Skip to main content
6 events
when toggle format what by license comment
Apr 22, 2016 at 7:07 comment added Antoine Ducros And I disagree with ACL: $F(t)$ cannot be real closed. For instance, $t$ will never be a square in $F(t)$.
Apr 22, 2016 at 7:00 comment added Antoine Ducros My construction is perhaps more explicit, and I can make it completely explicit (neither choice nor compactness involved): embed $F(t)$ into $F((s))$ with $s=1/t$ , and set $ K=\bigcup_n F((s^{1/n}))$. Equip $K$ with the ordering extending that of $F$ for which $s$ is positive and smaller than every positive rational number; this ordering extends that of $F(t)$, and makes $K$ a real closed field. Now you may take for $S$ the algebraic closure of $F(t)$ inside $K$.
Apr 22, 2016 at 6:11 comment added ACL Is it so different from the other answer? If you add a constant $t$ to the language with the axioms that it be greater than any integer $x$, then $t$ is transcendental and any model contains $F(t)$ — the simplifying point being that $F(t)$ is already real closed.
Apr 21, 2016 at 20:49 comment added Emil Jeřábek @GeraldEdgar While I agree with the general sentiment, the compactness theorem for countable theories does not need the axiom of choice.
Apr 21, 2016 at 20:44 comment added Gerald Edgar I think it is more satisfying to write down a concrete model like this. Rather than depend on "compactness" which gives you a model out in "Axiom of Choice Land".
Apr 21, 2016 at 20:26 history answered Antoine Ducros CC BY-SA 3.0