Proposition 4

Let A be a commutative ring filtered by an exhaustive filtration (An) and p an ideal of A. Suppose that the ideal $\operatorname{gr}(p) = \bigoplus_{n \in \mathbf{Z}} (p \cap A_n)/(p \cap A_{n+1})$ of the ring $\operatorname{gr}(A)$ is prime. Then the closure of p in A is a prime ideal.*

We know that gr(A/p) is isomorphic to gr(A)/gr(p) (no. 4, Proposition 2) and hence an integral domain; we conclude that $A / \bigcap_{n \in \mathbf{Z}} (p + A_n)$ is an integral domain (no. 3, Corollary to Proposition 1). Then the closure $\bigcap_{n \in \mathbf{Z}} (p + A_n)$ of $p$ is a prime ideal.

Let $A$ be a ring and $m$ a two-sided ideal of $A$; the topology defined on $A$ by the $m$-adic filtration (no. 1, Example 3) is called the m-adic topology; as the m-adic filtration is exhaustive, $A$ is a topological ring with this topology (Corollary to Proposition 3). Similarly, for every $A$-module $E$, the topology defined by the $m$-adic filtration is called the m-adic topology on $E$; $E$ is a topological $A$-module under this topology.

Let $m'$ be another two-sided ideal of $A$; for the $m'$-adic topology on $A$ to be finer than the $m$-adic topology, it is necessary and sufficient that there exist an integer $n > 0$ such that ${m'}^n \subset m$; the condition is necessary and, if it is fulfilled, ${m'}^{hn} \subset m^h$ for all $h > 0$ and hence the condition is sufficient. If $A$ is a commutative Noetherian ring, it amounts to the same to say that $V(m) \subset V(m')$ in the prime spectrum of $A$ (Chapter II, § 4, no. 3, Corollary 2 to Proposition 11 and § 2, no. 6, Proposition 15).