## Tag `08XN`

## 45.3. Injective modules

Some results about injective modules over rings.

Lemma 45.3.1. Let $R$ be a ring. Any product of injective $R$-modules is injective.

Proof.Special case of Homology, Lemma 12.24.3. $\square$Lemma 45.3.2. Let $R \to S$ be a flat ring map. If $E$ is an injective $S$-module, then $E$ is injective as an $R$-module.

Proof.This is true because $\mathop{\rm Hom}\nolimits_R(M, E) = \mathop{\rm Hom}\nolimits_S(M \otimes_R S, E)$ by Algebra, Lemma 10.13.3 and the fact that tensoring with $S$ is exact. $\square$Lemma 45.3.3. Let $R \to S$ be an epimorphism of rings. Let $E$ be an $S$-module. If $E$ is injective as an $R$-module, then $E$ is an injective $S$-module.

Proof.This is true because $\mathop{\rm Hom}\nolimits_R(N, E) = \mathop{\rm Hom}\nolimits_S(N, E)$ for any $S$-module $N$, see Algebra, Lemma 10.106.14. $\square$Lemma 45.3.4. Let $R \to S$ be a ring map. If $E$ is an injective $R$-module, then $\mathop{\rm Hom}\nolimits_R(S, E)$ is an injective $S$-module.

Proof.This is true because $\mathop{\rm Hom}\nolimits_S(N, \mathop{\rm Hom}\nolimits_R(S, E)) = \mathop{\rm Hom}\nolimits_R(N, E)$ by Algebra, Lemma 10.13.4. $\square$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 $(M, \varphi)$. $\square$

Example 45.3.6. Let $R$ be a reduced ring. Let $\mathfrak p \subset R$ be a minimal prime so that $K = R_\mathfrak p$ is a field (Algebra, Lemma 10.24.1). Then $K$ is an injective $R$-module. Namely, we have $\mathop{\rm Hom}\nolimits_R(M, K) = \mathop{\rm Hom}\nolimits_K(M_\mathfrak p, K)$ for any $R$-module $M$. Since localization is an exact functor and taking duals is an exact functor on $K$-vector spaces we conclude $\mathop{\rm Hom}\nolimits_R(-, K)$ is an exact functor, i.e., $K$ is an injective $R$-module.

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

- $E$ is an injective $R$-module, and
- 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. $\square$

Lemma 45.3.8. Let $R$ be a Noetherian ring. A direct sum of injective modules is injective.

Proof.Let $E_i$ be a family of injective modules parametrized by a set $I$. Set $E = \bigcup E_i$. To show that $E$ is injective we use Lemma 45.3.7. Thus let $\varphi : I \to E$ be a module map from an ideal of $R$ into $E$. As $I$ is a finite $R$-module (because $R$ is Noetherian) we can find finitely many elements $i_1, \ldots, i_r \in I$ such that $\varphi$ maps into $\bigcup_{j = 1, \ldots, r} E_{i_j}$. Then we can extend $\varphi$ into $\bigcup_{j = 1, \ldots, r} E_{i_j}$ using the injectivity of the modules $E_{i_j}$. $\square$Lemma 45.3.9. Let $R$ be a Noetherian ring. Let $S \subset R$ be a multiplicative subset. If $E$ is an injective $R$-module, then $S^{-1}E$ is an injective $S^{-1}R$-module.

Proof.Since $R \to S^{-1}R$ is an epimorphism of rings, it suffices to show that $S^{-1}E$ is injective as an $R$-module, see Lemma 45.3.3. To show this we use Lemma 45.3.7. Thus let $I \subset R$ be an ideal and let $\varphi : I \to S^{-1} E$ be an $R$-module map. As $I$ is a finitely presented $R$-module (because $R$ is Noetherian) we can find find an $f \in S$ and an $R$-module map $I \to E$ such that $f\varphi$ is the composition $I \to E \to S^{-1}E$ (Algebra, Lemma 10.10.2). Then we can extend $I \to E$ to a homomorphism $R \to E$. Then the composition $$ R \to E \to S^{-1}E \xrightarrow{f^{-1}} S^{-1}E $$ is the desired extension of $\varphi$ to $R$. $\square$Lemma 45.3.10. Let $R$ be a Noetherian ring. Let $I$ be an injective $R$-module.

- Let $f \in R$. Then $E = \bigcup I[f^n] = I[f^\infty]$ is an injective submodule of $I$.
- Let $J \subset R$ be an ideal. Then the $J$-power torsion submodule $I[J^\infty]$ is an injective submodule of $I$.

Proof.We will use Lemma 45.3.5 to prove (1). Suppose that $E \subset E' \subset I$ and that $E'$ is an essential extension of $E$. We will show that $E' = E$. If not, then we can find $x \in E'$ and $x \not \in E$. Let $J = \{a \in R \mid ax \in E'\}$. Since $R$ is Noetherian we can choose $x$ with $J$ maximal. Since $R$ is Noetherian we can write $J = (g_1, \ldots, g_t)$ for some $g_i \in R$. Say $f^{n_i}$ annihilates $g_ix$. Set $n = \max\{n_i\}$. Then $x' = f^n x$ is an element of $E'$ not in $E$ and is annihilated by $J$. By maximality of $J$ we see that $R x' = (R/J)x' \cap E = (0)$. Hence $E'$ is not an essential extension of $E$ a contradiction.To prove (2) write $J = (f_1, \ldots, f_t)$. Then $I[J^\infty]$ is equal to $$ (\ldots((I[f_1^\infty])[f_2^\infty])\ldots)[f_t^\infty] $$ and the result follows from (1) and induction. $\square$

Lemma 45.3.11. Let $A$ be a Noetherian ring. Let $E$ be an injective $A$-module. Then $E \otimes_A A[x]$ has injective-amplitude $[0, 1]$ as an object of $D(A[x])$. In particular, $E \otimes_A A[x]$ has finite injective dimension as an $A[x]$-module.

Proof.Let us write $E[x] = E \otimes_A A[x]$. Consider the short exact sequence of $A[x]$-modules $$ 0 \to E[x] \to \mathop{\rm Hom}\nolimits_A(A[x], E[x]) \to \mathop{\rm Hom}\nolimits_A(A[x], E[x]) \to 0 $$ where the first map sends $p \in E[x]$ to $f \mapsto fp$ and the second map sends $\varphi$ to $f \mapsto \varphi(xf) - x\varphi(f)$. The second map is surjective because $\mathop{\rm Hom}\nolimits_A(A[x], E[x]) = \prod_{n \geq 0} E[x]$ as an abelian group and the map sends $(e_n)$ to $(e_{n + 1} - xe_n)$ which is surjective. As an $A$-module we have $E[x] \cong \bigoplus_{n \geq 0} E$ which is injective by Lemma 45.3.8. Hence the $A[x]$-module $\mathop{\rm Hom}\nolimits_A(A[x], I[x])$ is injective by Lemma 45.3.4 and the proof is complete. $\square$

The code snippet corresponding to this tag is a part of the file `dualizing.tex` and is located in lines 134–381 (see updates for more information).

```
\section{Injective modules}
\label{section-injective-modules}
\noindent
Some results about injective modules over rings.
\begin{lemma}
\label{lemma-product-injectives}
Let $R$ be a ring. Any product of injective $R$-modules is injective.
\end{lemma}
\begin{proof}
Special case of Homology, Lemma \ref{homology-lemma-product-injectives}.
\end{proof}
\begin{lemma}
\label{lemma-injective-flat}
Let $R \to S$ be a flat ring map. If $E$ is an injective $S$-module,
then $E$ is injective as an $R$-module.
\end{lemma}
\begin{proof}
This is true because $\Hom_R(M, E) = \Hom_S(M \otimes_R S, E)$
by Algebra, Lemma \ref{algebra-lemma-adjoint-tensor-restrict}
and the fact that tensoring with $S$ is exact.
\end{proof}
\begin{lemma}
\label{lemma-injective-epimorphism}
Let $R \to S$ be an epimorphism of rings. Let $E$ be an $S$-module.
If $E$ is injective as an $R$-module, then $E$ is an injective $S$-module.
\end{lemma}
\begin{proof}
This is true because $\Hom_R(N, E) = \Hom_S(N, E)$ for any $S$-module $N$,
see Algebra, Lemma \ref{algebra-lemma-epimorphism-modules}.
\end{proof}
\begin{lemma}
\label{lemma-hom-injective}
Let $R \to S$ be a ring map. If $E$ is an injective $R$-module,
then $\Hom_R(S, E)$ is an injective $S$-module.
\end{lemma}
\begin{proof}
This is true because $\Hom_S(N, \Hom_R(S, E)) = \Hom_R(N, E)$ by
Algebra, Lemma \ref{algebra-lemma-adjoint-hom-restrict}.
\end{proof}
\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}
\begin{example}
\label{example-reduced-ring-injective}
Let $R$ be a reduced ring. Let $\mathfrak p \subset R$ be a minimal prime
so that $K = R_\mathfrak p$ is a field
(Algebra, Lemma \ref{algebra-lemma-minimal-prime-reduced-ring}).
Then $K$ is an injective $R$-module. Namely, we have
$\Hom_R(M, K) = \Hom_K(M_\mathfrak p, K)$ for any $R$-module
$M$. Since localization is an exact functor and taking duals is
an exact functor on $K$-vector spaces we conclude $\Hom_R(-, K)$
is an exact functor, i.e., $K$ is an injective $R$-module.
\end{example}
\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}
\begin{lemma}
\label{lemma-sum-injective-modules}
Let $R$ be a Noetherian ring. A direct sum of injective modules
is injective.
\end{lemma}
\begin{proof}
Let $E_i$ be a family of injective modules parametrized by a set $I$.
Set $E = \bigcup E_i$. To show that $E$ is injective we use
Lemma \ref{lemma-characterize-injective}.
Thus let $\varphi : I \to E$ be a module map from an ideal of $R$
into $E$. As $I$ is a finite $R$-module (because $R$ is Noetherian)
we can find finitely many elements $i_1, \ldots, i_r \in I$
such that $\varphi$ maps into $\bigcup_{j = 1, \ldots, r} E_{i_j}$.
Then we can extend $\varphi$ into $\bigcup_{j = 1, \ldots, r} E_{i_j}$
using the injectivity of the modules $E_{i_j}$.
\end{proof}
\begin{lemma}
\label{lemma-localization-injective-modules}
Let $R$ be a Noetherian ring. Let $S \subset R$ be a multiplicative
subset. If $E$ is an injective $R$-module, then $S^{-1}E$ is an
injective $S^{-1}R$-module.
\end{lemma}
\begin{proof}
Since $R \to S^{-1}R$ is an epimorphism of rings, it suffices
to show that $S^{-1}E$ is injective as an $R$-module, see
Lemma \ref{lemma-injective-epimorphism}.
To show this we use Lemma \ref{lemma-characterize-injective}.
Thus let $I \subset R$ be an ideal and let
$\varphi : I \to S^{-1} E$ be an $R$-module map.
As $I$ is a finitely presented $R$-module (because $R$ is Noetherian)
we can find find an $f \in S$ and an $R$-module map $I \to E$
such that $f\varphi$ is the composition $I \to E \to S^{-1}E$
(Algebra, Lemma \ref{algebra-lemma-hom-from-finitely-presented}).
Then we can extend $I \to E$ to a homomorphism $R \to E$.
Then the composition
$$
R \to E \to S^{-1}E \xrightarrow{f^{-1}} S^{-1}E
$$
is the desired extension of $\varphi$ to $R$.
\end{proof}
\begin{lemma}
\label{lemma-injective-module-divide}
Let $R$ be a Noetherian ring. Let $I$ be an injective $R$-module.
\begin{enumerate}
\item Let $f \in R$. Then $E = \bigcup I[f^n] = I[f^\infty]$
is an injective submodule of $I$.
\item Let $J \subset R$ be an ideal. Then the $J$-power torsion
submodule $I[J^\infty]$ is an injective submodule of $I$.
\end{enumerate}
\end{lemma}
\begin{proof}
We will use Lemma \ref{lemma-essential-extensions-in-injective}
to prove (1).
Suppose that $E \subset E' \subset I$ and that $E'$ is an essential
extension of $E$. We will show that $E' = E$. If not, then we can
find $x \in E'$ and $x \not \in E$. Let $J = \{a \in R \mid ax \in E'\}$.
Since $R$ is Noetherian we can choose $x$ with $J$ maximal.
Since $R$ is Noetherian we can write $J = (g_1, \ldots, g_t)$ for some
$g_i \in R$. Say $f^{n_i}$ annihilates $g_ix$. Set $n = \max\{n_i\}$.
Then $x' = f^n x$ is an element of $E'$ not in $E$ and is annihilated
by $J$. By maximality of $J$ we see that $R x' = (R/J)x' \cap E = (0)$.
Hence $E'$ is not an essential extension of $E$ a contradiction.
\medskip\noindent
To prove (2) write $J = (f_1, \ldots, f_t)$. Then
$I[J^\infty]$ is equal to
$$
(\ldots((I[f_1^\infty])[f_2^\infty])\ldots)[f_t^\infty]
$$
and the result follows from (1) and induction.
\end{proof}
\begin{lemma}
\label{lemma-injective-dimension-over-polynomial-ring}
Let $A$ be a Noetherian ring. Let $E$ be an injective $A$-module.
Then $E \otimes_A A[x]$ has injective-amplitude $[0, 1]$
as an object of $D(A[x])$. In particular, $E \otimes_A A[x]$
has finite injective dimension as an $A[x]$-module.
\end{lemma}
\begin{proof}
Let us write $E[x] = E \otimes_A A[x]$. Consider the short exact
sequence of $A[x]$-modules
$$
0 \to E[x] \to \Hom_A(A[x], E[x]) \to \Hom_A(A[x], E[x]) \to 0
$$
where the first map sends $p \in E[x]$ to $f \mapsto fp$ and the
second map sends $\varphi$ to $f \mapsto \varphi(xf) - x\varphi(f)$.
The second map is surjective because
$\Hom_A(A[x], E[x]) = \prod_{n \geq 0} E[x]$ as an abelian group and
the map sends $(e_n)$ to $(e_{n + 1} - xe_n)$ which is surjective.
As an $A$-module we have $E[x] \cong \bigoplus_{n \geq 0} E$
which is injective by Lemma \ref{lemma-sum-injective-modules}.
Hence the $A[x]$-module $\Hom_A(A[x], I[x])$ is injective by
Lemma \ref{lemma-hom-injective} and the proof is complete.
\end{proof}
```

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