Let $A$ be a ring (assume Noetherian if necessary). Then it is clear to me that we have $$ \sup_{\operatorname{ht}(\mathfrak{p}) = 0} \dim A/\mathfrak{p}\leq \dim A.$$
However, I can't seem to prove the other inequality, and here is why I think it is true. For the topological space $\operatorname{Spec} A$, we define the dimension of $\operatorname{Spec} A$ to be the maximum of the dimensions of its irreducible components. However, the dimension of an irreducible component is the dimension of $A/\mathfrak{p}$ for $\mathfrak{p}$ a minimal prime, so in essence the equality I'm asking for is a tautology.
Hence, in ring theoretic terms, I should be able to prove the other inequality trivially. Why am I confused, or is it not true?
Yes, since any prime ${\mathfrak p}$ in a commutative ring $R$ contains a minimal prime. For the proof, replacing $R$ by $R_{\mathfrak p}$ shows that it suffices to prove that any commutative ring has a minimal prime. This follows from Zorn's lemma, since for any linearly ordered family $\{{\mathfrak p}_i\}_i$ of primes in $R$ the intersection $\bigcap_i {\mathfrak p}_i$ is a prime, too.
Addendum From this, you can conclude that $\text{dim}\ A$ equals the supremum of the lengths of chains of prime ideals that begin with a minimal prime, and this is precisely $\sup\limits_{\text{ht}\ {\mathfrak p}=0} \text{dim}\ A/{\mathfrak p}$.