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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.

syusim
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Tuo
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1 Answers1

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When the field $F$ is having $2$ for characteristic

$$\sqrt{D} = \prod_{i<j}(\alpha_i-\alpha_j)$$ is preserved by all permutations as $-1=+1$.

mathcounterexamples.net
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    Another way to say this is that permutations still change the sign of $\sqrt{D}$ correctly, it's just that a characteristic $2$ field can't see it. – Andrew Dudzik Jun 17 '15 at 20:32