Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

If one has a finitely generated integral domain over a field, is it true that any two maximal chains of primes have equal length? If not, are there any other conditions that allows one to conclude that the maximal chains of primes have equal length?

Add: I added the condition from finitely generated algebra to finitely generated integral domain, based on the first answer I received.

share|cite|improve this question

No, it is not true: a counterexample is $k[x,y,z]/(xz,yz)$.

It is true however for a finitely-generated algebra $A$ (over a field $k$) without zero-divisors .
Indeed this follows from [Matsumura, Commutative Rings, Ch.5, (14.H)].

There he proves that $A$ is "catenary" (even "universally catenary", a stronger property) .
He also proves a formula which implies that all maximal ideals of $A$ have height $dim(A)$, and together these results show that all maximal chains of prime ideals have the same length, namely $dim(A)$.
[He defines "catenary" page 84]

share|cite|improve this answer
If you can, could you please explain why is it true in the second case? Thanks and regards, – Thome Feb 22 '12 at 23:35
Dear @Thome, I have now added an explanation and a reference. – Georges Elencwajg Feb 23 '12 at 10:22

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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