The Stacks project

Lemma 47.3.5. Let $R$ be a ring. Let $I$ be an injective $R$-module. Let $E \subset I$ be a submodule. The following are equivalent

  1. $E$ is injective, and

  2. for all $E \subset E' \subset I$ with $E \subset E'$ essential we have $E = E'$.

In particular, an $R$-module is injective if and only if every essential extension is trivial.

Proof. The final assertion follows from the first and the fact that the category of $R$-modules has enough injectives (More on Algebra, Section 15.55).

Assume (1). Let $E \subset E' \subset I$ as in (2). Then the map $\text{id}_ E : E \to E$ can be extended to a map $\alpha : E' \to E$. The kernel of $\alpha $ has to be zero because it intersects $E$ trivially and $E'$ is an essential extension. Hence $E = E'$.

Assume (2). Let $M \subset N$ be $R$-modules and let $\varphi : M \to E$ be an $R$-module map. In order to prove (1) we have to show that $\varphi $ extends to a morphism $N \to E$. Consider the set $\mathcal{S}$ of pairs $(M', \varphi ')$ where $M \subset M' \subset N$ and $\varphi ' : M' \to E$ is an $R$-module map agreeing with $\varphi $ on $M$. We define an ordering on $\mathcal{S}$ by the rule $(M', \varphi ') \leq (M'', \varphi '')$ if and only if $M' \subset M''$ and $\varphi ''|_{M'} = \varphi '$. It is clear that we can take the maximum of a totally ordered subset of $\mathcal{S}$. Hence by Zorn's lemma we may assume $(M, \varphi )$ is a maximal element.

Choose an extension $\psi : N \to I$ of $\varphi $ composed with the inclusion $E \to I$. This is possible as $I$ is injective. If $\psi (N) \subset E$, then $\psi $ is the desired extension. If $\psi (N)$ is not contained in $E$, then by (2) the inclusion $E \subset E + \psi (N)$ is not essential. hence we can find a nonzero submodule $K \subset E + \psi (N)$ meeting $E$ in $0$. This means that $M' = \psi ^{-1}(E + K)$ strictly contains $M$. Thus we can extend $\varphi $ to $M'$ using

\[ M' \xrightarrow {\psi |_{M'}} E + K \to (E + K)/K = E \]

This contradicts the maximality of $(M, \varphi )$. $\square$

Comments (1)

Comment #7938 by on

This proof seems to adapt to any Grothendieck abelian category. A reference for this general statement: Lurie, Spectral Algebraic Geometry, Lemma C.5.6.4.

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  • 2 comment(s) on Section 47.3: Injective modules

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