---
tag: YEJVKCY78E
kind: counterexample
name: Nagata's example of a Noetherian ring of infinite dimension
msc: [13C15, "13E05"]
uses: [0670T59HGE, MVGBS3J1P7, 0SDC934R52, 4C1AE84N5N, AN9SKEZ3VY, 00MHHZ9SNE, 87WH65FJB8]
sources:
  - url: https://en.wikipedia.org/wiki/Krull_dimension
---

## Counterexample

Let $k$ be a field and let $A = k[x_1, x_2, x_3, \ldots]$ be the [[polynomial ring|0SDC934R52]] in countably many variables.
Split the variables into consecutive blocks $B_1, B_2, B_3, \ldots$ where the block $B_i$ has exactly $i$ variables.
Let $\mathfrak{p}_i$ be the [[prime ideal|00MHHZ9SNE]] of $A$ generated by the variables of $B_i$.
Let $S = A \setminus \bigcup_{i \ge 1} \mathfrak{p}_i$, a [[multiplicative set|AN9SKEZ3VY]], and let $R = S^{-1} A$ be the [[localization|4C1AE84N5N]].
Then $R$ is a [[Noetherian ring|0670T59HGE]], but its [[Krull dimension|MVGBS3J1P7]] is infinite.

## Remarks

The maximal ideals of $R$ are exactly the ideals $\mathfrak{p}_i R$.
The localization of $R$ at $\mathfrak{p}_i R$ is a localization of a polynomial ring in $i$ variables over the field generated by the other variables.
So it is Noetherian of dimension $i$.
Each nonzero element of $R$ involves finitely many variables, so it lies in only finitely many maximal ideals.
A ring whose localizations at maximal ideals are all Noetherian and whose nonzero elements lie in finitely many maximal ideals is Noetherian.
The [[height|87WH65FJB8]] of $\mathfrak{p}_i R$ is $i$, so the dimension of $R$ is at least $i$ for every $i$.
So the statement "a Noetherian ring has finite Krull dimension" is false, although every Noetherian local ring has finite dimension.

## Explanations

### L1 human

In a Noetherian ring every ascending chain of ideals stops, so one might expect the length of chains of prime ideals to be bounded.
Each single chain of primes is finite here, but the chains can be as long as one likes.
The ring is glued from local pieces of dimension $1, 2, 3, \ldots$, each of them well behaved, and there is no common bound.