Extensions of degree two are Galois Extensions.

Question

This question from Allan Clark's "Elements of Abstract Algebra"

Show that an extension of degree 2 is Galois except possibly when the characteristic is 2. What is the case when the characteristic is 2?

Tips are helpful, a solution is ideal.

Thanks.

Answer 1

Let $L/K$ be a field extension of degree $2$. If $\alpha \in L \setminus K$ then $p(t)=\min_K(\alpha,t)$ has degree $2$. In particular $p(t)$ must split over $L$ since $p(t)=(t-\alpha)q(t)$, forcing $q(t)$ to be degree $1$. If the characteristic of $K$ is not equal to $2$ then $p^\prime(t) = 2t + \cdots \neq 0$, so $\alpha$ is separable over $L$. Thereby $L/K$ is Galois.

For a counterexample in the case of characteristic $2$ consider the splitting field of $p(x)=x^2-t \in \mathbb{F}_2(t)[x]$. It's not hard to see that $p(x)=(x+\sqrt{t})^2$ so the extension is purely inseparable and not Galois.