Exercise 7, § 6

Let $A$ be a noetherian ring and $\mathscr{F} = (\mathscr{F}_i)_{i\in\mathbf{Z}}$ a decreasing filtration of $A$ such that $\mathscr{F}_0 = A$, $\mathscr{F}_1 \neq A$ and such that the $A$-algebra $B = \sum_{i\in\mathbf{Z}} \mathscr{F}_i v^{-i}$ is of finite type.

We propose to show that one has $\dim(B) = \dim(A) + 1$ and to use this fact to generalize the result of § 6, No. 3, corollary of prop. 5. We shall use the results of the two preceding exercises. (a) Let $\mathfrak{m}$ be a maximal ideal of $B$. Now prove that one has

and deduce the inequality

(b) Using the fact that every element of the $A[v]$-module $A[v, v^{-1}]/A[v]$ is annihilated by a power of $v$, now prove that one has

and deduce that one has $\dim(B) \geq \dim(A) + 1$, whence finally $\dim(B) = \dim(A) + 1$ (one may use the fact that the ring $A[v, v^{-1}]$ is a ring of fractions of B).

(c) Now prove that the minimal prime ideals of B are the ideals of the form :

where $\mathfrak{p}$ is a minimal prime ideal of A.

(d) Now prove that one has $vB \neq B$ and that $v$ belongs to none of the minimal prime ideals of B. Deduce the inequality $\dim(B/vB) \leq \dim(A)$ (one may use § 1 and § 3, no 3, cor. 1 of prop. 4).

(e) Suppose now that there exists a maximal ideal $m$ of A containing $\mathscr{F}_1$ and such that $\mathrm{ht}(m) = \dim(A)$. Now prove that in this case one has $\dim(B/vB) \geq \dim(A)$ and therefore $\dim(B/vB) = \dim(A)$. (One may reduce to the case where A is a local ring with maximal ideal $m$, show that the ideal $\mathfrak{M}$ of B generated by $v$ and $\sum_{i \in \mathbf{Z}} (\mathscr{F}_i \cap m) v^{-i}$ is a maximal ideal of B having height $\dim(A) + 1$ and localize B at $\mathfrak{M}$.)

(f) If A is a local noetherian ring, and $\mathscr{F}$ a decreasing filtration of A such that $\mathscr{F}_0 = A$, $\mathscr{F}_1 \neq A$, one has the equality