I am reading "Cohen-Macaulay Rings" by Bruns & Herzog. I have difficulty in the proposition 1.4.1:
Proposition 1.4.1
Let $R$ be a Noetherian ring, and $M$ a finite $R$-module. Then,
(a) $M$ is torsionless if and only if
(i) $M_{\mathfrak{p}}$ is torsionless for all $\mathfrak{p}\in\mathop{\mathrm{Ass}}R$, and
(ii) $\mathop{\mathrm{depth}}M_{\mathfrak{p}}\geq 1$ for $\mathfrak{p}\in\mathop{\mathrm{Spec}}R$ with $\mathop{\mathrm{depth}}R_{\mathfrak{p}}\geq 1$.
In its proof, it is written that
As to the sufficiency of (a)(i) and (ii), note that $U_{\mathfrak{p}} = 0$ for all $\mathfrak{p}\in\mathop{\mathrm{Ass}}R$ by (i), and, by (ii), $\mathop{\mathrm{depth}}U_{\mathfrak{p}}\geq 1$ if $\mathop{\mathrm{depth}}R_{\mathfrak{p}}\geq 1$. It follows that $\mathop{\mathrm{Ass}}U = \varnothing$, hence $U = 0$.
Question.
In this proof, why does $\mathop{\mathrm{Ass}}U = \varnothing$ follow from those conditions?
I see that $\mathop{\mathrm{Ass}}R\subseteq \mathop{\mathrm{Spec}}R\setminus \mathop{\mathrm{Supp}}U$, and $\mathop{\mathrm{Ass}}U\subseteq \mathop{\mathrm{Supp}}U$. I think it suffices to show that $\mathop{\mathrm{Ass}}U\subseteq \mathop{\mathrm{Ass}}R$, but I have no idea.
Best Answer
Let $\mathfrak p$ be an associated prime of $U$. Then $\operatorname{depth}U_{\mathfrak p}=0$. It follows from (ii) that $\operatorname{depth}R_{\mathfrak p}=0$, that is, $\mathfrak p$ is an associated prime of $R$. From (i) we get $U_{\mathfrak p}=0$, a contradiction. (Recall that $\operatorname{Ass}U\subseteq\operatorname{Supp}U$.)