Question: Let $E$ be an algebraically closed extension field of a field $F$. Show that the algebraic closure $\bar F_E$ of $F$ in $E$ is algebraically closed.
Firstly, I conclude a thing, if $\bar F_E$ is properly contained in $E$, then it couldn't be algebraically closed, since an algebraically closed field cannot be properly contained. Therefore my idea is to prove $\bar F_E=E$.
Then I notice that $E$ is algebraically closed, then for each polynomial $f(x)\in E[x]$, it can be split completely into linear factors:
$$f(x)=(x-\alpha_1)\cdots(x-\alpha_n)\quad\text{ for $\alpha_i\in E$, $i=1,2,\cdots,n$. }$$
But since $\bar F_E\leq E$ obviously, the remaining things to prove is $E\leq \bar F_E$, so I want to choose an element $\alpha\in E$, and I want to prove it is algebraic over $F$, but I cannot find any connection between this and the completely factorization of $f(x)$. Any suggestion?
Best Answer
As @EclipseSun mentioned in the comment section, an algebraically closed field can be properly contained in another algebraically closed field.
Define $$\overline{F}_E=\{a\in E\mid a \;\text{is algebraic over $F $}\}.$$
Clearly $\overline{F}_E$ is algebraic over $F$.
Let $\alpha$ be a root of $f(x)\in \overline{F}_E[x]$. Then $\overline{F}_E(\alpha) $ is algebraic over $\overline{F}_E$ and $\overline{F}_E$ is algebraic over $F$. Hence $\overline{F}_E(\alpha)$ is algebraic over $F$(why?). So $\alpha\in \overline{F}_E$, since $\alpha $ is algebraic over $F$. Hence $\overline{F}_E$ is algebraically closed.
We need to show that $\overline{F}_E$ is actually a field. In other words, if $\alpha$ and $\beta$ are algebraic over $F $, then $\alpha+\beta,\alpha-\beta, \alpha\cdot\beta,\frac\alpha\beta(\beta\neq0)$ are algebraic over $F$. This is easy to prove if you recall that $F(\alpha,\beta)=F(\alpha)(\beta)$ and $\alpha$ is algebraic over $F$ iff $F(\alpha)/F $ is finite. I hope you can take this from here.