Let $E$ be an algebraic extension of $F$. If every polynomial in $F[x]$ splits in $E$, show that $E$ is algebraically closed.

This question appear in the Gallian's contemporary abstract algebra. There are some equivalent question: here, here and here. There are also some solution: here and here. I just rephrase these answer in my own answer.


Let $K$ be an extension field of $F$. Then $[K:F]$ is finite if and only if $K=F(a_1,a_2,...,a_n)$, where $a_1,a_2,...,a_n$ are algebraic over $F$.

Proof of the Question.
Suppose that $E$ is NOT algebraically closed. Then there exists a polynomial $f(x)\in E[x]$ which is irreducible over $E$ and $\deg{f(x)}\geq 2$. Let $\alpha\notin E$ be a root of $f(x)$ in the extension field $E(\alpha)\cong E[x]/\langle f(x)\rangle$ of $E$.

Suppose that $f(x)=a_n x^n+\cdots+a_1 x+a_0\in E[x]$. Since $E$ is an algebraic extension of $F$, $a_n, ...,a_1,a_0$ all are algebraic over $F$. Then by Lemma, $F(a_n,...,a_1,a_0)$ is a finite extension of $F$. Now, $f(x)\in F(a_n,...,a_1,a_0)[x]$ and $\alpha$ is a root of $f(x)$, we have $\alpha$ is algebraic over $F(a_n,...,a_1,a_0)$ and $F(a_n,...,a_1,a_0)(\alpha)$ is an finite extension of $F(a_n,...,a_1,a_0)$ by Lemma again. Then we have a tower of fields $$F\underbrace{\leq}_{<\infty}F(a_n,...,a_1,a_0)\underbrace{\leq}_{<\infty}F(a_n,...,a_1,a_0)(\alpha).$$ It follows that $F(a_n,...,a_1,a_0)(\alpha)$ is a finite extension of $F$ and $\alpha$ has the minimal polynomial $m(x)\in F[x]$ because a finite extension must be an algebraic extension.

By the hypothesis, $m(x)$ splits in $E[x]$. Suppose that $m(x)=(x-b_m)\cdots (x-b_2)(x-b_1)\in E[x]$. We know that $\alpha$ is a root of $m(x)$, so $\alpha=b_i$ for some $i\in \{1,2,...,m\}$. But which implies that $\alpha=b_i\in E$, a contradiction.


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