Ring of invariants It is known that if you have a reductive group acting on a regular algebra, then the ring of invariants is Cohen-Macaulay. Is this still true for Gorenstein algebras?
 A: That is not true.  Let $k$ be a field.  Let $R$ be the regular $k$-algebra, $R=k[x,y,z,w].$  Let $G = \text{Spec}\ k[s,s^{-1},t,t^{-1}]$ be the multiplicative $k$-group of rank $2$.  Let $G$ act on $\text{Spec}(R)$ via the following coaction, $$m^*:k[x,y,z,w] \to k[s,s^{-1},t,t^{-1},x,y,z,w], $$ $$x\mapsto sx, \ \ y \mapsto s^{-1}y, \  \ z\mapsto tz, \ \ w\mapsto t^{-1}w.$$  The invariant subring is $k[u,v]$, where $u$ equals $xy$ and $v$ equals $zw$.  
Now consider the ideal in $R$, $I=\langle x^2,yz\rangle$.  This ideal is generated by a regular sequence.  Thus the quotient algebra $R/I$ is Gorenstein, and even a complete intersection ring.  Moreover, $I$ is $G$-stabilized.  Thus, the action of $m$ on $\text{Spec}(R)$ restricts to an action of $m$ on the closed subscheme $\text{Spec}(R/I)$.  
Since $G$ is linearly reductive over $k$, the invariant ring $(R/I)^G$ equals $R^G/I^G$. Finally, $I^G$ equals $\langle u^2,uv\rangle$.  Thus the invariant ring $(R/I)^G$ is $k[u,v]/\langle u^2,uv \rangle$.  This is not a Cohen-Macaulay ring; it is not even $S_1$.
