I'm not sure how long this [iterative](https://mathoverflow.net/questions/43438/) [questions](https://mathoverflow.net/questions/43255/) can go on, but let me try again. Let's say $X$ is a Cohen-Macaulay scheme with an action of $\mathbb{G}\_m$ (i.e. if $X$ is affine, a grading on the coordinate ring). Are the schematic fixed points $X^{\mathbb{G}\_m}$ of $X$ Cohen-Macaulay? 



Are schematic fixed-points of a Cohen-Macaulay scheme Cohen-Macaulay?
Here is a counterexample. Consider the action of $\mathbb G\_{\rm m}$ on $\mathbb A^4$ defined by $t \cdot(x,y,z,w) = (x, y, tz, t^{-1}w)$, and let $X$ be the invariant closed subscheme with ideal $(xy, y^2 + zw)$; this is a complete intersection, hence it is Cohen-Macaulay. The fixed point subscheme is obtained by intersecting with the fixed point subscheme in $\mathbb A^4$, which is given by $z = w = 0$; hence it is the subscheme of $\mathbb A^2$ given by $xy = y^2 = 0$, which is of course the canonical example of a non Cohen-Macaulay scheme.


Developing this idea a little, one can show that any kind of horrible singularity can appear in the fixed point subscheme of a $\mathbb G\_{\rm m}$-action on a complete intersection variety.


