The Stacks project

Lemma 47.5.3. Let $R$ be a ring. Let $M$, $N$ be $R$-modules and let $M \to E$ and $N \to E'$ be injective hulls. Then

  1. for any $R$-module map $\varphi : M \to N$ there exists an $R$-module map $\psi : E \to E'$ such that

    \[ \xymatrix{ M \ar[r] \ar[d]_\varphi & E \ar[d]^\psi \\ N \ar[r] & E' } \]


  2. if $\varphi $ is injective, then $\psi $ is injective,

  3. if $\varphi $ is an essential injection, then $\psi $ is an isomorphism,

  4. if $\varphi $ is an isomorphism, then $\psi $ is an isomorphism,

  5. if $M \to I$ is an embedding of $M$ into an injective $R$-module, then there is an isomorphism $I \cong E \oplus I'$ compatible with the embeddings of $M$,

In particular, the injective hull $E$ of $M$ is unique up to isomorphism.

Proof. Part (1) follows from the fact that $E'$ is an injective $R$-module. Part (2) follows as $\mathop{\mathrm{Ker}}(\psi ) \cap M = 0$ and $E$ is an essential extension of $M$. Assume $\varphi $ is an essential injection. Then $E \cong \psi (E) \subset E'$ by (2) which implies $E' = \psi (E) \oplus E''$ because $E$ is injective. Since $E'$ is an essential extension of $M$ (Lemma 47.2.2) we get $E'' = 0$. Part (4) is a special case of (3). Assume $M \to I$ as in (5). Choose a map $\alpha : E \to I$ extending the map $M \to I$. Arguing as before we see that $\alpha $ is injective. Thus as before $\alpha (E)$ splits off from $I$. This proves (5). $\square$

Comments (0)

There are also:

  • 4 comment(s) on Section 47.5: Injective hulls

Post a comment

Your email address will not be published.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 toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

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 input field. As a reminder, this is tag 08Y4. Beware of the difference between the letter 'O' and the digit '0'.