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.






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.
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.







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







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.