How to define $\operatorname{dim}(\{0\})$ and $\operatorname{ht}(A)$?

53 Views Asked by At

Matsumura's "Commutative Algebra", Chapter 5, Page 72.

It follows from the definition that

$\operatorname{ht}(\mathfrak p)=\operatorname{dim}(A_{\mathfrak p})\quad (\mathfrak p\in \operatorname{Spec}(A))$,

and that, for any ideal $I$ of $A$,

$\operatorname{dim}(A/I)+\operatorname{ht}(I)\le \operatorname{dim}(A).\\$

If $I=A$, then we have $A/I=\{0\}$.

I want to know how to define $\operatorname{dim}(\{0\})$ and $\operatorname{ht}(A)$?