# The Stacks Project

## Tag 01D8

### 15.52. Injective modules

Some lemmas on injective modules.

Definition 15.52.1. Let $R$ be a ring. An $R$-module $J$ is injective if and only if the functor $\mathop{\mathrm{Hom}}\nolimits_R(-, J) : \text{Mod}_R \to \text{Mod}_R$ is an exact functor.

The functor $\mathop{\mathrm{Hom}}\nolimits_R(- , M)$ is left exact for any $R$-module $M$, see Algebra, Lemma 10.10.1. Hence the condition for $J$ to be injective really signifies that given an injection of $R$-modules $M \to M'$ the map $\mathop{\mathrm{Hom}}\nolimits_R(M', J) \to \mathop{\mathrm{Hom}}\nolimits_R(M, J)$ is surjective.

Before we reformulate this in terms of ${Ext}$-modules we discuss the relationship between $\mathop{\mathrm{Ext}}\nolimits^1_R(M, N)$ and extensions as in Homology, Section 12.6.

Lemma 15.52.2. Let $R$ be a ring. Let $\mathcal{A}$ be the abelian category of $R$-modules. There is a canonical isomorphism $\mathop{\mathrm{Ext}}\nolimits_\mathcal{A}(M, N) = \text{Ext}^1_R(M, N)$ compatible with the long exact sequences of Algebra, Lemmas 10.70.6 and 10.70.7 and the $6$-term exact sequences of Homology, Lemma 12.6.4.

Proof. Omitted. $\square$

Lemma 15.52.3. Let $R$ be a ring. Let $J$ be an $R$-module. The following are equivalent

1. $J$ is injective,
2. $\mathop{\mathrm{Ext}}\nolimits^1_R(M, J) = 0$ for every $R$-module $M$.

Proof. Let $0 \to M'' \to M' \to M \to 0$ be a short exact sequence of $R$-modules. Consider the long exact sequence $$\begin{matrix} 0 \to \mathop{\mathrm{Hom}}\nolimits_R(M, J) \to \mathop{\mathrm{Hom}}\nolimits_R(M', J) \to \mathop{\mathrm{Hom}}\nolimits_R(M'', J) \\ \phantom{0~} \to \mathop{\mathrm{Ext}}\nolimits^1_R(M, J) \to \mathop{\mathrm{Ext}}\nolimits^1_R(M', J) \to \mathop{\mathrm{Ext}}\nolimits^1_R(M'', J) \to \ldots \end{matrix}$$ of Algebra, Lemma 10.70.7. Thus we see that (2) implies (1). Conversely, if $J$ is injective then the $\mathop{\mathrm{Ext}}\nolimits$-group is zero by Homology, Lemma 12.24.2 and Lemma 15.52.2. $\square$

Lemma 15.52.4. Let $R$ be a ring. Let $J$ be an $R$-module. The following are equivalent

1. $J$ is injective,
2. $\mathop{\mathrm{Ext}}\nolimits^1_R(R/I, J) = 0$ for every ideal $I \subset R$, and
3. for an ideal $I \subset R$ and module map $I \to J$ there exists an extension $R \to J$.

Proof. We have seen the implication (1) $\Leftrightarrow$ (2) in Lemma 15.52.3. In this proof we will show that (1) $\Leftrightarrow$ (3) which is known as Baer's criterion.

Assume (1). Given a module map $I \to J$ as in (3) we find the extension $R \to J$ because the map $\mathop{\mathrm{Hom}}\nolimits_R(R, J) \to \mathop{\mathrm{Hom}}\nolimits_R(I, J)$ is surjective by definition.

Assume (3). Let $M \subset N$ be an inclusion of $R$-modules. Let $\varphi : M \to J$ be a homomorphism. We will show that $\varphi$ extends to $N$ which finishes the proof of the lemma. Consider the set of homomorphisms $\varphi' : M' \to J$ with $M \subset M' \subset N$ and $\varphi'|_M = \varphi$. Define $(M', \varphi') \geq (M'', \varphi'')$ if and only if $M' \supset M''$ and $\varphi'|_{M''} = \varphi''$. If $(M_i, \varphi_i)_{i \in I}$ is a totally ordered collection of such pairs, then we obtain a map $\bigcup_{i \in I} M_i \to J$ defined by $a \in M_i$ maps to $\varphi_i(a)$. Thus Zorn's lemma applies. To conclude we have to show that if the pair $(M', \varphi')$ is maximal then $M' = N$. In other words, it suffices to show, given any subgroup $M \subset N$, $M \not = N$ and any $\varphi : M \to J$, then we can find $\varphi' : M' \to J$ with $M \subset M' \subset N$ such that (a) the inclusion $M \subset M'$ is strict, and (b) the morphism $\varphi'$ extends $\varphi$.

To prove this, pick $x \in N$, $x \not \in M$. Let $I = \{f \in R \mid fx \in M\}$. This is an ideal of $R$. Define a homomorphism $\psi : I \to J$ by $f \mapsto \varphi(fx)$. Extend to a map $\tilde\psi : R \to J$ which is possible by assumption (3). By our choice of $I$ the kernel of $M \oplus R \to J$, $(y, f) \mapsto y - \tilde\psi(f)$ contains the kernel of the map $M \oplus R \to N$, $(y, f) \mapsto y + fx$. Hence this homomorphism factors through the image $M' = M + Rx$ and this extends the given homomorphism as desired. $\square$

In the rest of this section we prove that there are enough injective modules over a ring $R$. We start with the fact that $\mathbf{Q}/\mathbf{Z}$ is an injective abelian group. This follows from Lemma 15.51.1.

Definition 15.52.5. Let $R$ be a ring.

1. For any $R$-module $M$ over $R$ we denote $M^\vee = \mathop{\mathrm{Hom}}\nolimits(M, \mathbf{Q}/\mathbf{Z})$ with its natural $R$-module structure. We think of $M \mapsto M^\vee$ as a contravariant functor from the category of $R$-modules to itself.
2. For any $R$-module $M$ we denote $$F(M) = \bigoplus\nolimits_{m \in M} R[m]$$ the free module with basis given by the elements $[m]$ with $m \in M$. We let $F(M)\to M$, $\sum f_i [m_i] \mapsto \sum f_i m_i$ be the natural surjection of $R$-modules. We think of $M \mapsto (F(M) \to M)$ as a functor from the category of $R$-modules to the category of arrows in $R$-modules.

Lemma 15.52.6. Let $R$ be a ring. The functor $M \mapsto M^\vee$ is exact.

Proof. This because $\mathbf{Q}/\mathbf{Z}$ is an injective abelian group by Lemma 15.51.1. $\square$

There is a canonical map $ev : M \to (M^\vee)^\vee$ given by evaluation: given $x \in M$ we let $ev(x) \in (M^\vee)^\vee = \mathop{\mathrm{Hom}}\nolimits(M^\vee, \mathbf{Q}/\mathbf{Z})$ be the map $\varphi \mapsto \varphi(x)$.

Lemma 15.52.7. For any $R$-module $M$ the evaluation map $ev : M \to (M^\vee)^\vee$ is injective.

Proof. You can check this using that $\mathbf{Q}/\mathbf{Z}$ is an injective abelian group. Namely, if $x \in M$ is not zero, then let $M' \subset M$ be the cyclic group it generates. There exists a nonzero map $M' \to \mathbf{Q}/\mathbf{Z}$ which necessarily does not annihilate $x$. This extends to a map $\varphi : M \to \mathbf{Q}/\mathbf{Z}$ and then $ev(x)(\varphi) = \varphi(x) \not = 0$. $\square$

The canonical surjection $F(M) \to M$ of $R$-modules turns into a canonical injection, see above, of $R$-modules $$(M^\vee)^\vee \longrightarrow (F(M^\vee))^\vee.$$ Set $J(M) = (F(M^\vee))^\vee$. The composition of $ev$ with this the displayed map gives $M \to J(M)$ functorially in $M$.

Lemma 15.52.8. Let $R$ be a ring. For every $R$-module $M$ the $R$-module $J(M)$ is injective.

Proof. Note that $J(M) \cong \prod_{\varphi \in M^\vee} R^\vee$ as an $R$-module. As the product of injective modules is injective, it suffices to show that $R^\vee$ is injective. For this we use that $$\mathop{\mathrm{Hom}}\nolimits_R(N, R^\vee) = \mathop{\mathrm{Hom}}\nolimits_R(N, \mathop{\mathrm{Hom}}\nolimits_{\mathbf{Z}}(R, \mathbf{Q}/\mathbf{Z})) = N^\vee$$ and the fact that $(-)^\vee$ is an exact functor by Lemma 15.52.6. $\square$

Lemma 15.52.9. Let $R$ be a ring. The construction above defines a covariant functor $M \mapsto (M \to J(M))$ from the category of $R$-modules to the category of arrows of $R$-modules such that for every module $M$ the output $M \to J(M)$ is an injective map of $M$ into an injective $R$-module $J(M)$.

Proof. Follows from the above. $\square$

In particular, for any map of $R$-modules $M \to N$ there is an associated morphism $J(M) \to J(N)$ making the following diagram commute: $$\xymatrix{ M \ar[d] \ar[r] & N \ar[d] \\ J(M) \ar[r] & J(N) }$$ This is the kind of construction we would like to have in general. In Homology, Section 12.24 we introduced terminology to express this. Namely, we say this means that the category of $R$-modules has functorial injective embeddings.

The code snippet corresponding to this tag is a part of the file more-algebra.tex and is located in lines 12116–12378 (see updates for more information).

\section{Injective modules}
\label{section-injectives-modules}

\noindent
Some lemmas on injective modules.

\begin{definition}
\label{definition-projective}
Let $R$ be a ring. An $R$-module $J$ is {\it injective} if and only if
the functor $\Hom_R(-, J) : \text{Mod}_R \to \text{Mod}_R$ is
an exact functor.
\end{definition}

\noindent
The functor $\Hom_R(- , M)$ is left exact for any $R$-module $M$, see
Algebra, Lemma \ref{algebra-lemma-hom-exact}.
Hence the condition for $J$ to be injective really signifies that given
an injection of $R$-modules $M \to M'$ the map
$\Hom_R(M', J) \to \Hom_R(M, J)$ is surjective.

\medskip\noindent
Before we reformulate this in terms of ${Ext}$-modules we discuss the
relationship between $\Ext^1_R(M, N)$ and extensions as in
Homology, Section \ref{homology-section-extensions}.

\begin{lemma}
\label{lemma-relation-ext-ext}
Let $R$ be a ring. Let $\mathcal{A}$ be the abelian category of
$R$-modules. There is a canonical isomorphism
$\Ext_\mathcal{A}(M, N) = \text{Ext}^1_R(M, N)$
compatible with the long exact sequences of
Algebra, Lemmas \ref{algebra-lemma-long-exact-seq-ext} and
\ref{algebra-lemma-reverse-long-exact-seq-ext}
and the $6$-term exact sequences of
Homology, Lemma \ref{homology-lemma-six-term-sequence-ext}.
\end{lemma}

\begin{proof}
Omitted.
\end{proof}

\begin{lemma}
\label{lemma-characterize-injective}
Let $R$ be a ring. Let $J$ be an $R$-module.
The following are equivalent
\begin{enumerate}
\item $J$ is injective,
\item $\Ext^1_R(M, J) = 0$ for every $R$-module $M$.
\end{enumerate}
\end{lemma}

\begin{proof}
Let $0 \to M'' \to M' \to M \to 0$ be a short exact sequence of $R$-modules.
Consider the long exact sequence
$$\begin{matrix} 0 \to \Hom_R(M, J) \to \Hom_R(M', J) \to \Hom_R(M'', J) \\ \phantom{0\ } \to \Ext^1_R(M, J) \to \Ext^1_R(M', J) \to \Ext^1_R(M'', J) \to \ldots \end{matrix}$$
of Algebra, Lemma \ref{algebra-lemma-reverse-long-exact-seq-ext}.
Thus we see that (2) implies (1). Conversely, if $J$ is injective
then the $\Ext$-group is zero by
Homology, Lemma \ref{homology-lemma-characterize-injectives} and
Lemma \ref{lemma-relation-ext-ext}.
\end{proof}

\begin{lemma}
\label{lemma-characterize-injective-bis}
Let $R$ be a ring. Let $J$ be an $R$-module.
The following are equivalent
\begin{enumerate}
\item $J$ is injective,
\item $\Ext^1_R(R/I, J) = 0$ for every ideal $I \subset R$, and
\item for an ideal $I \subset R$ and module map $I \to J$
there exists an extension $R \to J$.
\end{enumerate}
\end{lemma}

\begin{proof}
We have seen the implication (1) $\Leftrightarrow$ (2) in
Lemma \ref{lemma-characterize-injective}. In this proof
we will show that (1) $\Leftrightarrow$ (3) which is known
as Baer's criterion.

\medskip\noindent
Assume (1). Given a module map $I \to J$ as in (3) we find
the extension $R \to J$ because the map
$\Hom_R(R, J) \to \Hom_R(I, J)$ is surjective
by definition.

\medskip\noindent
Assume (3). Let $M \subset N$ be an inclusion of $R$-modules.
Let $\varphi : M \to J$ be a homomorphism. We will show that $\varphi$
extends to $N$ which finishes the proof of the lemma.
Consider the set of homomorphisms $\varphi' : M' \to J$
with $M \subset M' \subset N$ and $\varphi'|_M = \varphi$.
Define $(M', \varphi') \geq (M'', \varphi'')$ if
and only if $M' \supset M''$ and $\varphi'|_{M''} = \varphi''$.
If $(M_i, \varphi_i)_{i \in I}$ is a totally
ordered collection of such pairs, then we obtain a map
$\bigcup_{i \in I} M_i \to J$ defined by $a \in M_i$
maps to $\varphi_i(a)$. Thus Zorn's lemma applies.
To conclude we have to show that if the pair
$(M', \varphi')$ is maximal then $M' = N$.
In other words, it suffices to show, given
any subgroup $M \subset N$, $M \not = N$ and
any $\varphi : M \to J$, then we can find
$\varphi' : M' \to J$ with $M \subset M' \subset N$
such that (a) the inclusion $M \subset M'$ is strict, and
(b) the morphism $\varphi'$ extends $\varphi$.

\medskip\noindent
To prove this, pick $x \in N$, $x \not \in M$.
Let $I = \{f \in R \mid fx \in M\}$. This is an ideal of $R$.
Define a homomorphism $\psi : I \to J$ by $f \mapsto \varphi(fx)$.
Extend to a map $\tilde\psi : R \to J$ which is possible by assumption (3).
By our choice of $I$ the kernel of
$M \oplus R \to J$, $(y, f) \mapsto y - \tilde\psi(f)$
contains the kernel of the map $M \oplus R \to N$,
$(y, f) \mapsto y + fx$. Hence this homomorphism factors
through the image $M' = M + Rx$ and this extends the given homomorphism
as desired.
\end{proof}

\noindent
In the rest of this section we prove that there are enough injective
modules over a ring $R$. We start with the fact that $\mathbf{Q}/\mathbf{Z}$
is an injective abelian group. This follows from
Lemma \ref{lemma-injective-abelian}.

\begin{definition}
\label{definition-simple-functors}
Let $R$ be a ring.
\begin{enumerate}
\item For any $R$-module $M$ over $R$ we denote
$M^\vee = \Hom(M, \mathbf{Q}/\mathbf{Z})$
with its natural $R$-module structure. We think
of {\it $M \mapsto M^\vee$} as a contravariant functor
from the category of $R$-modules to itself.
\item For any $R$-module $M$ we denote
$$F(M) = \bigoplus\nolimits_{m \in M} R[m]$$
the {\it free module} with basis given by the elements $[m]$ with
$m \in M$. We let $F(M)\to M$, $\sum f_i [m_i] \mapsto \sum f_i m_i$
be the natural surjection of $R$-modules.
We think of $M \mapsto (F(M) \to M)$ as a functor from
the category of $R$-modules to the category of
arrows in $R$-modules.
\end{enumerate}
\end{definition}

\begin{lemma}
\label{lemma-vee-exact}
Let $R$ be a ring.
The functor $M \mapsto M^\vee$ is exact.
\end{lemma}

\begin{proof}
This because $\mathbf{Q}/\mathbf{Z}$
is an injective abelian group by Lemma \ref{lemma-injective-abelian}.
\end{proof}

\noindent
There is a canonical map $ev : M \to (M^\vee)^\vee$
given by evaluation: given $x \in M$ we let
$ev(x) \in (M^\vee)^\vee = \Hom(M^\vee, \mathbf{Q}/\mathbf{Z})$
be the map $\varphi \mapsto \varphi(x)$.

\begin{lemma}
\label{lemma-ev-injective}
For any $R$-module $M$ the evaluation map
$ev : M \to (M^\vee)^\vee$ is injective.
\end{lemma}

\begin{proof}
You can check this using that $\mathbf{Q}/\mathbf{Z}$ is an injective
abelian group. Namely, if $x \in M$ is not zero, then let
$M' \subset M$ be the cyclic group it generates. There exists
a nonzero map $M' \to \mathbf{Q}/\mathbf{Z}$ which necessarily does
not annihilate $x$. This extends to
a map $\varphi : M \to \mathbf{Q}/\mathbf{Z}$
and then $ev(x)(\varphi) = \varphi(x) \not = 0$.
\end{proof}

\noindent
The canonical surjection $F(M) \to M$ of $R$-modules turns into
a canonical injection, see above, of $R$-modules
$$(M^\vee)^\vee \longrightarrow (F(M^\vee))^\vee.$$
Set $J(M) = (F(M^\vee))^\vee$. The composition of $ev$ with this
the displayed map gives $M \to J(M)$ functorially in $M$.

\begin{lemma}
\label{lemma-JM-injective}
Let $R$ be a ring. For every $R$-module $M$ the
$R$-module $J(M)$ is injective.
\end{lemma}

\begin{proof}
Note that $J(M) \cong \prod_{\varphi \in M^\vee} R^\vee$ as an $R$-module.
As the product of injective modules is injective, it suffices to
show that $R^\vee$ is injective. For this we use that
$$\Hom_R(N, R^\vee) = \Hom_R(N, \Hom_{\mathbf{Z}}(R, \mathbf{Q}/\mathbf{Z})) = N^\vee$$
and the
fact that $(-)^\vee$ is an exact functor by Lemma
\ref{lemma-vee-exact}.
\end{proof}

\begin{lemma}
\label{lemma-injectives-modules}
Let $R$ be a ring.
The construction above defines a covariant functor
$M \mapsto (M \to J(M))$ from the category of
$R$-modules to the category of arrows of $R$-modules
such that for every module $M$ the output
$M \to J(M)$ is an injective map of $M$ into
an injective $R$-module $J(M)$.
\end{lemma}

\begin{proof}
Follows from the above.
\end{proof}

\noindent
In particular, for any map of $R$-modules $M \to N$
there is an associated morphism $J(M) \to J(N)$
making the following diagram commute:
$$\xymatrix{ M \ar[d] \ar[r] & N \ar[d] \\ J(M) \ar[r] & J(N) }$$
This is the kind of construction we would like to have in general.
In Homology, Section \ref{homology-section-injectives}
we introduced terminology to express this. Namely,
we say this means that the category of $R$-modules
has functorial injective embeddings.

Comment #112 by Kiran Kedlaya on February 4, 2013 a 11:44 pm UTC

In the proof of Lemma 01DC, I think $J(M) \cong \prod_{m\in M} R^\vee$ is a typo for $J(M) \cong \prod_{m\in M^\vee} R^\vee$.

Comment #114 by Johan (site) on February 5, 2013 a 12:53 am UTC

Fixed. Thanks!

Comment #349 by Fan on November 25, 2013 a 3:16 am UTC

The first sentence in the last paragraph:

"This the kind of construction we would like to have in general." missing "is"?

Comment #354 by Johan (site) on November 25, 2013 a 6:52 pm UTC

@#349: Thanks! Fixed here.

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