Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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

Let $A$ be an integral domain of finite Krull dimension. Let $\mathfrak{p}$ be a prime ideal. Is it true that $$\operatorname{height} \mathfrak{p} + \dim A / \mathfrak{p} = \dim A$$ where $\dim$ refers to the Krull dimension of a ring?

Hartshorne states it as Theorem 1.8A in Chapter I (for the case $A$ a finitely-generated $k$-algebra which is an integral domain) and cites Matsumura and Atiyah–Macdonald, but I haven't been able to find anything which looks relevant in either. (Disclaimer: I know nothing about dimension theory, and very little commutative algebra.) If it is true (under additional assumptions, if need be), where can I find a complete proof?

It is obvious that $$\operatorname{height} \mathfrak{p} + \dim A/\mathfrak{p} \le \dim A$$ by a lifting argument, but the reverse inequality is eluding me. Localisation doesn't seem to be the answer, since localisation can change the dimension...

share|cite|improve this question
This is not true in general. The keyword is the catenary ring: – Soarer Jul 3 '11 at 6:36
up vote 29 down vote accepted

Yours is a very interesting and subtle question, which often generates confusion. First let us give a name to the property you are interested in: a ring $A$ will be said to satisfy (DIM) if for all $\mathfrak p \in \operatorname{Spec}(A)$ we have $$\operatorname{height}(\mathfrak p) +\dim A/\mathfrak p=\dim(A) \quad \quad (\text{DIM})$$
The main misconception is to believe that this follows from catenarity:
Fact 1: A catenary ring, or even a universally catenary ring, does not satisfy (DIM) in general.
Counterexample: Let $(R,\mathfrak m)$ be a discrete valuation ring whose maximal ideal has uniformizing parameter $\pi$, i.e. $\mathfrak m =(\pi)$. Let $A=R[T]$, the polynomial ring over $R$. The ring $A$ has dimension $2.$ Then for the maximal ideal $\mathfrak p=(\pi T-1)$, the relation (DIM) is false: $\operatorname{height}(\mathfrak p)+\dim A/\mathfrak p= 1+0=1\neq 2=\dim (A)$.
And this even though $A$ is as nice as can be: an integral domain, noetherian, regular, universally catenary,...

Happily here are two positive results:

Fact 2: A finitely generated integral algebra over a field satisfies (DIM) (and is universally catenary).
So, by the algebro-geometric dictionary, an affine variety $X$ has the pleasant property that for each integral subvariety $Y\subset X$ we have, as hoped, $\operatorname{dimension}(Y) + \operatorname{codimension}(Y)$ $=$ $\operatorname{dimension}(X).$

Fact 3: A Cohen-Macaulay local ring satisfies (DIM) (and is universally catenary).
For example a regular ring is Cohen-Macaulay. This "explains" why my counter-example above was not local.

The paradox resolved. How is it possible for a catenary ring $A$ not to satisfy (DIM)? Here is how. If you have an inclusion of two primes $\mathfrak p\subsetneq \mathfrak q$ catenarity says that you can complete it to a saturated chain of primes $\mathfrak p\subsetneq \mathfrak p_1\subsetneq \ldots \subsetneq \mathfrak p_{r-1} \subsetneq \mathfrak q$ and that all such completions will have length the same length $r$. Fine. But what can you say if you have just one prime $\mathfrak p$ ? Not much! The catenary ring $A$ may have dimension $\dim(A) > \operatorname{height}( \mathfrak p) +\dim(A/\mathfrak p)$ because it possesses a long chain of primes avoiding the prime $\mathfrak p$ altogether. In my counterexample above the only saturated chain of primes containing $\mathfrak p=(\pi T-1)$ is $0\subsetneq \mathfrak p$. However the ring $A$ has dimension 2 because of the saturated chain of primes $0\subsetneq (\pi) \subsetneq (\pi,T)$, which avoids $\mathfrak p$.

Addendum. Here is why the ideal $\mathfrak p$ in the counter-example is maximal. We have $A/\mathfrak p=R[T]/(\pi T-1)=R[1/\pi]=\operatorname{Frac}(R)$, since the fraction field of a discrete valuation ring can be obtained just by inverting a uniformizing parameter. So $A/\mathfrak p$ is a field and $\mathfrak p$ is maximal.

share|cite|improve this answer
Thanks for clarifying the catenary thing. I think the correct statement is that a "catenary, equidimensional (equicodimensional?) ring satisfies DIM. – Akhil Mathew Jul 3 '11 at 20:55
Dear Soarer, don't worry: that catenary implies (DIM) is one of the most widespread misconceptions I have ever met (and I spent quite some time trying to clear these issues for myself). – Georges Elencwajg Jul 3 '11 at 23:01
Equidimensional is defined in EGA 0-IV.14 as all irreducible components having the same dimension; equicodimensional is that all minimal irreducible closed sets have the same codimension. A noetherian space of finite dimension is biequidimensional if and only if the equality (DIM) is verified, if I am not mistaken. – Akhil Mathew Jul 4 '11 at 2:54
Ah, ok. Then I agree with your claim. Also, I agree that 0-IV.14 is mostly content-free. The real content, that (DIM) holds for integral domains finitely generated over a field, is in IV-5.2.1 (though catenary-ness is deduced from the fact that a regular local ring is universally catenary; the way I always thought of the result was via the transcendence degree additivity. (I certainly haven't read all of EGA.) – Akhil Mathew Jul 4 '11 at 13:23
I think it is also worth noticing that when a ring satisfies (DIM) for prime ideals, then the dimensional equality holds for any ideal: If $I$ is an arbitrary ideal and $\mathfrak{p}$ a prime ideal over $I$ with $ht(I) = ht(p)$, then the quotient map $R/I \to R/\mathfrak{p}$ is surjective, so the induced spectra map is injective, yielding $\dim R/I \ge \dim R/\mathfrak{p}$. So we have $\dim R/\mathfrak{p} \le \dim R/I \le \dim R - ht(I) = \dim R- ht(\mathfrak{p})$, and since DIM holds for prime ideals, the equality follows. – Sebastian Jul 31 '14 at 13:24

The statement with the hypotheses given in Hartshorne is true.

For a reference, see COR 13.4 on pg. 290 of Eisenbud's Commutative Algebra.

The general idea of proof is this: Consider a maximal chain of prime ideals in $A$ which includes the given prime $\mathfrak p$, the length of which is dim $A$ (see Thm A, pg. 290 of Eisenbud). It follows that $\dim A = \operatorname{height} \mathfrak p + \dim A/\mathfrak p$.

share|cite|improve this answer
You are specifying a prime $p$ here, while OP was asking about arbitrary prime. – Soarer Jul 3 '11 at 6:37
@Soarer: I'm not sure what you mean. Given an arbitrary prime $\mathfrak p$, construct a maximal chain of prime ideals which includes $\mathfrak p$. Every maximal chain of primes in $A$ has the same length, i.e. the Krull dimension of $A$, although this assertion itself is not trivial, but the full proof is in Eisenbud. – John M Jul 3 '11 at 6:59
Sorry I misunderstood you. In that case I think the whole point of the question is your assertion. – Soarer Jul 3 '11 at 7:06
@Soarer: You're right. I'm being unclear. I'll edit my answer. – John M Jul 3 '11 at 7:13

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.