This question refers to the Rees ring, and can be posed from first principles as follows.
Let $A$ be a Noetherian ring, $t$ an indeterminate over $A$ and define $u = t^{-1}$. Let $I$ be a proper ideal of $A$ generated by some $a_1, \dots,a_n \in A$. Then the Rees ring $R$ is defined to be $R = A[u,a_1t,\dots,a_nt]$. On the other hand, the associated graded ring of $A$ with respect to $I$ is defined as $G = \oplus_{m>0} I^m / I^{m+1}$. The relation between $R$ and $G$ is $G = R/uR$.
Matsumura (CRT, p.122) says that $\dim G \le \dim A$, but I can not see that from his arguments. Could somebody please explain why this is true?
1 Answer
Take a graded maximal ideal in $G=R/uR$, and write it as $M/uR$ with $M$ graded maximal in $R$ (containing $u$). Set $\mathfrak m=M\cap A$. Since $M$ contains $u$, $R/M$ is an $\mathbb N$-graded ring. But $R/M$ is also field, so it is isomorphic to its degree $0$ part, that is $A/\mathfrak m$. This shows that $$M=(\oplus_{i<0} At^i)\oplus\mathfrak m\oplus(\oplus_{i>0}I^it^i),$$ and therefore $M=(\mathfrak m',u)$.
Now you can follow the arguments from the book in order to prove that $\operatorname{ht}M/uR=\operatorname{ht}\mathfrak m$. Since $\operatorname{ht}\mathfrak m\le\dim A$ you get $\dim G\le\dim A$.
