---
tag: RE5T59YZ46
kind: definition
name: Injective module
msc: [13C11, 16D50]
defines: 85H7PRHDPZ
uses: [0310REJJY8, 0NZ8JA25WK, 0NRYHJPQG2, JB7ZBFBA9C, 0CMD0BCH8T]
sources:
  - url: https://en.wikipedia.org/wiki/Injective_module
---

## Definition

Let $R$ be a [[commutative ring|0310REJJY8]] and let $E$ be an [[$R$-module|0NZ8JA25WK]].
The module $E$ is *injective* if it satisfies the following extension property.
Let $M$, $N$ be $R$-modules, $i \colon M \to N$ an injective $R$-[[linear map|0NRYHJPQG2]], and $f \colon M \to E$ an $R$-linear map.
Then there is an $R$-linear map $g \colon N \to E$ with $g \circ i = f$.

## Remarks

The following are equivalent for an $R$-module $E$:

1. $E$ is injective.
2. Every injective linear map $E \to N$ has a left inverse, that is, it splits.
3. The functor $\Hom_R(-, E)$ takes [[exact sequences|JB7ZBFBA9C]] to exact sequences.
4. ([[Baer's criterion|MR5HNV0Q07]]) Every linear map from an ideal $I$ of $R$ to $E$ extends to $R$.

Over a [[principal ideal domain|0CMD0BCH8T]], a module $E$ is injective exactly when it is divisible: $rE = E$ for every nonzero $r \in R$.
The same definition works over any ring, commutative or not.

Example: $\Q$ is an injective $\Z$-module.
By Baer's criterion it suffices to extend a map $f \colon n\Z \to \Q$ with $n \ne 0$; with $q = f(n)$, the map $g(x) = xq/n$ on $\Z$ satisfies $g(n) = q$.

Non-example: $\Z$ is not an injective $\Z$-module.
Take $i \colon \Z \to \Z$, $x \mapsto 2x$, and $f$ the identity of $\Z$.
An extension $g$ would satisfy $1 = f(1) = g(2) = 2g(1)$, which has no solution in $\Z$.