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Let $A$ be a Noetherian ring, $\mathfrak{p},\mathfrak{q}\subset A$ distinct prime ideals of the same height, $N$ an $A_\mathfrak{p}$-module of finite length. Then is it true that $$\operatorname{Hom}_A(N,E(A/\mathfrak{q}))=0,$$ where $E(A/\mathfrak{q})$ is the injective hull of $A/\mathfrak{q}$?

The main trouble here is that $N$ may not be finitely generated over $A$. If it were, I would use formulas like \begin{eqnarray} \operatorname{Ass}\operatorname{Hom}_A(N,E(A/\mathfrak{q}))&=&\operatorname{Supp}_AN\cap\operatorname{Ass}E(A/\mathfrak{q})\ &=&V(\mathfrak{p})\cap\lbrace\mathfrak{q}\rbrace\ &=&\emptyset. \end{eqnarray}

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Presumably $E$ stands for injective envelope? – Stephen Mar 4 '12 at 19:04
@Steve Yes; Sorry, I edited that in. – ashpool Mar 4 '12 at 19:31
up vote 2 down vote accepted

Suppose $\phi\in\operatorname{Hom}_A(N,E(A/\mathfrak{q}))$, and $\phi(n)=e\neq0$. We will derive a contradiction.

Since $N$ is of finite length over $A_{\mathfrak{p}}$, $\mathfrak{p}^kN=0$ for some $k$. Since $\mathfrak{p}\not\subset\mathfrak{q}$, $\exists a\in\mathfrak{p}^k\backslash\mathfrak{q}$, so $a$ acts as a unit on $E(A/\mathfrak{q})$. Then $$0=\phi(an)=ae\neq0,$$ a contradiction.

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