Example of Field Extension $E/F$ with $Char(F)=2$ and $[E:F]=2$, but is not Galois

Question

I understand that for a field extension $E/F$, if $Char(F)\neq 2$ and $[E:F]=2$ then it must be a Galois Extension. I have proved this, but I am having trouble finding a counterexample when the characteristic requirement is dropped.

My first thought was something like $\mathbb{F}_{2}(\sqrt{t})/\mathbb{F}_{2}(t)$.

I am not sure, however, if this is a degree 2 extension.

Regardless, the minimum irreducible polynomial of $\sqrt{t}$ over $\mathbb{F}_{2}(t)$ is $f(x) = x^{2}-t = 0$. Clearly, $(f,f')=f\neq1$ and, in $\mathbb{F}_{2}(\sqrt{t})$, $f$ splits into $(x-\sqrt{t})^{2}$, so it is purely inseparable.

Is this enough to show that $\mathbb{F}_{2}(\sqrt{t})/\mathbb{F}_{2}(t)$ is an inseparable extension? And is it indeed degree 2?

Answer 1

Yes, nice counterexample.

The degree of the extension $\Bbb F_2(\sqrt t):\Bbb F_2(t)$ is indeed $2$, because -- as you wrote -- the minimum irreducible polynomial of $\sqrt t$ over $\Bbb F_2(t)$ is $f(x)=x^2-t$, which has degree $2$.

Answer 2

Another way, equivalent to yours, let's determine $Aut$($F_{2}$($\sqrt{t})/F_{2}(t))$. As you suggested, $\sqrt{t}$ is the only root of $x^{2}-t \in F_{2}(t)[x]$ and hence $Aut$($F_{2}$($\sqrt{t})/F_{2}(t))$=$\lbrace$ identity $\rbrace$<2.