Symmetrical polynomial under even permutation

\mathbf{Problem:} Let F be of characteristic \neq 2 and also let H be the subgroup of G=\text{Gal }E/F(p_1,p_2,\cdots,p_n) corresponding to the alternating group, that is the set of \zeta(\pi), \pi\in A_n. Show that \text{Inv}H=F(p_1,p_2,\cdots,p_n,\Delta) where \displaystyle\Delta=\prod\limits_{i<j}(x_i-x_j).

\mathbf{Proof:} It is a basic fact that \zeta(\pi)(\Delta)=\Delta when \pi is an even permutation and \zeta(\pi)(\Delta)=-\Delta when \pi is an odd permutation.

So all \pi\in H preserve \Delta and p_1,p_2,\cdots,p_n, which means F(p_1,p_2,\cdots,p_n,\Delta)\subset \text{Inv}H.

On the other hand no odd permutation can preserve \Delta, then \text{Gal }E/F(p_1,p_2,\cdots,p_n,\Delta)\leq H.

Combining these two, we know \text{Inv}H=F(p_1,p_2,\cdots,p_n,\Delta).

\text{Q.E.D}\hfill \square

\mathbf{Remark:} Jacobson p244.

Advertisements
Post a comment or leave a trackback: Trackback URL.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: