Is $\operatorname{height} \mathfrak{p} + \dim A / \mathfrak{p} = \dim A$ true?

Question

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...

Answer 1

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.

Answer 2

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

Answer 3

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].

Answer 4

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 $\dim \mathcal{O}_{X,\eta} + \dim Y = \dim X$.

Answer 5

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.