# The Stacks Project

## Tag 08XU

Lemma 45.3.7. Let $R$ be a ring. Let $E$ be an $R$-module. The following are equivalent

1. $E$ is an injective $R$-module, and
2. given an ideal $I \subset R$ and a module map $\varphi : I \to E$ there exists an extension of $\varphi$ to an $R$-module map $R \to E$.

Proof. The implication (1) $\Rightarrow$ (2) follows from the definitions. Thus we assume (2) holds and we prove (1). First proof: The lemma follows from More on Algebra, Lemma 15.52.4. Second proof: Since $R$ is a generator for the category of $R$-modules, the lemma follows from Injectives, Lemma 19.11.6.

Third proof: We have to show that every essential extension $E \subset E'$ is trivial, see Lemma 45.3.5. Pick $x \in E'$ and set $I = \{f \in R \mid fx \in E\}$. The map $I \to E$, $f \mapsto fx$ extends to $\psi : R \to E$ by (2). Then $x' = x - \psi(1)$ is an element of $E'$ whose annihilator in $E'/E$ is $I$ and which is annihilated by $I$ as an element of $E'$. Thus $Rx' = (R/I)x'$ does not intersect $E$. Since $E \subset E'$ is an essential extension it follows that $x' \in E$ as desired. $\square$

The code snippet corresponding to this tag is a part of the file dualizing.tex and is located in lines 248–256 (see updates for more information).

\begin{lemma}
\label{lemma-characterize-injective}
Let $R$ be a ring. Let $E$ be an $R$-module. The following are equivalent
\begin{enumerate}
\item $E$ is an injective $R$-module, and
\item given an ideal $I \subset R$ and a module map $\varphi : I \to E$
there exists an extension of $\varphi$ to an $R$-module map $R \to E$.
\end{enumerate}
\end{lemma}

\begin{proof}
The implication (1) $\Rightarrow$ (2) follows from the definitions.
Thus we assume (2) holds and we prove (1).
First proof: The lemma follows from
More on Algebra, Lemma \ref{more-algebra-lemma-characterize-injective-bis}.
Second proof: Since $R$ is a generator for the category of $R$-modules,
the lemma follows from
Injectives, Lemma \ref{injectives-lemma-characterize-injective}.

\medskip\noindent
Third proof: We have to show that every essential extension $E \subset E'$
is trivial, see Lemma \ref{lemma-essential-extensions-in-injective}.
Pick $x \in E'$ and set $I = \{f \in R \mid fx \in E\}$.
The map $I \to E$, $f \mapsto fx$ extends to $\psi : R \to E$ by (2).
Then $x' = x - \psi(1)$ is an element of $E'$ whose annihilator in
$E'/E$ is $I$ and which is annihilated by $I$ as an element of $E'$.
Thus $Rx' = (R/I)x'$ does not intersect $E$. Since $E \subset E'$
is an essential extension it follows that $x' \in E$ as desired.
\end{proof}

Comment #2764 by Dario Weißmann on August 10, 2017 a 8:35 pm UTC

This result is also stated in Lemma 19.2.6, is this intentional? In Lemma 19.2.6 (or rather directely above) this is called "criterion of Baer". I think it's quite nice that the result has a name, although "Baer's criterion" sounds better.

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