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Apr 1, 2010 at 19:29 history edited Victor Miller CC BY-SA 2.5
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Apr 1, 2010 at 18:21 comment added Torsten Ekedahl This looks to be in contradiction with the result that says that the invariant ring is not Cohen-Macaulay as a polynomial ring is certainly Cohen-Macaulay. Also according to the Math Review the condition is that the subspace $\{gv-v\}$ be $1$-dimensional not of codimension $1$. (The space of covariants is the quotient of $V$ by that space.)
Apr 1, 2010 at 17:36 history answered Victor Miller CC BY-SA 2.5