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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.

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1 Answer 1

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]

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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

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