# 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.

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?

• Are you assuming $E$ is an algebraic extension of $F$? – Eclipse Sun Jan 21 '19 at 0:51
• @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? – kelvin hong 方 Jan 21 '19 at 0:58
• 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. – Eclipse Sun Jan 21 '19 at 1:03
• 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! – kelvin hong 方 Jan 21 '19 at 1:06
• $\mathbb{C}$ is contained in $\mathbb{C}(x)$, the field of rational functions on $\mathbb{C}$. – Eclipse Sun Jan 21 '19 at 1:13

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.

• 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. – kelvin hong 方 Jan 21 '19 at 7:00