I worked through the following exercise in a textbook:
Suppose
- $L:K$ is a field extension with $[L:K] = 2$.
- Every element of $L$ has a square root in $L$.
- Every polynomial of odd degree in $K[x]$ has a root in $K$.
- char $K \ne 2$
Let $f$ be an irreducible polynomial in $K[x]$, let $M:L$ be a splitting field for $f$ over $L$, and let $G = \Gamma(M:K)$.
Then
- $|G| = 2^n$
- If $n > 1$, then there is an irreducible quadratic in $L[x]$.
- $L$ is algebraically closed.
- The complex numbers are algebraically closed.
Later I found out this (mostly algebraic) proof of the FTA is by Artin and Schreier.
But why is char $K \ne 2$ needed?
The next exercise in the textbook addresses this condition saying to consider the splitting field of all polynomials of odd degree over $\mathbb{Z}_2$ to show that the condition is needed.
The point:
When $\operatorname{char}(L)\neq 2$, every quadratic polynomial over $L$ can be written as $(x - a)^2 - b$ for some $a, b \in L$.
Consequently, in that situation, the hypothesis "every element of $L$ has a square root in $L$" implies that all quadratic polynomials over $L$ are reducible, i.e. there is no quadratic extension of $L$.
This is not true if $\operatorname{char}(L) = 2$, as can be seen with the example given in "the next exercise" (recall that every element in a finite field of characteristic $2$ is a square).