# 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

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Dear Jonathan, fields of characteristic 2 are a little strange but I can't answer your question "is there a nicer reason why we don't like them?" , because personally I do like them! –  Georges Elencwajg Dec 12 '11 at 12:46

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.

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Dear @prime , here is another joke : all prime numbers are odd... and $2$ is the oddest of them all! –  Georges Elencwajg Dec 12 '11 at 12:54
Thanks - once I realised that your $\delta$ was my $\Delta$, it made perfect sense. –  Jonathan Dec 12 '11 at 13:11

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 $Gal(L/K) \subset A_2=\lbrace Id \rbrace$.

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Thanks for this example. Is there a particular reason why you've written $\mbox{Gal}(L/K) \subset A_2$, as opposed to $\mbox{Gal}(L/K) \leq A_2$? (my concern here is that my notes used the subset notation also) –  Jonathan Dec 12 '11 at 13:12
Dear @Jonathan, I use the inclusion sign for subgroups out of cultural habit. I think the notation $\leq$ for "subgroup" is more usual in the anglophone sphere than in continental Europe. That said, I think this use of $\leq$ is a very good idea from a didactical point of view and I certainly would not criticize it. –  Georges Elencwajg Dec 12 '11 at 13:31
Great, thanks Georges. –  Jonathan Dec 12 '11 at 13:43

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 $K$.

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