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 15.54).

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 (0)

There are also:

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

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 08XS. Beware of the difference between the letter 'O' and the digit '0'.