussion. Suppose that is as in the theorem, and let . Membership of is decided by the vanishing of some set of polynomials, but by Hilbert's basis theorem the ideal generated by these polynomials is generated by some finite collection . Choose a matrix such that is actually diagonal, and consider now the ring generated by all the entries of , , , their inverses and all the coefficients of the polynomials . This is a finitely-generated ring, and Theorem 3 takes place over . For each maximal ideal of , consider the natural reduction map . By the first part of Proposition 1 the image is a finite field. This reduction map fairly obviously preserves the Jordan-Chevalley decomposition, that is to say the Jordan-Chevalley components of are and . By Theorem 4 (applied to the subgroup of generated by ), each of and is a power of , and hence and for integers . But the polynomials vanish on the whole of the group , and in particular at the elements and . Therefore for . To conclude, we use the second part of Proposition 1, which asserted that the intersection of all maximal ideals is zero. Together with the preceding line this implies that for , which of course means that both and lie in the group .

It remains to prove Theorem 4. We leave the details as an exercise to the reader, using the following three observations: (1) If then has finite order; (2) if the order of an element is coprime to then is semisimple; (3) if the order of an element is a power of then is unipotent. Using this it is not hard to exhibit and as powers of .

### General discussion

There are connections between ideas of this kind and logic/model theory, but this author is not qualified to discuss them. More details may be found in the "Princeton Companion" article by David Marker, where explicit mention of the Ax-Grothendieck theorem is made on p642.