
  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...

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. 
