Associated primes of quotient module

Question

Let $R$ be a Noetherian local ring of Krull dimension $d$, $M$ a finitely generated module over $R$. Suppose $\dim M=d$ and $K$ is a submodule of $M$ maximal with respect to the property that $\dim K\leq d-1$, then can we deduce that $\operatorname{Ass}_R(M/K)\subseteq \operatorname{Ass}(R)$?

Answer 1

I think we can.

Consider a minimal primary decomposition of $K$, say $K=Q_1\cap\cdots\cap Q_n$ with $P_i=\sqrt{Q_i}$. Since $\dim K<d$ it follows that $\dim M/K=d$. On the other hand, one knows that $\dim M/K=\max_{1\le i\le n}\dim R/P_i.$
If $\dim R/P_i=d$, then $P_i$ is minimal, and therefore $P_i\in\operatorname{Ass}(R)$.
If $\dim R/P_i<d$, then consider $K'=\cap_{j\ne i}Q_j$; we have $K\subsetneq K'$ and since $\dim K'=\max(\dim K,\dim K'/K)<d$ we obtain a contradiction. (Note that $K'/K=K'/K'\cap Q_i\simeq(K'+Q_i)/Q_i\subseteq M/Q_i$, so $\operatorname{Ass}(K'/K)\subseteq\operatorname{Ass}(M/Q_i)=\{P_i\}$, hence equality.)