Height and coheight of an ideal

Question

Given an ideal $\mathfrak{a}$, Matsumura defined the height of $\mathfrak{a}$ as: $$\text{ht}(\mathfrak{a})=\inf_{\mathfrak{p}\in V(\mathfrak{a})}\text{ht}(\mathfrak{p})$$ He states that: $$\text{ht}(\mathfrak{a})+\dim(A/\mathfrak{a})\leq \dim(A)$$ Any ideas on how to show this? I know I have to use the correspondence theorem. I'm just having a hard time writting it properly with the definitions of height and dimension. Thanks in advance!

Answer 1

Hints.
$\operatorname{ht}\mathfrak{p}+\dim A/\mathfrak{p}\leq \dim A$ for any prime ideal $\mathfrak p$ (why?).
$\dim A/\mathfrak a=\sup_{\mathfrak{p}\in V(\mathfrak{a})}\dim A/\mathfrak p$.