Let $M$ be a module and $I$ an injective module.
We know that $I/M$ is not injective in general. But can we assume some conditions on $M$ such that $I/M$ is injective?
Thank you!
Let $M$ be a module and $I$ an injective module.
We know that $I/M$ is not injective in general. But can we assume some conditions on $M$ such that $I/M$ is injective?
Thank you!
Copyright © 2021 JogjaFile Inc.
A module $J$ is injective iff $\mathrm{Ext}^1(N,J)=0$ for all modules $N$. In your case, we can look at the long exact sequence of Ext associated to the short exact sequence $0\to M\to I\to I/M\to 0$: $$\mathrm{Ext}^1(N,I)\to \mathrm{Ext}^1(N,I/M) \to \mathrm{Ext}^2(N,M) \to \mathrm{Ext}^2(N,I)$$ Since $I$ is injective, the first and last terms are $0$, so $\mathrm{Ext}^1(N,I/M)\cong \mathrm{Ext}^2(N,M)$. We thus conclude that $I/M$ is injective iff $\mathrm{Ext}^2(N,M)=0$ for all $N$. That is, $I/M$ is injective iff $M$ has injective dimension $\leq 1$.
Of course, this is somewhat tautological when you unravel the definition of "injective dimension" or $\mathrm{Ext}^2(N,M)=0$ in terms of an injective resolution of $M$. But I don't think there's any really better criterion you can give in general, and in particular this criterion makes it clear that the injectivity of $I/M$ depends only on $M$, not on $I$.