# If $\ell_{A_\mathfrak{p}}(N)<\infty$, then is it true that $\operatorname{Hom}_A(N,E(A/\mathfrak{q}))=0$?

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}

-

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

## 1 Answer

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.

-
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 21, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.7926022410392761, "perplexity_flag": "head"}