Section 47.2 (08XI): Essential surjections and injections

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.

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 $A$ .
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.

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

$M \subset N$ is an essential extension,
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 Weißmann on May 07, 2017 at 11:18

The statement of 45.2.4 (2) should read: for all $x \in N$ nonzero ...

Comment #2572 by Johan on May 25, 2017 at 19:14

Thanks Dario. Fixed here.

Comment #3442 by Sebastian Bozlee on July 27, 2018 at 20:21

The second-to-last sentence of the proof of Lemma 08XM should read: By (2) we can find $f \in R$ with $fx \in M$ and $fx \neq 0$ .

I also have to ask if a compatible system of essential injections $E_i \to M$ yielding an essential injection $E \to M$ 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.

Comment #3496 by Johan on August 05, 2018 at 12:01

Thanks for the fix. See changes here.

Lemma 47.2.3 is stated as used later, so I think it is fine as is.

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The Stacks project

47.2 Essential surjections and injections

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