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In this thread on the axioms of $\mathbb Q$ it's stated that a field having characteristic zero can be written down in first-order logic. The definition in the logic lecture notes I work with (by Stephen G. Simpson) give the standard definition, which isn't a formal first order statement:

$$"(1+1+\ \dots\ +1+1)_{n\ \text{times}}\neq0"$$

Now in this Wikipedia page on periodic groups (and in the reference given there, I looked it up), it is stated that a group being periodic

$$"\forall x.\, ((x=e) \lor (x\circ x=e) \lor ((x\circ x)\circ x=e) \lor \ldots)"$$

can't be stated in first-order logic. Both would be "for all $n\in \mathbb N$" statements and it makes my question how the characteristic zero axiom can be a first-order statement and secondly if we can formally pull off $\mathbb Q$ without first formalize $\mathbb N$.

My question is: What is the first order axiom characterizing a field having characteristic zero?

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    $\begingroup$ It's an axiom scheme, not a single axiom. $\endgroup$ – Thomas Andrews Dec 27 '14 at 17:20
  • $\begingroup$ I'm not quite sure that [peano-axioms] is suitable here. $\endgroup$ – Asaf Karagila Dec 27 '14 at 17:23
  • $\begingroup$ @ThomasAndrews: Okay, then what is the crucial difference between the two cases? For group example, it says "It is not possible to get around this infinite disjunction by using an infinite set of axioms: the compactness theorem implies that no set of first-order formulae can characterize the torsion groups." $\endgroup$ – Nikolaj-K Dec 27 '14 at 17:23
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    $\begingroup$ Because you can't create an axiom scheme that means that. The key is the $\lor$ in the group axiom, which means you can't break it up into individual finite axioms. $\endgroup$ – Thomas Andrews Dec 27 '14 at 17:27
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There is no single axiom doing that. But rather a schema of axioms stating that the characteristic is not positive of any possible value.

To see that this is the case, consider the ultraproduct of all prime fields of finite order, with a free ultrafilter. The result is a characteristics $0$ field. And any statement true in that field is true in infinitely many finite fields. So there cannot be a single axiom stating that the characteristics is $0$.

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  • $\begingroup$ I only encountered ultra-filtered before in a sketch of Cohens prove. Is the ultra-filter argument here related to the proves with the (completness?) theorem, where one says that if it holds for all parts, it holds for the whole? I read about that in the reference for the periodic group wikipedia article. $\endgroup$ – Nikolaj-K Dec 27 '14 at 17:34
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    $\begingroup$ The use of the ultraproduct can be replaced by the Compactness Theorem. $\endgroup$ – André Nicolas Dec 27 '14 at 17:38
  • $\begingroup$ Right, that was it, thx. / I'm going to accept this answer - I learned different things from all 3 posted so far, but this one was here first. $\endgroup$ – Nikolaj-K Dec 27 '14 at 17:40
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What one means by declaring $$(1+1+\cdots+1+1)_{n\text{ times}}\neq0$$ to be an axiom (schema!) is that each of the formulas $$ \begin{align} 1 &\neq 0 \\ 1+1 &\neq 0 \\ 1+1+1 &\neq 0 \\ 1+1+1+1 &\neq 0 \end{align} $$ and so forth, is separtely an axiom.

As Asaf argues, the resulting theory of fields of characteristic 0 is not finitely axiomatizable, but there is nothing intrinsically wrong with having a theory with infinitely many axioms, as long as there's a definite way to determine whether a given formula is an axiom or not.

Axiom schemas are also used in the usual first-order theories for basic arithmetic (Peano Arithmetic with its induction axiom schema) or set theory (ZFC with axiom schemas for selection and replacement).

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  • $\begingroup$ But Peano is also not finitely axiomatizable, since it includes the (far more complicated) axiom scheme of induction. $\endgroup$ – Thomas Andrews Dec 27 '14 at 17:29
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    $\begingroup$ With axiom schemas and in particular infinite sets of axioms, is there any way to work around the appearent use of naive set theory in the meta-logic? Can you say what set-like features of these collections is used by logicans? $\endgroup$ – Nikolaj-K Dec 27 '14 at 17:31
  • $\begingroup$ @ThomasAndrews: Agreed. I mentioned those examples to show that there isn't anything particularly esoteric about theories that are not finitely axiomatizable. $\endgroup$ – hmakholm left over Monica Dec 27 '14 at 17:31
  • $\begingroup$ You can use naive computability theory instead, @NikolajK. Basically, you need to be able to check, "Is this an axiom?" (It is theoretically possible to have an infinite non-computable set of axioms, and that will require some naive set theory.) $\endgroup$ – Thomas Andrews Dec 27 '14 at 17:33
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    $\begingroup$ @NikolajK That's an entirely different question, and definitely harder than proper use of comments can permit. $\endgroup$ – Thomas Andrews Dec 27 '14 at 17:44
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The problem with the group theory case is that there is no natural way to break up an infinite $\lor$ into separate axioms. But the infinite axiom scheme for $\mathbb Q$ is representing an infinite $\land$, so we can break it up into separate axioms.

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  • $\begingroup$ Thanks. Could I say that in the second case, the postulates in the would-be-axiom schema would somehow have to cross-check each other? $\endgroup$ – Nikolaj-K Dec 27 '14 at 17:32
  • $\begingroup$ Not really. It's just that an infinite disjunction can't be broken up. You don't know whether $x\cdot x=e$ or $x\cdot x\cdot x=e$, etc. Any axiom would have to assert all of these possibilities at once. $\endgroup$ – Thomas Andrews Dec 27 '14 at 17:35

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