This is a homework problem and I have some questions regarding it (not looking for solution):
Let $\tau$ be an endomorphism of a field $F$ that is algebraically closed.
i.e. $\tau: F\rightarrow F$
If $F$ algebraic over $\tau(F)$, show that $\tau$ is surjective.
Initially, I contemplated a direct approach of assuming $u\in F$, such that $u$ satisfies
$a_0+a_1u+\dots+a_{r-1}u^{r-1}+u^r=0$, where $a_i\in\tau(F)$.
But I was unable to show that $u\in\tau(F)$ in the end.
Is it possible to solve the problem this way?
My current proof:
Let $K$ be the fixed field under $\tau$.
Let $u_1\in F\setminus K$.
Since $F$ is algebraic over $\tau(F)$ and $K\subseteq\tau(F)$, we have $F$ algebraic over $K$.
Then there exists an irreducible polynomial $f$ of $u_1$ over $K$.
Let deg $f=n$. Since $F$ is algebraically closed, the roots are $\lbrace u_1,\dots,u_n\rbrace\in F$
Hence the splitting field of $f=K(u_1,\dots,u_n)$.
Under $\tau$, we have $\tau(f)=f$, whence the roots are still $\lbrace u_1,\dots,u_n\rbrace$.
This shows that $\tau(K(u_1,\dots,u_n))=K(u_1,\dots,u_n)$.
i.e the splitting field of $f$ is isomorphic under $\tau$.
This may happen only if $\tau(\lbrace u_1,\dots,u_n\rbrace)=\lbrace u_1,\dots,u_n\rbrace$, a bijection.
Hence $u_1\in\tau(K)$, i.e. $\tau$ is surjective.
Is the proof correct?
I am also wondering if the problem is intended to illustrate some properties of an algebraically closed field that I think is not captured by my approach.
My only observation is that since a field homomorphism is necessarily injective, this problem shows that any endomorphism of an algebraically closed field is an automorphism.
I appreciate if someone can give me a hint on how to solve the problem in a way that gives more insight to the structure of an algebraically closed field. (If this makes sense)