$E$ is any arbitrary field with char$\neq 2$ . And $F$ is an extension of $E$ such that $[E:F]=2$ . Then $E/F$ is Galois extension .
Since the degree is $2$ , the irreducible polynomial of any element of $E$ is quadratic : $$ax^{2}+bx+c$$
Now giving $x$ a translation by $\ \ ({-}{{a}\over {2}})\ \ $ reduces the polynomial to the form _:_$$x^{2}-d$$
So $d$ and $-d$ are two roots of the polynomial and it splits over $E$. Hence $E/F$ is Galois .
But nowhere in the proof did I use the fact that $char(F)\neq 2$ or the concept of characteristic of fields at all . Then is the proof correct $?$ .
If so then why do we have to mention that char($F$)$\neq2$ $?$ What purpose does it serve $?$
I really need clarification on this part .
Thanks for any help.
In characteristic $2$, there are two possibilities for an irreducible quadratic $X^2+aX+b$. If $a=0$, then the extension is inseparable, and thus non-Galois. If $a\ne0$, then the extension is separable, and Galois, ’cause if $z$ is one root, the other root is $a+z$, so your factorization over $k(a)$ is $(X+z)(X+a+z)$.