# $\operatorname{height} \mathfrak{p} + \dim A / \mathfrak{p} = \dim A$

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...

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This is not true in general. The keyword is the catenary ring: en.wikipedia.org/wiki/Catenary_ring. – Soarer Jul 3 '11 at 6:36

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.

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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 $\dim A = \operatorname{height} \mathfrak p + \dim A/\mathfrak p$.

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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