Example 7

If $\mathfrak{n}$ is a nilideal of $A$, $\mathrm{Spec}(A)$ is homeomorphic to $\mathrm{Spec}(A/\mathfrak{n})$ (II, § 4, No. 3, remark). Therefore one has $\dim(A/\mathfrak{n}) = \dim(A)$; in particular, one has $\dim(A) = \dim(A_{\mathrm{red}})$ where $A_{\mathrm{red}}$ is the quotient of the ring $A$ by its nilradical.