1
$\begingroup$

Let $S_2$ be a finite group of order $2$ and let $S_2$ act on $k[x,y]$ by interchanging $x$ and $y$, where $k=\overline{k}$. Then since

$ R = \left( \dfrac{k[x,y]}{(x+y)} \right)^{S_2} = \dfrac{k[x+y,xy]}{(x+y)} \cong k[xy], $

$\mbox{Spec} \;R =k$.

On the other hand,

$ Q = \left( \dfrac{k[x,y]}{(x-y)} \right)^{S_2} = \dfrac{k[x+y,xy]}{(x-y)}. $

Can we simplify $Q$ further as in the first example, and isn't $\mbox{Spec}\; Q$ isomorphic to a line?

$\mathbf{Remark}:$ the $S_2$ invariant functions on $k[x,y]/(x+y)$ is certainly $k[xy]$ (the entire line is $S_2$-invariant) but I don't think the RHS of $Q$ is correct: the $S_2$-invariant functions on the line defined by $k[x,y]/(x-y)$ should really be a point. Don't you agree?

$ \mathbf{General \; case:} $ Let $A=k[x_1, \ldots, x_n]$ and let $I$ be an ideal of $A$. Give some action of $S_k$ on $A$. Then do we have $ \left( \dfrac{A}{I} \right)^{S_k} = \dfrac{A^{S_k} }{I} $ or is it $ \left( \dfrac{A}{I} \right)^{S_k} = \dfrac{A^{S_k} }{IA^{S_k}}? $

$ $

  • 0
    Thanks for clarifying that Steve!2012-06-14

0 Answers 0