Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $$ \left( \dfrac{A}{I} \right)^{S_k} = \dfrac{A^{S_k} }{IA^{S_k}}? $$

$$ $$

share|improve this question
    
I am guessing that one can consider $$0\rightarrow I \rightarrow A \rightarrow A/I\rightarrow 0$$ and then take the $S_2$-invariant pieces: $$0\rightarrow I^{S_k} \rightarrow A^{S_k} \rightarrow \left( A/I\right)^{S_k}\rightarrow 0.$$ Is exactness still preserved in the second short exact sequence? –  math-visitor Jun 13 '12 at 19:50
    
I'm not familiar with everything in this question, but don't we have $x+y\equiv 2x\bmod x-y$ hence $$\frac{k[x+y,xy]}{(x-y)}\cong k[x]~?$$ –  anon Jun 13 '12 at 20:12
    
Hi anon, that may possibly be true but geometrically speaking (trying to make sense of this both geometrically and algebraically), I am not certain at the moment... –  math-visitor Jun 13 '12 at 20:19
    
You might be doing something wrong: quotienting out $x-y$ means $S_2$ acts trivially, since $x=y$ in such a ring! In general, taking the invariant subalgebra of a group action is left exact, but not right exact (that way $\Ext$ groups lie...). –  user641 Jun 14 '12 at 5:34
1  
You are confusing isomorphism and equality. Your first comment above about $Q$ is correct. Now $k[x,y]/(x+y)$ is indeed just $k[y]$; but $S_2$ acts by sending $y$ to $-y$; thus the fixed points are $k[y^2]\subset k[y]$. This is isomorphic to $k[y]$, but they are not equal. –  user641 Jun 14 '12 at 13:26

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.