Suppose $\mathfrak{p},\mathfrak{q} \in \mathrm{Ass}(M)$ such that $\mathfrak{p}\subsetneq \mathfrak{q}$. Does there always exist $\mathfrak{p}'\in \mathrm{Ass}(M)$ such that $\mathfrak{p}\subsetneq \mathfrak{p}'\subset\mathfrak{q}$ and $\mathrm{height}(\mathfrak{p}')=\mathrm{height}(\mathfrak{p})+1$? Assume that the ring is Noetherian or that the module is finitely generated if necessary.
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Try $R=k[x_1,\ldots, x_n]/(x_1^2, x_1x_2,\ldots, x_1x_n)$ and $M=R$.
Mohan
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Did you mean $\mathfrak{p}$ and $\mathfrak{q}$ to be $(x_1)$ and $(x_1,\dots, x_n)$ respectively? – Jehu314 Jun 11 '20 at 15:42
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@Jehu314 Yes. Since $n$ can be arbitrary. – Mohan Jun 11 '20 at 18:18