Let be a perfect field and an algebraic field extension, such that every non-constant polynomial has a root in . Show that is algebraically closed.

Suppose is a non-constant polynomial. Let be a splitting field of .

Since is perfect, is a separable extension of . Primitive element theorem implies that any finite separable extension is simple extension. That is such that . Suppose has a minimal polynomial over . By assumption, has a root . Since there exists an isomorphism by , then contains a splitting field of over .

, adjoin a root of , we get . Since and are algebraic. is also algebraic. Hence, is a root of some non-constant polynomial . By the previous proof, splits over . We conclude that and .

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