## 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 $N' \cap M \not= 0$.
$\square$

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