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

Thanks in advance.

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.