Is $\mathbb{F}_p(X)/\mathbb{F}_p(X^2)$ Galois extension?

Question

Let $E:=\mathbb{F}_p(X)/F:=\mathbb{F}_p(X^2)$ be a field extension where $p$ is a prime. Is this a Galois extension?

My attampt: $$ \text{irr}(X,F)=x^2-X^2\in F[x]$$ Thus, $E/F$ is algebraic. Furthermore, $x^2-X^2=(x-X)(x+X)$. Thus, $E/F$ is a normal extension. We conclude that the extension is Galois.

In the solution they claim that $E/F$ is Galois iff $p\ne 2$ but I don't understand why.

Answer 1

If $p=2$, $p(x) = x^2-X^2 = (x-X)^2$. Hence $p$ is not a separable polynomial and $E/F$ is not Galois.

If $p \neq 2$, then $X \neq -X$ in that case so the extension $E/F$ is separable. It is also normal as if $y$ is a root, then other one is $-y$.