Exercise 2.4.2

Proof. Here, the field $F$ have characteristic $\ne 2$.

Let $f\in F[x_1,\cdots ,x_n]$ such that $\tau \cdot f=-f$ for all transpositions $\tau \in S_n$.

If $\sigma \in A_n$ is an even permutation, then $\sigma$ is product of an even number of permutations :

As the group $S_n$ acts on $F[x_1,\cdots ,x_n]$, $\sigma \cdot f={\tau }_1\cdot ({\tau }_2\cdot (\cdots ({\tau }_{2k}\cdot f)\cdots ))={(-1)}^{2k}f=f$. Therefore $f$ is invariant under $A_n$ and so the theorem 2.4.4 applies:

There exist $A,B\in F[{\sigma }_1,\cdots ,{\sigma }_n]$ such that

Therefore $-f=\tau \cdot f=\tau \cdot A+(\tau \cdot B)(\tau \cdot \sqrt{\mathrm{\Delta }})=A-B\sqrt{\mathrm{\Delta }}$ (by 2.31).

So $f=A+B\sqrt{\mathrm{\Delta }}$ and $f=-A+B\sqrt{\mathrm{\Delta }}$, thus $2A=0$. Since the characteristic is not 2, $A=0$, therefore

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