# Prime ideals of infinite depth in Noetherian rings

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 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?

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.
• @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. Jan 16, 2015 at 0:54
• "The resulting ring $R$ is noetherian because the localizations at each maximal ideal are noetherian" and something more. Jan 18, 2015 at 22:46