13
$\begingroup$

Let $(R,\frak m)$ be a Noetherian local ring, and let $X$ be a set of ideals in $R$. Assume $\bigcap_{I \in X} I = 0$. Is there some sequence $\{I_n\}_{n \in \mathbb N}$, with $I_n \in X$ for all $n$, such that $\bigcap_n I_n =0$?

The above is true if $R$ is complete, by a slight alteration of the argument in Chevalley's Theorem (lemma 7 in Chevalley, On the theory of local rings, Ann. Math. 44 (4), 1943). I want to know whether completeness is necessary.

$\endgroup$
1
  • 2
    $\begingroup$ My previous answer was wrong (thanks to @Pace Nielsen for pointing that out). I will leave with the following remark: by Krull's intersection theorem, it suffices to show that for each $n$, there exists some countable intersection landing in $\mathfrak m^n$. Remarkably, this does not immediately follow from the fact that $R/\mathfrak m^n$ is Artinian, because intersections do not behave well with respect to quotients (even finite intersections don't). $\endgroup$ Commented Apr 1, 2017 at 4:40

1 Answer 1

12
$\begingroup$

Yes it's true.

First, in a (commutative) noetherian ring $A$, every chain of ideals is well-ordered by reverse inclusion. The supremum of ordinal types of such chains is denoted $o(A)$.

A particular case of Theorem 2.12 in [Bass71] is that for $A$ of finite Krull dimension $d$, we have $o(A)\le\omega^d$; hence $o(A)<\omega^\omega$ ($\omega^\omega$ is the countable ordinal $\sup_{d<\omega}\omega^d$). This holds in particular if $A$ is noetherian local.

Now if $A$ is a ring such that there is some family of ideals $(I_x)$ whose intersection is not achieved by any countable subfamily, then it is quite immediate that there is a descending chain of ordinal type $\omega_1$: indeed, by induction one defines for $\alpha<\omega_1$ $x_{\alpha}$ such that $I_{x_\alpha}$ does not contain $\bigcap_{\beta<\alpha}I_{x_\beta}$, and hence for $J_\alpha=\bigcap_{\beta<\alpha}I_{x_\alpha}$, we obtain $(J_{\alpha})_{\alpha<\omega_1}$ as the desired chain.

But actually Bass' main result (his Theorem 1.1) says that any chain of submodules of a f.g. module over an arbitrary noetherian ring, is countable. This shows that the result holds for arbitrary (commutative) noetherian rings and not only local ones.

The case of countable Krull dimension (encompassing the case of noetherian local rings, and finitely generated algebra over fields) is quite easy to understand using ordinal length, as in Gulliksen [Gull73].

[Bass71] H. Bass. Descending chains and the Krull ordinal of commutative Noetherian rings, Journal of Pure and Applied Algebra Volume 1, Issue 4, December 1971, Pages 347-360 (Sciencedirect link -poor scan)

[Gull73] T. Gulliksen. A theory of length for Noetherian modules. Journal of Pure and Applied Algebra Volume 3, Issue 2, June 1973, Pages 159-170. (Sciencedirect link)

$\endgroup$
2
  • 2
    $\begingroup$ This is a beautiful answer, especially the part about dropping the local condition. Just commenting to add that if we assume the ring is non-commutative, there can be descending chains of arbitrary ordinal length, even when the ring is both a left and right PID. So commutativity cannot be dropped. $\endgroup$ Commented Apr 1, 2017 at 18:07
  • 2
    $\begingroup$ One think I like in this application of Bass' result: so far I viewed Bass' paper as just superseded by Gulliksen's one (the ordinal length is more flexible than the invariant $o(A)$ on ordinal types of descending chains). But this makes me change my mind since at least we get here a natural application of the invariant $o(A)$, which is not in terms of chains of submodules, and which does not seem to immediately follow from the theory developed from ordinal length (at least, for noetherian rings of uncountable Krull dimension). $\endgroup$
    – YCor
    Commented Apr 1, 2017 at 23:57

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .