Cohen-Macaulay/Gorenstein passing from associated graded to general fiber Let $R$ be a filtered commutative ring over a field $k$. Let $A$ denote the Rees algebra and $A_0:=\operatorname{gr}_F(R)$ be the associated graded ring.  $\operatorname{Spec}(A)$ gives rise to a flat family over $\mathbb{A}^1$ with fiber over zero $\operatorname{Spec}(A_0)$ and general fiber $\operatorname{Spec}(R)$. Assume that $A$(and hence $R$ and $A_0$) is finitely generated. 
Question 1) Is it true that $\operatorname{Spec}(A_0)$ being Cohen-Macaulay implies that $\operatorname{Spec}(R)$ is Cohen-Macaulay?
Question 2) Is there a nice condition for when $A_0$ being Gorenstein implies that $R$ is Gorenstein? How about in the case when the filtration is associated to some ideal I? 
 A: *

*Yes. Call the total space of the family $X$ and let $X_0=\operatorname{Spec} R_0$. Let $x\in X_0$ and let $t$ be the coordinate on $\mathbb{A}^1$. Consider the exact sequence $0\to\mathcal{O}_X\stackrel{t}{\to} \mathcal{O}_X \to \mathcal{O}_{X_0}\to 0$. Taking stalks at $x$, we get that $0\to \mathcal{O}_{X,x}\stackrel{t}{\to} \mathcal{O}_{X,x} \to \mathcal{O}_{X_0,x}\to 0$ is exact, ie $\mathcal{O}_{X,x}/t\cong \mathcal{O}_{X_0,x}$, so a regular sequence in $\mathcal{O}_{X_0,x}$ gives rise to a regular sequence in $\mathcal{O}_{X,x}$ by appending $t$. (More generally, this strategy works for any surjective flat morphism of schemes $f:X\to Y$. $X$ is CM iff $Y$ is CM and each fiber $X_y$ is CM). This shows that every point in $X_0$ considered as a point of $X$ is Cohen-Macaulay, and since Cohen-Macaulayness is an open condition, there exists an open set $U$ containing $X_0$ such that $U$ is Cohen-Macaulay. But by the dilation action coming from $\mathbb{A}^1$, we know that $U$ is in fact the whole space.

*Yes, via a similar strategy to the last time. In full generality, let $f:X\to Y$ be a flat surjective morphism of schemes. $X$ Gorenstein is equivalent to $Y$ Gorenstein and $X_y$ Gorenstein for all $y\in Y$. See for instance http://projecteuclid.org/euclid.kjm/1250523903 theorem 1. Apply with $X$ same as in 1 and $Y=\mathbb{A}^1$ to get that each point of $X_0$ considered as a point in $X$ is Gorenstein, combine with Gorenstein being an open condition, and the same reasoning about the dilation action.
A: The purpose of this answer is to provide certain clarifications to KReiser's answer

*

*In general, it is not true that the general fiber must be Cohen Macaulay see Example 3.6 of https://projecteuclid.org/euclid.kjm/1250521977
for an example where $R=k[x,y]/(x^2,xy)$, the filtration is associated to the maximal ideal $(x,1-y)$, and $A_0=k[y]$.


*In order to deduce the CM property or Gorenstein property of the special fiber using the approach outlined in KReiser's answer (or in commutative algebra terms the approach of the above article) one needs that every $\mathbb{G}_m$ orbit has a limit point in the special fiber.
The easiest situation to do this is when the descending filtration lives in nonpositive degrees and $F_0=k\cdot 1$. In the case of filtrations associated to an ideal $I$(the setting of the aforementioned article), this is rather rare and requires that $I$ is contained in the Jacobson radical of $R$.
