MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).

## Making the fixed field algebraically closed

K is a field and f is an automorphism of K. We're also given that the order of f, as an element of the group Aut(K), is not finite. Is it always possible to find a field L which contains K and an automorphism of L, say g, such that g = f over K and the fixed field of g is algebraically closed in L?

-

I think if you take $K=\overline{\mathbb{Q}}$, and $f$ almost any element of $\mathrm{Gal}(K/\mathbb{Q})$ (if $f^2 \neq \mathrm{Id}$, $f$ cannot be of finite order), you get a counter-example.

-

It is in fact never possible if the fixed field $K^f$ is not algebraically closed in $K$, since in that case, $K \cap L^g = K^f$, and $K$ (hence $L$) contains a nontrivial algebraic extension.

If you were to add the assumption that $K^f$ is algebraically closed in $K$, then the answer is yes. We can take $L=K$.

-