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