Does every infinite field contain a countably infinite subfield?
It's easy to see that every field $K$ contains either the rational numbers $\Bbb Q$ (when $K$ has characteristic $0$) or a finite field $\Bbb F_p$ (when $K$ has characteristic $p$). Thus, in the characteristic $0$ case, the answer is an easy "yes."
But if $K$ is infinite and has characteristic $p>0$, does the fact that $K \supset \Bbb F_p$ allow us to conclude that $K$ has a countably infinite subfield?
Best Answer
Yes. If $S$ is a countably infinite subset of $K$ then the subfield generated by $S$ is countable.
Since the question got six upvotes before any answers, maybe this is not obvious. Details:
Let $S_0=S$. Let $S_{n+1}$ consist of the elements of $S_n$ together with all the $x+y$, $x-y$, $xy$ and $x^{-1}$ for $x,y\in S_n$. Then $\bigcup_{n=0}^\infty S_n$ is a countable subfield.
(Yes, the same argument shows that $K$ has a subfield of any infinite cardinality less than or equal to the cardinality of $K$.)