Projective module

Remarks

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

1. $P$ is projective.
2. Every surjective linear map $M \to P$ has a right inverse, that is, it splits.
3. $P$ is a direct summand of a free module: there is a module $Q$ with $P \oplus Q$ free.
4. The functor $\Hom_R(P, -)$ takes exact sequences to exact sequences.

The same definition works over any ring, commutative or not.

Example: every free module $F$ with basis $(e_i)$ is projective. Given $g$ and $f$, choose $m_i \in M$ with $g(m_i) = f(e_i)$ and let $h(e_i) = m_i$; then $g \circ h$ and $f$ agree on a basis, so they are equal. Over $R = \Z/6\Z$, the module $\Z/2\Z$ is projective but not free, since $R \cong \Z/2\Z \oplus \Z/3\Z$ as $R$-modules.

Non-example: over $R = \Z$, the module $\Z/2\Z$ is not projective. Take $g \colon \Z \to \Z/2\Z$ the reduction and $f$ the identity of $\Z/2\Z$. A lift $h \colon \Z/2\Z \to \Z$ must be zero, because $2h(1) = h(0) = 0$ and $\Z$ has no nonzero element of order $2$; then $g \circ h = 0 \ne f$.