## Tag `08XS`

Chapter 45: Dualizing Complexes > Section 45.3: Injective modules

Lemma 45.3.5. Let $R$ be a ring. Let $I$ be an injective $R$-module. Let $E \subset I$ be a submodule. The following are equivalent

- $E$ is injective, and
- 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 15.52).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 $(M, \varphi)$. $\square$

The code snippet corresponding to this tag is a part of the file `dualizing.tex` and is located in lines 226–237 (see updates for more information).

```
\begin{lemma}
\label{lemma-essential-extensions-in-injective}
Let $R$ be a ring. Let $I$ be an injective $R$-module. Let $E \subset I$
be a submodule. The following are equivalent
\begin{enumerate}
\item $E$ is injective, and
\item for all $E \subset E' \subset I$ with $E \subset E'$ essential
we have $E = E'$.
\end{enumerate}
In particular, an $R$-module is injective if and only if every essential
extension is trivial.
\end{lemma}
\begin{proof}
The final assertion follows from the first and the fact that the
category of $R$-modules has enough injectives
(More on Algebra, Section \ref{more-algebra-section-injectives-modules}).
\medskip\noindent
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 $E = E'$.
\medskip\noindent
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.
\medskip\noindent
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 $(M, \varphi)$.
\end{proof}
```

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