Relationship between discriminant of a separable polynomial and its Galois group Probably a straightforward question:
Let $f \in K[X]$ be a polynomial with disctinct roots $ \alpha_1, ..., \alpha_d$ in a splitting field $L$. Set $\Delta = \Pi_{i <j} (\alpha_i - \alpha_j) $. Then the discriminant $D$ of $f$ is $D = \Delta^2 = (-1)^{d(d-1)/2} \Pi_{i \neq j} (\alpha_i - \alpha_j) $.
I can see that $D$ is fixed by $\mbox{Gal}(L/K)$, but that $\Delta$ isn't necessarily. In fact, provided $\mbox{char}(K) \neq 2$, if $f$ is irreducible and separable of degree $d$, then $\mbox{Gal}(L/K)$ is a subgroup of $A_d$ iff $\Delta$ is fixed under $\mbox{Gal}(L/K)$ iff $D$ is a square in $K$. 

My query is: why can't we have $\mbox{char}{K} =2$? I imagine it's something fairly obvious. 
It seems this restriction is often required, which leads me to also ask: Why do results often fail when the characterstic is 2? Is it merely because these are the most trivial fields, or is there a nicer reason why we don't like them?
EDIT: Come to think of it, that results I'm thinking of that fail if $\mbox{Char}(K) = 2$ are related to quadratics in some way, so I no longer expect an answer for my second question.
Thanks
 A: Suppose $K=\mathbb F_2\subset L=\mathbb F_4$ and $f(X)=X^2+X+1$,  a separable polynomial of degree $d=2$.
Then $\Delta=1$ is a square in $K$ and is fixed under $Gal(L/K)=: \lbrace Id, \sigma \rbrace$.
However it is not true that $Gal(L/K) \subset A_2=\lbrace Id \rbrace$.
A: In general we have: $\sigma(\delta)=sgn(\sigma)\delta$. Thus if the field has characteristic equal to 2, then $\delta$ is always fixed by the action of the Galois group. That is why it tells us nothing about the signature of elements in $\text{Gal}(G/K)$.
PS: I once heard someone joke that some people didn't consider $2$ as a prime number.
A: In characteristic 2, $(\alpha_i - \alpha_j) = (\alpha_i + \alpha_j)$, so $\Delta$ is always fixed by $Gal(L/K)$, so $D$ is always a square in $K$.
In characteristic other than 2, let $L = K[X_1 \ldots X_n]$, Galois theory says that $L^{A_n}$ is a degree 2 extension of 
$L^{S_n}$, and Kummer theory tells you that $L^{A_n} = L^{S_n}[P- \sigma P]$ and $(P - \sigma P)^2 \in L^{S_n}$ for any element $P \in L^{A_n} - L^{S_n}$ (if you pick $P = \Delta/2$ you obtain $P- \sigma P = \Delta$). So to check if your Galois Group is included in $A_n$, you just have to check if $\Delta(\alpha_1 \ldots \alpha_n)$ is in $K$, and it is if and only if $D(\alpha_1 \ldots \alpha_n)$ is a square in $K$
In characteristic 2, square roots don't behave at all like you're used to. For example if $K$ is algebraically closed and of characteristic $2$, square root is the reciprocal of the Frobenius automorphism $x \mapsto x^2$.
You don't have this luxury of "every degree 2 extension is obtained by adjoining a square root" so to check if your Galois Group is included in $A_n$, you pick a generator $P$ of $L^{A_n}$ over $L^{S_n}$ and check if $P(\alpha_1 \ldots \alpha_n)$ is in $K$.
