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


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