The Stacks project

47.2 Essential surjections and injections

We will mostly work in categories of modules, but we may as well make the definition in general.

Definition 47.2.1. Let $\mathcal{A}$ be an abelian category.

  1. An injection $A \subset B$ of $\mathcal{A}$ is essential, or we say that $B$ is an essential extension of $A$, if every nonzero subobject $B' \subset B$ has nonzero intersection with $A$.

  2. A surjection $f : A \to B$ of $\mathcal{A}$ is essential if for every proper subobject $A' \subset A$ we have $f(A') \not= B$.

Some lemmas about this notion.

Lemma 47.2.2. Let $\mathcal{A}$ be an abelian category.

  1. If $A \subset B$ and $B \subset C$ are essential extensions, then $A \subset C$ is an essential extension.

  2. If $A \subset B$ is an essential extension and $C \subset B$ is a subobject, then $A \cap C \subset C$ is an essential extension.

  3. If $A \to B$ and $B \to C$ are essential surjections, then $A \to C$ is an essential surjection.

  4. Given an essential surjection $f : A \to B$ and a surjection $A \to C$ with kernel $K$, the morphism $C \to B/f(K)$ is an essential surjection.

Proof. Omitted. $\square$

Lemma 47.2.3. Let $R$ be a ring. Let $M$ be an $R$-module. Let $E = \mathop{\mathrm{colim}}\nolimits E_ i$ be a filtered colimit of $R$-modules. Suppose given a compatible system of essential injections $M \to E_ i$ of $R$-modules. Then $M \to E$ is an essential injection.

Proof. Immediate from the definitions and the fact that filtered colimits are exact (Algebra, Lemma 10.8.8). $\square$

Lemma 47.2.4. Let $R$ be a ring. Let $M \subset N$ be $R$-modules. The following are equivalent

  1. $M \subset N$ is an essential extension,

  2. for all $x \in N$ nonzero there exists an $f \in R$ such that $fx \in M$ and $fx \not= 0$.

Proof. Assume (1) and let $x \in N$ be a nonzero element. By (1) we have $Rx \cap M \not= 0$. This implies (2).

Assume (2). Let $N' \subset N$ be a nonzero submodule. Pick $x \in N'$ nonzero. By (2) we can find $f \in R$ with $fx \in M$ and $fx \not= 0$. Thus $N' \cap M \not= 0$. $\square$


Comments (4)

Comment #2538 by Dario WeiƟmann on

The statement of 45.2.4 (2) should read: for all nonzero ...

Comment #3442 by Sebastian Bozlee on

The second-to-last sentence of the proof of Lemma 08XM should read: By (2) we can find with and .

I also have to ask if a compatible system of essential injections yielding an essential injection was intended in Lemma 08XL. Both statements are true, but it feels odd not to invoke the universal property of the colimit to get the map between E and M.


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 08XI. Beware of the difference between the letter 'O' and the digit '0'.