## Algebraically closed and Primitive element theorem

$\mathbf{Problem:}$ Let $F$ be a perfect field and $F\subset E$ an algebraic field extension, such that every non-constant polynomial $f(x)\in F[x]$ has a root in $E$. Show that $E$ is algebraically closed.

$\mathbf{Proof:}$ Suppose $f(x)\in F[x]$ is a non-constant polynomial. Let $K$ be a splitting field of $f$.
Since $F$ is perfect, $K$ is a separable extension of $F$. Primitive element theorem implies that any finite separable extension is simple extension. That is $\exists\,\alpha\in K$ such that $K\in F(\alpha)$. Suppose $\alpha$ has a minimal polynomial $g(x)$ over $F$. By assumption, $g(x)$ has a root $\gamma\in E$. Since there exists an isomorphism $\eta:F(\alpha)/ F\to F(\gamma)/ F\subset E$ by $\eta(\alpha)=\gamma$, then $E$ contains a splitting field of $f(x)$ over $F$.
$\forall\, f\in E[x]$, adjoin a root $\xi$ of $f(x)$, we get $E(\xi)/E$. Since $E(\xi)/E$ and $E/F$ are algebraic. $E(\xi)/F$ is also algebraic. Hence, $\xi$ is a root of some non-constant polynomial $h(x)\in F[x]$. By the previous proof, $h(x)$ splits over $E$. We conclude that $\xi\in E$ and $E(\xi)=E$.
$\text{Q.E.D}\hfill \square$

$\mathbf{Remark:}$