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).