9
$\begingroup$

Hello,

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.

$\endgroup$

2 Answers 2

14
$\begingroup$

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.

$\endgroup$
5
  • 4
    $\begingroup$ 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, 2011 at 22:00
  • $\begingroup$ Tanmay, thanks, yes, that is a much better way to see that such $a$ exist. $\endgroup$ Apr 14, 2011 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, 2011 at 22:55
  • 2
    $\begingroup$ Francis, if by "morphism" you mean elementary embedding, then there is no such prime $M$, since your theory is not complete. $\endgroup$ Apr 15, 2011 at 0:10
  • $\begingroup$ Welcome back, Joel! $\endgroup$ Apr 15, 2011 at 2:25
8
$\begingroup$

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).

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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