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

Let A be an integral domain of finite Krull dimension. Let p be a prime ideal. Is it true that heightp+dimA/p=dimA
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 heightp+dimA/pdimA 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

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 pSpec(A) we have height(p)+dimA/p=dim(A)(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,m) be a discrete valuation ring whose maximal ideal has uniformizing parameter π, i.e. m=(π). Let A=R[T], the polynomial ring over R. The ring A has dimension 2. Then for the maximal ideal p=(πT1), the relation (DIM) is false:
height(p)+dimA/p=1+0=12=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 YX we have, as hoped, dimension(Y)+codimension(Y) = 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 p 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.

Attribution
Source : Link , Question Author : Zhen Lin , Answer Author : Georges Elencwajg

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