Can "Taking algebraic closure" be made into a functor?

Solution 1:

No, this is not possible. For instance, let $K$ be any field with a automorphism $f:K\to K$ whose order is finite and greater than $2$. Then $A(f):A(K)\to A(K)$ would be an automorphism of the same order extending $f$. But no such automorphism exists: by the Artin-Schreier theorem, any finite-order automorphism of an algebraically closed field has order at most $2$.

Or without using any big theorems, you can find problems just looking at finite extensions. For instance, if $f$ is the Frobenius automorphism of $\mathbb{F}_{p^2}$ then $F(f)$ is an extension to an algebraic closure which still has order $2$. Since $\mathbb{F}_{p^4}$ is normal over $\mathbb{F}_{p}$, $F(f)$ restricts to an automorphism of $\mathbb{F}_{p^4}$, which must be the Frobenius squared in order to have order $2$. But the Frobenius squared does not restrict to $f$ on $\mathbb{F}_{p^2}$, so this is a contradiction.