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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?

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    $\begingroup$ Are you assuming $E$ is an algebraic extension of $F$? $\endgroup$ Jan 21, 2019 at 0:51
  • $\begingroup$ @EclipseSun I think no, it is the assumption of the question to make me think to that direction of proving. Is it $\bar F_E$ cannot properly contained in $E$ a wrong idea? $\endgroup$ Jan 21, 2019 at 0:58
  • $\begingroup$ An algebraically closed field, for example $\mathbb{C}$ can be properly contained in another algebraically closed field, if you allow that extension to be transcendental. $\endgroup$ Jan 21, 2019 at 1:03
  • $\begingroup$ But isn't that all transcendental numbers are already in $\mathbb R$ hence in $\mathbb C$? I never heard before $\mathbb C$ can be properly contained in other field. Wow! $\endgroup$ Jan 21, 2019 at 1:06
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    $\begingroup$ $\mathbb{C}$ is contained in $\mathbb{C}(x)$, the field of rational functions on $\mathbb{C}$. $\endgroup$ Jan 21, 2019 at 1:13

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Let $F$ be a field. An extension $K/F$ is called an algebraic closure of $F$ if $K$ is algebraically closed and $K/F$ is an algebraic extension.

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.

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    $\begingroup$ Thank you, the only thing I missed is the part $\bar F_E(\alpha)$ is algebraic over $F$ implies $\alpha\in\bar F_E$. After filling up this part the solution is completed. $\endgroup$ Jan 21, 2019 at 7:00

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