# $F/K$ algebraic and every nonconstant polynomial in $K[X]$ has a root in $F$ implies $F$ is algebraically closed.

Let $F/K$ be an algebraic extension of fields in characteristic zero. If $F/K$ is normal, and every nonconstant polynomial $f \in K[X]$ has a root in $F$, then $F$ is algebraically closed. This is obvious, because if $u$ is algebraic over $F$ (let us agree to fix some algebraic closure $\overline{K}$ of $K$ containing $F$ with $u \in \overline{K}$), then it is algebraic over $K$ with minimal polynomial $f \in K[X]$, which by hypothesis must have one and therefore all of its roots in $F$. In particular $u \in F$.

I have heard that the above assertion remains true if we drop the assumption that $F/K$ is normal. But it seems to be a nontrivial result. Can anyone get me started on how to prove this?

Thoughts so far: It's enough to show that every algebraic extension of $K$ is $K$-isomorphic to a subfield of $F$. Certainly this is true for every finite extension $E$ of $K$; we're in characteristic zero, so $E = K(v)$ for some $v \in E$. If $f \in K[X]$ is the minimal polynomial of $v$ over $K$, then $f$ has some root $u \in F$, whence $E$ is $K$-isomorphic to the subfield $K(u)$ of $F$.

So, I'm thinking next use some kind of Zorn's Lemma argument?

• Isn't it simpler, and perhaps more constructive to show that every polynomial in $F$ is obtained by composition of polynomials in $K$? – Asaf Karagila Jul 27 '14 at 2:30
• @Asaf Say $F=\mathbb C$ and $K=\mathbb R$, and $p(x)=x-i$. What do you mean by "composition" here? – Andrés E. Caicedo Jul 27 '14 at 2:38
• @Andres: I mean that in the sense that we can write $x^2+1$ instead of $i$. I am being a bit inaccurate here, mainly as a consequence of writing from a cellphone, but I hope my meaning is clear. In model theoretic terms, all the objects which can be described with formulas with parameters from $F$ can be described with parameters from $K$. – Asaf Karagila Jul 27 '14 at 2:48
• @Asaf Now you have to argue that just because $p\in K[x]$ has a root in $F$ then actually $p$ splits in $F$. – Andrés E. Caicedo Jul 27 '14 at 3:03
• (@Andres: François Dorais commented with a lovely forcing based construction of an algebraic closure somewhere on my blog recently.) – Asaf Karagila Jul 27 '14 at 3:20

Let $E/F$ be a finite extension (which is separable as the characteristic is zero), we will prove that $E=F$, hence $F$ is algebraically closed: Let $a$ be a primitive element of $E/F$. Then $a$ is algebraic over $K$, so there exists a finite Galois extension $N/K$ containing $a$, in particular $$E\subseteq NF.$$ (One can take $N$ to be the Galois closure of $K(a)$, for example.) Let $f\in K[x]$ be the minimal polynomial of a primitive element of $N$ over $K$. Since $N/K$ is Galois, any of the roots of $f$ generates $N$ over $K$.
Now, by assumption $f$ has a root in $F$, so $N\subseteq F$, hence $E\subseteq NF=F$. This proves that $F$ has no non-trivial extensions, hence $F$ is algebraically closed.