# If every polynomial of degree $2$ and odd degree has a root in $k[x]$, then $k$ is algebraically closed.

Let $k$ be a field of characteristic zero. Assume that every polynomial in $k[X]$ of odd degree and every polynomial in $k[X]$ of degree two has a root in $k$. Show that $k$ is algebraically closed.

What we need to show is that these two assumptions implies that given $f\in k[X]$ with deg $f = 2^nm,$ $m$ an odd integer, $f$ has a root in $k[X]$. We probably want to use induction on $n$. The case $n=0$ is good by assumption. Now assume $n>0$, and that $f$ is irreducible. And I don't know what to do after this point.

Here is the standard proof, due to Artin. Let $f \in k[X]$ be a polynomial, and let $K/k$ be its splitting field. It suffices to show that $K = k$. Since the characteristic of $k$ is $0$, $K$ is normal and separable, hence Galois, so let $G = \mathrm{Gal}(K/k)$. Let $H$ be a Sylow $2$-subgroup of $G$, and let $K^{H}$ be the fixed field of $H$. Then $[K^{H}:k] = [G:H] = m$ for some $m$ odd. Since $k$ has no nontrivial extensions of odd degree, $m = 1$, so $G = H$ and hence is a $2$-group. If $G$ is nontrivial, then $G$ contains a normal subgroup of index $2$, i.e. $K$ contains a subfield which is a quadratic extension of $k$. This is impossible by hypothesis, so $K = k$.