Algebraic extension of perfect field in which every polynomial has a root is algebraically closed

Solution 1:

The trick is to use the primitive element theorem. For any $f\in F[x]$, let $L$ be a splitting field of $f$. Then $L$ is finite over $F$ and separable (since $F$ is perfect), so it has a primitive element $\beta$. We then see that $f$ splits over any extension of $F$ that has a root of the minimal polynomial of $\beta$. In particular, $f$ splits over $K$, as desired.

Solution 2:

In case you didn't know, the condition that $F$ is perfect is unnecessary: if $F$ is an arbitrary field and $K/F$ is an algebraic extension such that each nonconstant polynomial in $F[x]$ has a root in $K$ (or, equivalently, each irreducible polynomial in $F[x]$ has a root in $K$) then $K$ is algebraically closed. This is a theorem of Robert Gilmer. See here or Theorem 2 here.