My question is about the non-standard models of the integers. If we add to the Peano's axioms $P$ of arithmetic the following axioms for a fixed constant $c$: $c \neq 0$, $c \neq 1$, $c \neq 1+1$, $c \neq 1+1+1$, etc... and $c=ab \implies a=1~~ou~~b=1$. We obtain a system of axioms $S$.

$S$ is consistent, by compacity. If $S$ is not consistent, a finite number of axioms in $S$ (a subset $S'$ of $S$) are not consistent, say axioms of $P$ and $c \neq 0, c \neq 1, c \neq 1+1, ...,c \neq \underbrace{1+1+1+...+1+1}_{k ~~\times}$. So we can consider a prime $p$ greater than $k$. We consider the standard model of $\mathbb{N}$ and we put $c=p$, to obtain a model of $S'$. Hence, $S'$ is consistent. Contradiction. So, $S$ is consistent.

$c$ is prime in a model $M$ of $S$, and $c$ is a non-standard integer. We can consider the field $F=\{x \in M | x< c\}$ obtained by setting $a~+_F~ b=(a~+_M~b)\mod c$, $a~\times_F~ b=(a~\times_M~b)\mod c$. We have an inverse for $a$ if $a\mod c \neq 0$.

$F$ is an infinite field. Which field is it isomorphic to ? Is $F$ algebraic over $\mathbb{Q}$ ($\mathbb{Q}$ is included in the field $F$ ) ?

Thanks in advance.


First, you haven't actually specified a particular field, since the field $F$ that you have will depend on your choice of $c$ and of $M$. For example, different nonstandard models can seriously affect even the cardinality of the field $F$ that you produce, so they are not all the same. (A Lowenheim-Skolem argument shows that $F$ can be found as you describe of any desired infinite cardinality.)

But to answer your question, none of these fields is algebraic over $\mathbb{Q}$. To see this, let $a$ be any nonstandard integer in $M$ whose finite powers are bounded below $c$ in $M$ (see below). It follows that any polynomial over $\mathbb{N}$ evaluated at $a$ is still less than $c$ in $M$. So the $\mod c$ part of the field operations of $F$ never arise when evaluating a polynomial over $\mathbb{N}$ at $a$. Thus, the problem reduces to showing that if $p$ is a nontrivial polynomial over $\mathbb{Z}$, then $p(a)\neq 0$ for nonstandard $a$ in $M$, and this follows because the basic eventually-unbounded asymptotic behavior of such polynomials is provable in your theory. Thus, $a$ is transcendental over $\mathbb{Q}$ in your field $F$.

Edit. Finally, here is a quick-and-dirty way to see that there is such a nonstandard element $a$, whose finite powers are bounded by $c$ in $M$. Let $a$ be the nearest nonstandard integer to $c^{1/N}$, where $N=\sqrt{\log c}$ as interpreted discretely in $M$. Since $N$ is nonstandard, it follows that the finite powers of $a$ are below $c$, and since $\log a\equiv\frac 1N\log c$, it follows by the choice of $N$ that $a$ is nonstandard. But I expect that there is an easier method.

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    $\begingroup$ Joel, wouldn't overspill work? For every $n$ in $\mathbb{N}$, $n^{n} < c$, so by overspill, there exists a nonstandard $a$ such that $a^{a} < c$, from which $a^{m} < c$ for all $m$ in $\mathbb{N}$. $\endgroup$ – tci Apr 14 '11 at 22:00
  • $\begingroup$ Tanmay, thanks, yes, that is a much better way to see that such $a$ exist. $\endgroup$ – Joel David Hamkins Apr 14 '11 at 22:53
  • $\begingroup$ Thanks, Tanmay. Is there a canonical model $M$ of $S$: for every model $M^′$, there is a morphism $M \rightarrow M^′ $? If this model $M$ exists, which field is $F$ isomorphic to ? $\endgroup$ – user12806 Apr 14 '11 at 22:55
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    $\begingroup$ Francis, if by "morphism" you mean elementary embedding, then there is no such prime $M$, since your theory is not complete. $\endgroup$ – Joel David Hamkins Apr 15 '11 at 0:10
  • $\begingroup$ Welcome back, Joel! $\endgroup$ – Kevin O'Bryant Apr 15 '11 at 2:25

The residue fields $F=M/cM$ of nonstandard primes in models of Peano arithmetic have interesting properties which were investigated by Macintyre. In particular, every such field is pseudofinite (i.e., an infinite model of the first-order theory of finite fields, in other words: a pseudo-algebraically closed field having exactly one extension of each finite degree).

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