Monthly Archives: August 2019

A special case of Krull’s intersection theorem

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In this post I will prove a special case of Krull’s intersection theorem, which can be proved without invoking the Artin-Rees lemma. This result is useful in the study of discrete valuation rings. The following proof is from Serre’s Local Fields.

Proposition: Let R, \frak{m} be a noetherian local domain where \frak{m} = (\pi). Then \bigcap_{n \geq 0} \frak{m}^n = 0.

Proof: Suppose y \in \bigcap_{n \geq 0} \frak{m}^n. Then y = \pi^n x_n for all n \geq 0. So for all n \pi^nx_n = \pi^{n+1}x_{n+1}. Since R is a domain, x_n = \pi x_{n+1}.

Therefore consider the ascending chain (x_0) \subseteq (x_1) \subseteq \cdots. This eventually stabilizes for high enough n since R is noetherian, so for some n, x_{n+1} = cx_n. Thus x_{n+1} = c\pi x_{n+1}, so (1-c\pi)x_{n+1} = 0. But 1-c\pi is a unit, so x_{n+1} = 0, so y = 0. \square

This theorem holds more generally even if R is not assumed to be a domain, but the proof is more complicated (but still among the same lines).

Proposition: Let R, \frak{m} be a noetherian local ring where \frak{m} = (\pi) and \pi is not nilpotent. Then \bigcap_{n \geq 0} \frak{m}^n = 0.

Proof: Let \frak{u} be the ideal of elements that kill some power of \pi. We will use variables u_1, u_2, \dots to refer to elements of \frak{u}. Since R is noetherian, \frak{u} must be finitely generated, so all elements of \frak{u} kill \pi^N for some fixed N.

Now suppose y \in \bigcap_{n \geq 0} \frak{m}^n. \pi^nx_n = \pi^{n+1}x_{n+1}, so \pi^n(x_n - \pi x_{n+1}). Thus x_n - \pi x_{n+1} \in \frak{u}.

Consider the ascending chain \frak{u} + (x_0) \subseteq \frak{u} + (x_1) \subseteq \cdots. Since R is noetherian it must eventually stablize, so for some n, x_{n+1} can be written as u_1 + cx_n. But recall that x_n = u_2 + \pi x_{n+1}. So x_{n+1} = u_1 + c(u_2 + \pi x_{n+1}) = u_3 + c\pi x_{n+1} so (1-c\pi)x_{n+1} = u_3. 1-c\pi is a unit since \frak{m} = (\pi), and R is local, so x_{n+1} \in \frak{u}. If we force n to be large enough to surpass N, then \pi^{n+1}x_{n+1} = 0, so y = 0. \square