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

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.

• Do you mean $p(X)\in K[X]$? Or "has a root in $K$"? Jan 29, 2016 at 20:09
• Sorry, that was a typo. The latter was meant. Jan 29, 2016 at 20:12
• 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) Jan 29, 2016 at 20:27
• 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 $F$. Jan 29, 2016 at 20:34
• I think the proof of lemma 1 here should do the trick. Note that the perfect assumption on $F$ is not required. Jan 29, 2016 at 20:57

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.
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.