I'm not sure how long this iterative questions can go on, but let me try again. Let's say $X$ is a CohenMacaulay 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 $X$ CohenMacaulay?

2$\begingroup$ I am not sure what the usual business is, but it is not true that if $\mathbb G_{\rm m}$ acts on a variety $X$ with a fixed point $p$, then this induces an action of $\mathbb G_{\rm m}$ on $\mathop{\rm Spec}\mathcal O_{X,p}$. $\endgroup$– AngeloCommented Oct 25, 2010 at 21:39

$\begingroup$ Is this an issue of not being able to find a $\mathbb{G}_m$invariant affine open containing $p$? $\endgroup$– Ben Webster ♦Commented Oct 25, 2010 at 22:23

$\begingroup$ Those invariant affines are not cofinal in all neighborhoods. Algebraically, the localization of C[x] at the ideal (x) does not admit a grading...does it? $\endgroup$– David TreumannCommented Oct 25, 2010 at 22:34

$\begingroup$ Ah, right. That was complete nonsense. Removed. I think you probably you can reduce to the graded local case, but let me not worry about that. $\endgroup$– Ben Webster ♦Commented Oct 25, 2010 at 23:03

$\begingroup$ Let $T^1$ act on ${\mathbb A}^3$ with weights $0,1,1$. Then on the ${\mathbb P}^2$, the fixedpoint set is not equidimensional. So you can only hope to have a local statement. $\endgroup$– Allen KnutsonCommented Oct 26, 2010 at 0:39
2 Answers
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 CohenMacaulay. 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 CohenMacaulay 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.

1$\begingroup$ Murphy's law strikes again! $\endgroup$– Ben Webster ♦Commented Oct 27, 2010 at 6:50
Edit: the following does not answer Ben's question. It gives an example of the subring fixed by $G_m$ being not CM, while the question asked about the subscheme of fixed points, see the comments for more details.
Let $R$ be the (homogenous) cone of a curve $C$ of genus $g>0$, for example $R=\mathbb C[x,y,z]/(x^3+y^3+z^3)$. Let $S=R[u,v]$, $X=\text{Spec}(S)$ and $G_m$ acts by
$a.(x,y,z,u,v) = (ax,ay,az,a^{1}u, a^{1}v)$.
Then $A= S^{G_m}$ would be a homogenous coordinate ring for $Y= C\times \mathbb P^1$, so it is not CohenMacaulay (if $A$ is CM, it would mean that $H^1(Y,\mathcal O_Y)=0$, impossible, see here for an explanation).
(I learned this idea from Hochster, let me try to find a reference)

$\begingroup$ Hailong this is the categorical quotient, not schematic fixed points. The schematic fixed points here are a single point with reduced structure and thus CohenMacaulay. $\endgroup$– Ben Webster ♦Commented Oct 25, 2010 at 23:12

$\begingroup$ Hmm, sorry, I always thought of this notation as the invariants. What do you mean by schematic fixed pts? Are they just literally the pts of $X$ fixed by the group action? $\endgroup$ Commented Oct 26, 2010 at 0:10

$\begingroup$ @Hailong: you applied the fixedpoint functor to the algebra rather than to the space. That is, the notation itself has a consistent meaning. (I'm not surely how to reasonable define schematic fixed points; in characteristic zero, with a connected group, one could look where the generating vector fields vanish.) $\endgroup$ Commented Oct 26, 2010 at 0:43

$\begingroup$ Dear Hailong: No ad hoc constructions/definitions are required. For any scheme $X$ over a ring $k$ and action on $X$ by a $k$gp scheme $G$, define functor $X^G$ on $k$algebras as follows: for $k$algebra $R$, $X^G(R)$ is set of $x \in X(R)$ fixed by the $G_R$action on $X_R$ (i.e., for any $R$algebra $R'$, $x$ viewed in $X(R')$ is $G(R')$invariant). Is this represented by a closed subscheme of $X$? Yes, provided $X$ is locally of finite type and separated over $k$ and $G$ is affine and fppf over $k$ with connected fibers. (See Prop. A.8.10(1)ff. in "Pseudoreductive groups" for details.) $\endgroup$– BCnrdCommented Oct 26, 2010 at 1:28

1$\begingroup$ @ Allen and BCnrd: thank you! I will leave my answer, so people might benefit from your explanations and avoid my mistakes! $\endgroup$ Commented Oct 26, 2010 at 2:07