Exercise 2, § 2

Let $A$ be a Macaulay local ring, and $\mathfrak{p} \in \operatorname{Spec}(A)$. Prove that one has $\dim(\widehat{A}/\mathfrak{q}) = \dim(A/\mathfrak{p})$ for all $\mathfrak{q} \in \operatorname{Ass}(\widehat{A}/\mathfrak{p}\widehat{A})$. In particular, the ring $\widehat{A}/\mathfrak{p}\widehat{A}$ has no embedded prime ideal.