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...

Martin Sleziak
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Zhen Lin
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    This is not true in general. The keyword is the catenary ring: http://en.wikipedia.org/wiki/Catenary_ring. –  Jul 03 '11 at 06:36
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    As George's answer shows, "catenary" is not enough. – Manos Oct 01 '20 at 05:58
  • This is explained in Matsumura's Commutative algebra book though. See (14.H) COROLLARY 3. – Zero Apr 29 '21 at 16:39

5 Answers5


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.

Georges Elencwajg
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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 03 '11 at 20:55
  • Dear Akhil,I don't know about equidimensional rings, although I more or less guess what it means . I have a suspicion that this would be a little tautological, but I'd love to be shown wrong: what family of rings not finitely generated over a field would have this property? – Georges Elencwajg Jul 03 '11 at 22:21
  • Thanks for the this answer! Seems like I had a big misunderstanding of this issue.. –  Jul 03 '11 at 22:50
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    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 03 '11 at 23:01
  • @Georges: Dear Georges, am I missing something here? Consider the variety which is a disjoint union of a line and a plane. This has dimension two. If we pick the subvariety consisting of a point in the line, then the dimension is zero and the codimension is one, no? I certainly agree that the equality holds for irreducible varieties though. – Akhil Mathew Jul 04 '11 at 02:51
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    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 04 '11 at 02:54
  • Dear Akhil, you cannot take a point plus a line because a variety is reduced and irreducible in my context, since it corresponds in the algebro-geometric dictionary to the finitely generated integral algebra of Fact 2. As for EGA 0-IV.14, I've just had a look. It seems to be a compendium of definitions and soft results with two-line proofs. The authors remark themselves that some results are even valid in any ordered set. So, again, I don't see how these ideas could yield rings satisfying (DIM). Anyway, thanks for the reference and your interest. (By the way, have you read all of EGA?) – Georges Elencwajg Jul 04 '11 at 08:20
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    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 04 '11 at 13:23
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    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
  • Incidentally, this nice counterexample is already stated in EGA IV, Remarque 5.2.5(i). – fherzig Jun 23 '21 at 14:59
  • Any reference for Fact 3? – Cyclicduck Jun 25 '21 at 02:42
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    @Cyclicduck [The Stacks Project](https://stacks.math.columbia.edu/tag/00N7) Lemma 10.104.4 – Georges Elencwajg Jun 25 '21 at 17:06

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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John M
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  • You are specifying a prime $p$ here, while OP was asking about arbitrary prime. –  Jul 03 '11 at 06:37
  • @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 03 '11 at 06:59

Although this is an old question, I thought it was worth mentioning a recent paper by Heinrich that corrects the statements in EGA$0_{\text{IV}}$ mentioned in the comments.

Let us start with some definitions (following [Heinrich, Def. 1.2, Prop. 4.1]):

Definition. Let $X$ be a topological space which is $T_0$, noetherian, and finite dimensional.

  1. The space $X$ is biequidimensional if all maximal chains of irreducible closed subsets of $X$ have the same length.
  2. The space $X$ is weakly biequidimensional if it is equidimensional, equicodimensional, and catenary.

The often cited result from EGA$0_{\text{IV}}$ is the following:

Claim [EGA$0_{\text{IV}}$, Prop. 14.3.3]. Let $X$ be a topological space which is $T_0$, noetherian, and finite dimensional. The following are equivalent:

  1. The space $X$ is biequidimensional.
  2. The space $X$ is weakly biequidimensional.
  3. The space $X$ is equicodimensional and for every inclusion of irreducible closed subsets $Y \subseteq Z$ in $X$, we have $$\dim(Z) = \dim(Y) + \operatorname{codim}(Y,Z).$$
  4. The space $X$ is equicodimensional and for every inclusion of irreducible closed subsets $Y \subseteq Z$ in $X$, we have $$\operatorname{codim}(Y,X) = \operatorname{codim}(Y,Z) + \operatorname{codim}(Z,X).$$

This is not quite correct, as was found independently by Gabber and by Chen (see [ILO, Exp. XV, §2.4, footnote (i) on p. 196]), and also by Heinrich [Heinrich].

Gabber and Heinrich both noted that (1), (3), and (4) are equivalent (see [Heinrich, Lem. 2.3] for a proof), and Heinrich showed that these conditions imply (2) [Heinrich, Lem. 2.1]. Gabber and Heinrich both gave examples where (2) does not imply (3); we reproduce Heinrich's here:

Example [Heinrich, Ex. 3.7]. The ring $A$ obtained by localizing the ring $$\frac{k[v, w, x, y]}{(vy, wy)}$$ away from the union $(v,w,x,y-1) \cup (v,w,y)$ is weakly biequidimensional but does not satisfy (3): setting $Y = V(v,w,x,y-1) \subsetneq V(v,w) = Z$, we have $$\dim(Z) = 2 > 0 + 1 = \dim(Y) + \operatorname{codim}(Y,Z).$$ See [Heinrich, Ex. 3.7] for details.

A preprint by Emerton and Gee gives a correct variant of the Claim above; see [Emerton and Gee, Lem. 2.32]. The basic difference is that the Claim is true if $X$ is assumed to be irreducible. Gabber also gives a variant where (2) is replaced by "$X$ is catenary and equidimensional and its irreducible components are equicodimensional" [ILO, Exp. XV, §2.4, footnote (i) on p. 196].

Takumi Murayama
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The following is a useful situation where $A$ is not necessarily an integral domain. It generalizes Fact 2 from Georges's answer and specializes (1) $\rightarrow$ (3) from Takumi's answer.

Let $A$ be a finitely generated algebra over a field $k$. Then $A = S/I$ where $S$ is a polynomial ring over $k$ and $I$ an ideal of $S$.

Definition 1. $A$ is called equidimensional if all minimal primes over $I$ have the same height.

Theorem 1. Suppose $A$ is an equidimensional $k$-algebra. Then $\operatorname{height} P + \dim A/P = \dim A$ for any $P \in \operatorname{Spec} A$.

Here is the scheme-theoretic picture. Let $X$ be a $k$-scheme.

Definition 2. $X$ is called equidimensional if its irreducible components have all the same dimension.

Theorem 2. Let $X$ be an equidimensional $k$-scheme locally of finite type. Let $Y$ be an irreducible closed subscheme with generic point $\eta$. Then $\dim \mathcal{O}_{X,\eta} + \dim Y = \dim X$.

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If $A$ is a finitely generated $k$-algebra, and $A$ is an integral domain, then for any prime $\mathfrak{p}\subset A$, $\operatorname{ht}\mathfrak{p} + \dim(A/\mathfrak{p}) = \dim A$.

This is proved in [Bosch, Section 3.3, Proposition 8] using induction on $\dim A$ and the going-down theorem.

  • Please read the other answers before posting. If you only wanted to add a new reference to a well known result, then wait untill you can comment and post it as a comment. – user26857 Apr 17 '22 at 19:41