### 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.

The proof of the first part of this theorem 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: Jordan Normal Form 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 decomposition of is a factorization into commuting semisimple and unipotent parts and . Semisimple is the same thing as diagonalizable, 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 normal form exists and is unique.

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

Let us give the deduction of Theorem 2, as this is 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 2 takes place *over *. For each maximal ideal of , consider the natural reduction map . This fairly obviously preserves Jordan normal form, that is to say the Jordan components of are and . By Theorem 3 (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 Theorem (ref), 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 .

### Example 2: The Ax-Grothendieck theorem

This discussion is taken from the blog article.

### General discussion

This trick, and much more, is covered in the article by Serre (ref). That article was blogged on by Tao (ref), and in the comments following that discussion one may find a proof of Principle (ref).

14 weeks 2 daysago1 year 37 weeksago2 years 9 weeksago2 years 26 weeksago2 years 26 weeksago