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 read on Ax's article that the elementary theory of finite fields is decidable if one assumes the continuum hypothesis to be true. What about if one assumes the hypothesis to be false?

share|cite|improve this question
What “Ax's article” are you referring to? – Harald Hanche-Olsen Dec 8 '09 at 18:22
Ax, James The elementary theory of finite fields. Ann. of Math. (2) 88 1968 239–271. – amateur algebraist Dec 8 '09 at 20:05

I'm not sure why you say this. I am right now looking at the following paper of Ax:

MR0229613 (37 #5187) Ax, James The elementary theory of finite fields. Ann. of Math. (2) 88 1968 239--271.

The first sentence is: "In this paper, we prove the decidability of the theory of finite fields and the theory of p-adic fields." He goes on in the introduction to explain exactly what this means: given a statement E in the language of rings, he gives an algorithm for determining the set of prime powers q such that E holds in the finite field of order q. He doesn't say anything about dependence on CH, and I don't see how this could possibly be the case ("decidability given CH" is a far cry from decidability!).

The only mention of CH in the introduction to the paper is in Theorem B, which states that a certain isomorphism of ultraproducts holds given CH. This makes more sense, since cardinality questions come into play in the isomorphism of elementarily equivalent objects.

share|cite|improve this answer
I was wondering whether there is a ring isomopphism $\prod_p\mathbb{F}_p/\bigoplus \mathbb{F}_p\approx \prod_p \mathbb{F}_{p^p}/\bigoplus \mathbb{F}_{p^p}$ if we don't assume continuum hypothesis. Probably this theorem or is not mandatory to prove the decidability result even if CH is assumed. But it would be nice to know why this does/doesn't hold if CH is not assumed. – amateur algebraist Dec 8 '09 at 20:05
Ax makes clear that CH is not needed to prove the decidability result: see the bottom of page 253. I am not a model theorist, but my take on the isomorphism is that it does require CH at least in the sense that the given proof doesn't go through without it. CH is used to get that ultraproducts in question are saturated models: c.f. Chang and Keisler's Model Theory, Corollary 6.1.2. – Pete L. Clark Dec 8 '09 at 20:39

Let me explain how Ax eliminates his use of CH in the proof of decidability.

The point is that if a given proof of decidability of any first order theory in a countable language uses CH, then it may be uniformly omitted. Thus, the decidability of a theory like this can never depend on CH or on the Axiom of Choice.

Theorem. The following are equivalent, for any first order theory T in a countable language.

  1. ZFC+CH proves that T is decidable.

  2. ZFC proves that T is decidable.

  3. ZF proves that T is decidable

In particular, if a given proof of decidability uses CH+AC, then these uses can be omitted.

Proof. Clearly, each statement implies the previous, since the theories are progressively weaker. Suppose now that we have a proof that t is decidable, but the proof uses ZFC+CH. The assertion that T is decidable is a statement of arithmetic, since it says "there is a computer program, which accepts all and only the theorems of T and rejects all other strings." This statement has complexity Sigma^0_3(T), which is to say, that it is fairly low in the arithmetic hierarchy (relative to the set of Goedel codes of T).

Let us now establish that T is decidable by arguing merely in ZF. Assume that ZF holds in the set-theoretic universe V. Goedel proved that ZFC+CH holds in the constructible universe L. This argument relativizes to show that ZFC+CH also holds in L[T], the constructible universe relative to T. (Note, the theory T of the question above is actually in L, so in this case, L[T]=L.) Thus, T is decidable in L[T]. Since this assertion is arithmetic, it has the same truth value in L[T] as in V, since these two set-theoretic models have the same arithmetic. So T is decidable. QED

One can push the argument much lower than ZF, down to something like second order number theory. Of course, what we would really like is to prove the decidability of T in PA, but I don't think that this is always possible. (Although PA proves all true existential statements, the complexity of the assertion that T is decidable rises above this.)

The more general fact is that arithmetic statements can never depend on AC or CH, because if one proves them in ZFC+CH, then they would be true in L, and hence true in V, where one might have only ZF. In fact, the same argument shows via the Shoenfield Absoluteness theorem that a Sigma^1_2 statement is provable in ZFC+CH if and only it is provable in ZF.

An instance of this would be: Fermat's Last Theorem is provable in ZFC+CH if and only if it is provable in ZF. In particular, any use of the Axiom of Choice in Wiles' proof can be uniformly eliminated.

Edit: I changed the presentation to be clearer.

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.