The Stacks Project


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

  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 15.52).

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

  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. $\square$

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.

  1. Let $f \in R$. Then $E = \bigcup I[f^n] = I[f^\infty]$ is an injective submodule of $I$.
  2. 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}

    Comments (0)

    There are no comments yet for this tag.

    Add a comment on tag 08XN

    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 lower-right corner).

    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 box. So in case this where tag 0321 you just have to write 0321. Beware of the difference between the letter 'O' and the digit 0.

    This captcha seems more appropriate than the usual illegible gibberish, right?