Is every isomorphism of an algebraically closed field onto one of its subfields an automorphism?

Question

I've just been reading about the Isomorphism Extension Theorem, and I think I can make the following argument:

Let $F$ be an algebraically closed field, and let $\sigma$ be an isomorphism of $F$ onto a subfield of $F$. Then $\sigma^{-1}$ is an isomorphism of $\sigma[F]$ onto $F$, and since $F$ is algebraic over $\sigma[F]$, by the Isomorphism Extension Theorem, we can extend $\sigma^{-1}$ to an isomorphism $\tau$ of $F$ onto a subfield of $\overline F = F$. But $\sigma^{-1}$ is already onto $\overline F = F$, so it cannot be extended any further, and thus we must have $\sigma[F] = F$; that is, $\sigma$ must be an automorphism of $F$.

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Update: I see from the comments that my reasoning was faulty because $F$ need not be algebraic over $\sigma[F]$. However, I think that the following revision works: Given any field $F$, an extension of the identity map on $F$ to an isomorphism of $\overline F$ onto one of its subfields is an automorphism of $\overline F$.

Proof: Let $\iota:F\rightarrow F$ be the identity map on $F$, and extend $\iota$ to an isomorphism $\tau$ of $\overline F$ onto a subfield of $\overline F$. Since $F\subseteq\tau[\overline F]$, we see that $\overline F$ is algebraic over $\tau[\overline F]$, and thus by the Isomorphism Extension Theorem, the isomorphism $\tau^{-1}:\tau[\overline F]\rightarrow \overline F$ can be extended to an isomorphism of $\overline F$ onto a subfield of $\overline F$. But $\tau^{-1}$ is already onto $\overline F$, so its domain must already be all of $\overline F$; that is, $\tau[\overline F] = \overline F$.

Answer 1

If $X$ is a set and $F$ a field, let $\overline{F(X)}$ denote the algebraic closure of the field $F(X)$ of rational functions with coefficients in $F$ and variables from the set $X$.

It is very easy to see that if $X$ is an infinite set, $x_0\in X$ and $\phi:X\to X\setminus\{x_0\}$ a bijection, there is a morphism of fields $f:\overline{F(X)}\to\overline{F(X)}$ such that $f(x)=\phi(x)$ for all $x\in X$. The image of $f$ is $\overline {F(X\setminus\{x_0\})}$, so $f$ is not surjective.