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)


Since $a_i\in\tau(F)$, select $b_i\in F$ such that $\tau(b_i)=a_i$. Let $v\in F=\bar F$ be a solution of $b_0+b_1v+\cdots+b_nv^n=0$. Then $\tau(v)$ is a solution of $a_0+a_1X+\cdots+a_nX^n=0$ hewnce $X-\tau(v)$ is a linear factor of $a_0+a_1X+\cdots+a_nX^n$ in $\tau(F)[X]$. Thus if $a_0+a_1X+\cdots+a_nX^n$ was the minimal (hence irreducible) polynomial of $u$ we conclude ($n=1$ and) $u=\tau(v)$.

  • $\begingroup$ Thank you for the answer. Since $b_0+b_1x+\dots+b_nx^n$ has $n$ solutions, how do you know that $\tau(v)=u$ and not the other roots? I am guessing that I need to extend this and show that there are $n$ roots before and after $\tau$, so the bijection ensures a preimage. Am I right? $\endgroup$ – Yong Hao Ng Nov 14 '12 at 2:48
  • $\begingroup$ @YongHaoNg It is enough for $a_0+\cdots + a_nX^n$ to have any root in $\tau(F)$ because we may assume the polynomial irreducible. I'll edit accordingly. $\endgroup$ – Hagen von Eitzen Nov 14 '12 at 7:34
  • $\begingroup$ This is a very clear explanation, thank you! $\endgroup$ – Yong Hao Ng Nov 14 '12 at 7:52

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