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: AxGrothendieck 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: JordanChevalley 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 Jordanâ€“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 uppertriangular matrix with ones on the diagonal. It is a wellknown theorem of linear algebra that the JordanChevalley 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 finitelygenerated 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 JordanChevalley decomposition, that is to say the JordanChevalley 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 AxGrothendieck theorem is made on p642.