TITLE: Discriminant of Polynomials (Galois Theory)
QUESTION [5 upvotes]: So I'm reading Dummit and Foote and they define the discriminant of $x_{1},...,x_{n}$ by
$$D=\prod_{i<j}(x_{i}-x_{j})^2$$
and the discriminant of a polynomial to be the discriminant of the roots. 
They say that a permutation $\sigma \in S_{n}$ is in the alternating group $A_{n}$ iff $\sigma$ fixes the product $\sqrt{D}$.

It follows by the Fundamental Theorem of Galois Theory that if $F$ has characterstic different from 2 then $\sqrt{D}$ generates the fixed field of $A_{n}$ and generates a quadratic extension of $K$.

I am confused where characteristic $2$ comes into this.

REPLY [2 votes]: When the field $F$ is having $2$ for characteristic
$$\sqrt{D} = \prod_{i<j}(\alpha_i-\alpha_j)$$ is preserved by all permutations as $-1=+1$.