Does the coinvariant algebra ever admit nilpotent elements?

60 Views Asked by At

Here's a simple question, presumably the answer is well known to certain people.

Let $k$ be a field and let $A$ be a finitely generated $k$-algebra which is reduced (i.e. it admits no nilpotent elements except zero). Suppose that $G$ is a group acting by automorphisms on $A$. Then we can define the coinvariant algebra $A_G$ which is the quotient of $A$ by the ideal generated by elements $\{g \cdot a - a\mid g\in G, a\in A\}.$

Question: is $A_G$ reduced?

If the answer is negative then seeing various examples would be great. If there is an example of an infinite group such that $A_G$ is nonreduced, then what about finite groups? I am especially interested in the case where $G$ is an algebraic group acting locally finitely and completely reducibly on $A$, so examples or general results in this context will be especially well received.

Many thanks in advance.

1

There are 1 best solutions below

1
On

This may not be very interesting for you.

Take $A=k[x,y]$, polynomial ring in two variables. Take the free abelian group generated by the automorphism, $x\mapsto x+y^2, y\mapsto y$. You can easily check that your ideal is then $y^2A$.