3
$\begingroup$

I'm struggling with the definition of depth of prime ideals given in Atiyah's book:

The depth of a prime ideal $p$ is longest strictly increasing chain of prime ideals starting at $p$. Clearly $\text{depth }p=\text{dim } A/ p$

The remark just below reads:

The depth of a prime ideal, even in a Noetherian ring, may be infinite (unless the ring is local)

And they refer to the construction of a Noetherian ring of infinite Krull dimension done by Nagata.

Problem is, that in the definition of Noetherian ring, one of the equivalent conditions is given by the ascending chain condition, which would imply that any chain (not even of primes) is stationary, i.e would stop for an n big enough.

This really looks like a contradiction to me, yet after reading some questions/answers (e.g Noetherian rings and prime ideals or Does every Noetherian domain have finitely many height 1 prime ideals?) I am sure that there must be some kind of logic flaw in my argumentation/thoughts... or is it just because Atiyah decides to use infinite instead of arbitrarily long? Or is the definition of big enough n to be thought of as a supremum (i.e asymptotically stationary)? can someone help me out?

Many thanks in advance

$\endgroup$
0

1 Answer 1

2
$\begingroup$

There is a classic example, due to Nagata, of a noetherian domain $R$ such that $R$ has infinite Krull dimension, so that $(0)$ is a prime of infinite depth. You can find details in a number of places using Google ("nagata noetherian infinite krull dimension" works fine). I'll give a sketch below.

We start with a polynomial ring $k[X_1, X_2, \cdots]$ in infinitely many variables, and localize at the set of primes $\{(X_1), (X_2,X_3), (X_4,X_5,X_6), \cdots\}$. The resulting ring $R$ is noetherian because the localizations at each maximal ideal are noetherian (and each ideal is contained in only finitely many maximal ideals), but the maximal ideals have arbitrarily large height.

Note that any particular chain of ideals has finite length.

$\endgroup$
4
  • $\begingroup$ Thank you for the effort Slade, yet I had seen the example before and I was trying to find out were my thoughts had a flaw. But could you maybe help me out with the (unless the ring is local) part? $\endgroup$
    – b00n heT
    Jan 15, 2015 at 21:47
  • $\begingroup$ @b00nheT The flaw in your thoughts is this: noetherian implies that every chain has finite length, but not that every chain has length $\leq n$ for some fixed $n$. The reason that this issue doesn't exist for local rings is Krull's height theorem: the height of an ideal in a noetherian ring is bounded by the size of a generating set. In particular, an ideal in a noetherian ring always has finite height. If there are only finitely many maximal ideals, this gives a fixed bound on the length of any ideal chain. $\endgroup$
    – Slade
    Jan 16, 2015 at 0:54
  • 1
    $\begingroup$ "The resulting ring $R$ is noetherian because the localizations at each maximal ideal are noetherian" and something more. $\endgroup$
    – user26857
    Jan 18, 2015 at 22:46
  • $\begingroup$ @user26857 I've added what I think is a sufficient condition. You're right, noetherian at each maximal isn't enough. $\endgroup$
    – Slade
    Jan 19, 2015 at 4:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.