Let $R$ be a Noetherian commutative ring. Let $E(M)$ denote the injective hull of $M$. I want to show that $E(M)_\mathfrak{p}\simeq E(M_\mathfrak{p})$ for any $\mathfrak{p}\in \text{Spec}(R)$.

To do so I was trying to use the characterization of the injective hull, $M\subseteq E$ is an essential extension and $E$ is an injective $R$-module. It is easy to see that $M_\mathfrak{p}\subseteq E(M)_\mathfrak{p}$, and that the latter is an injective $R_\mathfrak{p}$-module. But I'm struggling to see that the extension is essential. My naive attempt: Let $0\neq \frac{x}{s}\in E(M)_\mathfrak{p}$, where $s\not\in\mathfrak{p}$. Since $M\subseteq E(M)$ is essential, there is $r\in R$ such that $rx\in M$ and $rx\neq 0$. Then: $$\dfrac{r}{1}\cdot \dfrac{x}{s}=\dfrac{rx}{s}\in M_\mathfrak{p}$$ but I'm not assure $rx/s\neq 0$. Is this a good strategy? Is there another way of proving the result I'm looking for? Thanks in advance.


1 Answer 1


In Lemma 6 of Chapter 18 of 'Commutative ring theory' Matsumura proves the following.

Lemma: Let $A$ be a Noetherian ring, $S \subset A$ a multiplicative set, $M$ and $A$-module and $N \subset M$ a submodule. Assume that $M$ is an essential extension of $N$; then $M_S$ is an essential extension of $M_S$.


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