# Equivalence of Definitions of Projective Module

## Theorem

Let $A$ be a ring.

Let $M$ be an $A$-module.

The following are equivalent:

(1) $M$ is a projective module, that is, $M$ is a projective object in the category of left $A$-modules.
(2) $M$ is a direct summand of a free module.
(3) Every short exact sequence of the form:
$\xymatrix{ 0 \ar[r] & X \ar[r]^f & Y \ar[r]^g & M \ar[r] & 0 }$

splits, that is, there is a homomorphism $s : M \to Y$ with $g \circ s = \operatorname {id}_M$.

## Proof

### (1) implies (2)

By Surjection by Free Module there is a free module $Y$ and a surjection $g : Y \to M$.

By Epimorphism of modules iff surjective $g$ is an epimorphism.

By Definition:Projective Object applied to $\operatorname {id}_M$, there is a homomorphism $s : M \to Y$ with $g \circ s = \operatorname {id}_M$.

We have $Y = \map {\operatorname {im} } s \oplus \map \ker g$.

$\Box$

### (2) implies (1)

Assume, that there is an $A$-module $Q$, such that $M \oplus Q$ is free.

Let $f : Y \to Z$ be an epimorphism.

Let $h : M \to Z$ be a homomorphism.

Let $S$ be a basis of $M \oplus Q$.

By Epimorphism of modules iff surjective $f$ is surjective.

Hence for all $s \in S$, there is some $y_s \in Y$ with $\map f {y_s} = \map h {\map {\pr_1} s}$, where $\pr_1: M \oplus Q \to M$ denotes projection to $M$.

By Universal Property of Free Modules there is a unique homomorphism $t : M \oplus Q \to Y$, such that $f \circ t = h \circ \pr_1$.

We have $f \circ (t \circ i_1) = h \circ \pr_1 \circ i_1 = h$, where $i_1 : M \to M \oplus Q$ is the inclusion of the first summand.

It follows, that $t \circ i_1$ is the desired lift of $f$.

$\Box$

### (3) implies (2)

By Surjection by Free Module there is a free module $Y$ and a surjection $g : Y \to M$.

There is a short exact sequence: $\xymatrix{ 0 \ar[r] & \map \ker g \ar[r] & Y \ar[r]^g & M \ar[r] & 0 }$

By Structure of Split Exact Sequence there is an isomorphism $Y \cong M \oplus \map \ker g$.

$\Box$