showing Cohen-Macaulay property is preserved under a ring extension

Question

Let $R$ be an $\mathbb{N}$-graded Noetherian ring, with $R_0$ local Artinian. Assume also that $R$ is finitely generated over $R_0$ by elements of degree $1$. Let $M$ be a Cohen-Macaulay $R$-module. Let $Y$ be an indeterminate over $R_0$ and define the local ring $R_0(Y)=R_0[Y]_S$, where $S$ is the multiplicative set of $R_0[Y]$ which consists of all polynomials that contain at least one unit in their coefficients. Finally, define $R' = R \otimes_{R_0} R_0(Y)$ and $M' = M \otimes_{R_0} R_0(Y)$. Notice that the local ring homomorphism $R_0 \rightarrow R_0(Y)$ is flat and that its fibre is $k(Y)$, where $k$ is the residue class field of $R_0$.

Question 1: How can we see that $M'$ is a Cohen-Macaulay $R'$-module?

Question 2: How can we see that the Krull dimensions of $M,M'$ are equal?

PS: If we have a local homomorphism $(R,m) \rightarrow (S,n)$ of Noetherian rings and a finite $S$-module $N$ that is flat over $R$, then for a finite $R$-module $M$, $M \otimes_R N$ is CM over $S$ if and only if $M$ is CM and $N/mN$ is CM over $S$. What confuses me in the present situation is that there are more than two rings present i.e. $R, R_0, R_0(Y)$ and in fact $R$ is not even local.

Answer 1

Set $S_0=R_0(Y)$. The ring extension $R_0\subset S_0$ is local and flat. Then the ring extension $R\subset R'=S_0\otimes_{R_0}R$ is (faithfully) flat and we know that $M$ is a Cohen-Macaulay $R$-module of dimension $d$. The question is the following:

Let $R\to R'$ be a flat ring homomorphism of noetherian rings, and $M$ a Cohen-Macaulay $R$-module of dimension $d$. When $M'=R'\otimes_RM$ is a Cohen-Macaulay $R'$-module of dimension $d$?

If $R,R'$ are noetherian local rings and the homomorphism is also local, then $M'$ is a Cohen-Macaulay $R'$-module of dimension $d$ iff $R'/mR'$ is artinian (here $m$ denotes the maximal ideal of $R$).

In general, we proceed as follows: let $P'\in\operatorname{Supp}M'$. Then $p=P'\cap R\in\operatorname{Supp}M$, and the homomorphism $R_p\to R'_{P'}$ is faithfully flat. Since $M_p$ is Cohen-Macaulay we have only to look at the fibre $R'_{P'}/pR'_{P'}$. If all these fibres are artinian we are done.

In our particular case the fibre is $R[Y]_P/pR[Y]_P$ where $P$ is a prime ideal in $R[Y]$ lying over $p$. Thus this is a localization of $(R/p)[Y]$ at a prime ideal lying over $(0)$, and therefore it's Cohen-Macaulay. (For a better understanding let me change the notation: $S$ is an integral domain and $Q\subset S[Y]$ is a prime ideal with $Q\cap S=(0)$. Then $S[Y]_Q$ is isomorphic to a localization of $K[Y]$, where $K$ is the field of fractions of $S$, so it's Cohen-Macaulay.)

Although I find it very clear why $M$ and $M'$ have the same Hilbert functions (via the quoted result 1.2.25), and therefore $\dim M'=\dim M$, I still don't get it how to obtain this from above. Maybe I'm missing something about that fibre.

Edit. Alternative answer for question 1.

We have $R'=R\otimes_{R_0}R_0[Y]_S=R[Y]_S$. Then $M'=R'\otimes_RM=M[Y]_S$ (here $M[Y]$ stands for $R[Y]\otimes_RM$). Now apply Theorem 2.1.9 and Theorem 2.1.3(b) and obtain that $M'$ is Cohen-Macaulay.