Let $M$ be an $A$-module. Is its injective hull affected by whether I regard $M$ as an $A$-module or $A/\mbox{Ann}(M)$-module ?

## 2 Answers

I'll follow up on what Karl said with an example closer to my own experience. Let $\mathbb{Z}$ be the ring of integers and $p$ a positive prime. Then $\mathbb{Z}/p\mathbb{Z}$ is injective as a $\mathbb{Z}/p\mathbb{Z}$ - module, being a vector space over a field, whence $\mathbb{Z}/p\mathbb{Z}$ is its own injective envelope (hull) as a $\mathbb{Z}/p\mathbb{Z}$ module. However, the injective envelope of $\mathbb{Z}/p\mathbb{Z}$ as an abelian group is $\mathbb{Z}(p^{\infty})$, which gives witness to Karl's statement that the injective envelope over $A$ can be much larger than the injective envelope over $A/\mathrm{ann}(M)$. You can play this game with $A$ any commutative Noetherian ring with 1, ann($M$) = any maximal ideal of $R$, and $M = A/I$ where $I$ is the chosen maximal ideal. Karl's example presents very limited choice for $I$ since $k[[x]]$is local. I think Proposition 2.27 and Lemma 4.24 of "Injective Modules" by Sharpe and Vamos present enough to figure out what is going on in the general case.

Yes, take $A = k[[x]]$ and $M = A/(x)$. Then as a $k = A/(x) = A/\text{Ann}(M)$-module, the injective hull of $k$ is $k$. As an $A$-module, the injective hull is much much bigger.