Since this question turned out to be trivial, I'm now asking this strengthened version:

Is there a finitely axiomatized first order theory $T$ in the language of rings such that its finite models are exactly the fields $\Bbb F_p$ with $p$ prime (but no $\Bbb F_q$ with $q$ a proper prime power is a model of $T$)?

  • $\begingroup$ I don't know anything about model theory, so I could be totally off topic, but I think you can characterize $\mathbb{F}_p$ among the finite fields as the only ones who do not have any non trivial automorphisms. $\endgroup$ – Joel Cohen Jun 20 '13 at 9:08
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    $\begingroup$ @JoelCohen: But that is a second order statement. You are only allowed to quantify over variables, not over functions. $\endgroup$ – Dominik Jun 20 '13 at 9:55

EDIT: This answer is only a proof of a special case of the given question (the case when the infinite models of the given theory are exactly the infinite fields). It does not solve the general question.

Theorem. Let $\mathcal{M}$ be the class of all fields $F$ such that either $F$ is infinite or $F$ is isomorphic to $\Bbb{F}_p$ for some prime $p$. Then $\mathcal{M}$ has no finite FOL-axiomatization.

Proof. For any prime $p$ and any integer $n \geq p$, let $\sigma_{p,n}$ be the translation of the following statement of ring theory into a FOL-sentence:

If $R$ has cardinality $n$ and $p \cdot 1_R = 0_R$, then for every ring element x, x is either $0_R$, $1_R$, $2 \cdot 1_R$, $\ldots$, or $(p-1) \cdot 1_R$.

Let $\Sigma$ be the union of the first-order field axioms and the set of all FOL-sentences $\sigma_{p,n}$ for some prime number $p$ and for some integer $n \geq p$. Then the class of all models of $\Sigma$ is clearly $\mathcal{M}$.

Suppose now that the class $\mathcal{M}$ has a finite FOL-axiomatization $\{ \rho_1, \ldots, \rho_m \}$. Let $\rho$ be the conjunction of the sentences $\rho_i$ for $i = 1, \ldots, m$. Then since $\Sigma$ is also an axiomatization of $\mathcal{M}$, $\rho$ follows logically from $\Sigma$. Thus, $\rho$ follows logically from a finite subset $\Sigma_0$ of $\Sigma$ by the compactness theorem of first-order logic. Then $\Sigma_0$ is a finite axiomatization of the class $\mathcal{M}$.

Since $\Sigma_0$ is finite and there are infinitely many primes, we can choose a prime $\hat{p}$ and an integer $\hat{n} \geq \hat{p}$ such that for every prime $p \geq \hat{p}$ and for every $n \geq \hat{n}$, $\sigma_{p,n}$ is not an element of $\Sigma_0$. Then, clearly, every field $\Bbb{F}_{p^n}$ for $p \geq \hat{p}$ and $n \geq \hat{n}$ is a model of $\Sigma_0$ and must therefore be an element of $\mathcal{M}$. But this contradicts our initial assumption for $\mathcal{M}$. Therefore $\mathcal{M}$ has no finite FOL-axiomatization.


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    $\begingroup$ In order for $\mathcal M$ to be the class of models of $\Sigma$, you need to replace in the description of $\mathcal M$ the clause "$F$ is infinite" with the more restrictive "$F$ has characteristic $0$". (Your result doesn't yet solve the original question, but seems close.) $\endgroup$ – Andrés E. Caicedo Jun 21 '13 at 16:41
  • $\begingroup$ @AndresCaicedo: Thanks for your hint: I corrected my proof now. $\endgroup$ – Steve Jun 21 '13 at 17:35
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    $\begingroup$ Can this method be adapted to answer the actual question? $\endgroup$ – Chris Eagle Jun 21 '13 at 18:31
  • $\begingroup$ @ChrisEagle: Where do you see a problem? $\endgroup$ – Steve Jun 21 '13 at 18:40
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    $\begingroup$ The question isn't about your class $\mathcal{M}$. It just asks whether there's any finite theory whose finite models are the $\Bbb{F}_p$: there's no restriction on what the infinite models are allowed to look like. $\endgroup$ – Chris Eagle Jun 21 '13 at 18:41

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