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.

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