Vtyjkyuk

Suppose $E/F$ is an algebraic extension, where every polynomial over $F$ has a root in $E$. It's not clear to me why $E$ is actually algebraically closed.

I attempted the following, but I don't think it's correct:

I let $f$ be an irreducible polynomial in $E[X]$. I let $alpha$ be a root in some extension, so $f=m_alpha,E$. Since $alpha$ is algebraic over $E$, it is also algebraic over $F$, let $m_alpha,F$ be it's minimal polynomial. I now let $K$ be a splitting field of $m_alpha,F$, which is a finite extension since each root has finite degree over $F$.

If $m_alpha,F$ is separable, then $K/F$ is also separable, so as a finite, separable extension, we can write $K=F(beta)$ for some primitive element $beta$. By assumption, $m_alpha,F$ has a root in $E$, call it $r$. Then we can embed $F(beta)$ into $r$ by mapping $beta$ to $r$. It follows that $m_alpha,F$ splits in $E$. Since $fmid m_alpha,F$, we must also have the $f$ is split in $E$.

But what happens if $m_alpha,F$ is not separable? In such case, $F$ must have characteristic $p$. I know we can express $m_alpha,F=g(X^p^k)$ for some irreducible, separable polynomial $g(X)in F[X]$. But I'm not sure what follows after that.

NB: I say $E$ is algebraically closed if every nonconstant polynomial in $E[X]$ has a root in $E$.

This is a recent qual question at my school. :)

If $F$ is not perfect, say it has characteristic $p$. Let $F_perf$ be the extension of $F$ obtained by adjoining a root for $x^p^n-a$ for every $n in mathbb N, a in F$. Since these polynomials are purely inseparable, $F_perf$ embeds (via a unique isomorphism) into $E$, and we may identify $F_perf$ with its isomorphic copy in $E$.

Two facts for you to verify: $F_perf$ is a perfect field, and every element of $F_perf$ is a root of $x^p^n-a$ for some $n in mathbb N$ and $a in F$.

Now let $f$ be a polynomial with coefficients in $F_perf$, say $f(x) = x^k + a_k-1 x^k-1 + cdots + a_0$. Then for sufficiently large $n$, $a_i^p^n in F$ for all $i$. So $$f^p^n(x) = x^k p^n + a_k-1^p^n x^(k-1)p^n + cdots + a_0^p^n in F[x],$$ which has a root in $E$ by assumption. But $f$ and $f^p^n$ have the same roots, hence $f$ has a root in $E$. So without loss of generality, we may assume $F = F_perf$, and Bruno's answer gets us the rest of the way.

Note: feel free to ignore the warzone in the comments; it's not really relevant anymore.

If $F$ is perfect, we can proceed like this. Let $f$ be a polynomial with coefficients in $F$. Let $K/F$ be a splitting field for $f$. Then $K=F(alpha)$ for some $alpha in K$. Let $g$ be the minimal polynomial of $alpha$ over $F$. Then $g$ has a root in $E$ by assumption, hence $E$ contains a copy of $F(alpha)$, i.e. a splitting field for $f$.

Thus every $fin F[X]$ splits in $E$. Now I claim that $E$ is algebraically closed. Let $E'/E$ be an algebraic extension and let $beta in E'$. By transitivity, $beta$ is algebraic over $F$; let $h(X)$ be its minimal polynomial over $F$. By the above, $h$ splits in $E$, and therefore $beta in E$. Thus $E$ is algebraically closed.