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Let $F$ be a perfect field, i.e. every irreducible polynomial over $F$ has distinct roots in the algebraic closure of $F$. Suppose that $K$ is an algebraic extension of $F$ with the property that every non-constant $p(X) \in F[X]$ has a root in $K$.

I want to show that $K$ is algebraically closed, i.e. that $K = \overline{F}$, the algebraic closure of $F$.

If $K$ were a normal extension of $F$, this would follow immediately: Let $\alpha \in \overline{F}$ and $m(X)\in F[X]$ its minimal polynomial. By our assumption, $m(X)$ has at least one zero in $K$. Since $K$ is a normal extension of $F$, it thus contains all other roots of $m(X)$ including $\alpha$.

I don't have an idea as to how to show that $K$ is a normal extension.

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  • $\begingroup$ Do you mean $p(X)\in K[X]$? Or "has a root in $K$"? $\endgroup$ – Gregory Grant Jan 29 '16 at 20:09
  • $\begingroup$ Sorry, that was a typo. The latter was meant. $\endgroup$ – KDuck Jan 29 '16 at 20:12
  • $\begingroup$ Is the assumption that $F$ is perfect actually necessary? Do you know of an example where the conclusion fails without this assumption? (this is slightly off topic, but I'm curious) $\endgroup$ – Wojowu Jan 29 '16 at 20:27
  • $\begingroup$ I don't know of an example to show that the assumption that $F$ is perfect is necessary, and I think it would be hard to construct. For a non-perfect field we would have to take something like $F = \mathbb{F}_p(t)$ and construct $K$ by adjoining only certain roots of irreducible polynomials in $F$. $\endgroup$ – KDuck Jan 29 '16 at 20:34
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    $\begingroup$ I think the proof of lemma 1 here should do the trick. Note that the perfect assumption on $F$ is not required. $\endgroup$ – Stahl Jan 29 '16 at 20:57
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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.

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