MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I would like to see other examples of (ideas of) proofs and results in the same vein as the proof of the Ax-Grothendieck theorem. To explain what I mean by "in the same vein", I will quote from the corresponding wikipedia article:

This method of proof is noteworthy in that it is an example of the idea that finitistic algebraic relations in fields of characteristic $0$ translate into algebraic relations over finite fields with large characteristic.Thus, one can use the arithmetic of finite fields to prove a statement about $\mathbb{C}$ even though there is no non-trivial homomorphism from any finite field to $\mathbb{C}$.The proof thus uses model theoretic principles to prove an elementary statement about polynomials.The proof for the general case uses a similar method.

Edit(after the comment by Angelo to Martin's answer below): answers along the line of "using finite fields for problems concerning infinite fields" are also welcome.

share|cite|improve this question

There is a fantastic paper by Serre: How to use finite fields for problems concerning infinite fields, arXiv.

share|cite|improve this answer
Serre's paper is great, but does not use the model-theoretic approach that is described in the Wikipedia article. – Angelo Feb 11 '12 at 12:50

Mori's bend and break technique which is used to show the existence of rational curves in higher dimensional varieties, is a famous example of this.

share|cite|improve this answer

Since in your edit you say that you don't insist on the use of model theory:

A very nice and famous example is Deligne-Illusie's proof of the degeneration of the Hodge to deRahm spectral sequence. Deligne, P.; Illusie, L. Relèvements modulo p^2 et décomposition du complexe de de Rham. Invent. Math. 89 (1987), no. 2, 247–270.

share|cite|improve this answer

A not so famous example is that Mazorchuk and I used model theory to prove that if a finite semigroup S has a faithful representation of degree d over $\mathbb C$ then it must have a faithful representation of degree d over some finite field. The proof is that there is a first order sentence saying that S has a faithful representation of degree d. Thus it follows by model theory that if S has a faithful representation of degree d over $\mathbb C$ then S has a faithful representation over an algebraically closed field of characteristic p. But the image of S under the rep will live in matrices over a finite field.

This allowed us to use results about finite matrix monoids to obtain lower bounds for the minimum faithful degree of a complex representation over $\mathbb C$.

share|cite|improve this answer

See the papers:"Transfer Principle for the Fundamental Lemma" by Loeser, Cluckers and Hales (with an application to the fundamental lemma), "Constructible exponential functions, motivic Fourier transform and transfer principle" by Loeser and Cluckers and "Transfer principles for integrability and boundedness conditions for motivic exponential functions" by R. Cluckers, J. Gordon, I. Halupczok which all use and devleop transfer principle from finite characteristic to zero characteristic and vice versa.

share|cite|improve this answer
What about telling us the title, author (and, if necessary, a short description) of these papers, rather than just links? It is better posting style to have the paper title as the link text, as humans don't generally care about raw URLs. – Jacques Carette Feb 11 '12 at 14:36
Sorry I don't know how to make a link. – user16974 Feb 11 '12 at 17:09
@Ali: [foo](url) <a href="url">foo</a> where url is the fully qualified url starting with "http://" – Dima Sustretov Feb 13 '12 at 17:48
@Dima: Thank you! – user16974 Feb 14 '12 at 9:04

I saw a talk two weeks ago by Michel Brion that uses reduction to finite fields to prove that every algebraic semigroup (not necessarily affine) has an idempotent. Over a finite field it is clear because finite semigroups have idempotents.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.