The Stacks Project


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

  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 $(M, \varphi)$. $\square$

    The code snippet corresponding to this tag is a part of the file dualizing.tex and is located in lines 183–194 (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}

    Comments (0)

    There are no comments yet for this tag.

    Add a comment on tag 08XS

    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?