My question is somewhat similar to Dimension of a Finite Field Extension is a Power of 2 but there's a small complication.

Claim 1: Let $K$ be a field of characteristic $\neq 2$, and suppose that every polynomial of odd degree in $K[x]$ has a root in $K$. Let $L/K$ be a finite extension. Then $[L:K] = 2^n$ for some $n$.

Claim 2: Let $K,L$ be as above. If $L = L^2$, then $L$ is algebraically closed.

For Claim 1, I have the following reasoning, but it doesn't quite get me all the way there. Since every odd degree polynomial has a root, it is reducible into a linear factor and even degree polynomial, so all irreducible polynomials are even degree. Since $L/K$ is finite, it is finitely generated, so $L = K(\alpha_1,\ldots, \alpha_n)$ for some $\alpha_1,\ldots, \alpha_n \in L$. Then we have a tower of fields $$ K \subset K(\alpha_1) \subset K(\alpha_1,\alpha_2) \subset \ldots \subset K(\alpha_1,\ldots, \alpha_n) = L $$ In each step of the tower, $K(\alpha_1, \ldots, \alpha_{i+1}) = K(\alpha_1, \ldots, \alpha_i)$ and the irreducible polynomial of $\alpha_{i+1}$ is even degree, so each extension is even. Thus $[L:K]$ is even.

If I could show that each step of the tower is of degree 2, then I would have proved claim 1, but I don't know why that would be the case. Is there a reason that we can't have irreducible quartics over $K$?


Suppose every odd degree equation has a solution. Let $L/K$ be a finite extension. Go to a Galois closure $M/K$ with group $G$. It has a Sylow 2-subgroup $H$. Consider the fixed field $M^H$. This has odd degree over $K$, so $M^H=K$ and $H=G$. Thus $|G|$ is a power of $2$ and $|M:K|$ and $|L:K|$ are powers of $2$.

Every group of 2-power order's maximal subgroups have index 2, so $L/K$ can be represented as a tower of quadratic extensions.

This argument may break down due to inseparability. I'll let you worry about that!


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