example of a projective module which is not free

Let $R_{1}$ and $R_{2}$ be two nontrivial, unital rings and let $R=R_{1}\\oplus R_{2}$ . Furthermore let $\\pi_{i}:R\\to R_{i}$ be a projection for $i=1,2$ . Note that in this case both $R_{1}$ and $R_{2}$ are (left) modules over $R$ via

\\cdot:R\\times R_{i}\\to R_{i};

(r,s)\\cdot x=\\pi_{i}(r,s)x,

where on the right side we have the multiplication in a ring $R_{i}$ .

Proposition. Both $R_{1}$ and $R_{2}$ are projective $R$ -modules, but neither $R_{1}$ nor $R_{2}$ is free.

Proof. Obviously $R_{1}\\oplus R_{2}$ is isomorphic (as a $R$ -modules) with $R$ thus both $R_{1}$ and $R_{2}$ are projective as a direct summands of a free module.

Assume now that $R_{1}$ is free, i.e. there exists $\\mathcal{B}=\\{e_{i}\\}_{i\\in I}\\subseteq R_{1}$ which is a basis. Take any $i_{0}\\in I$ . Both $R_{1}$ and $R_{2}$ are nontrivial and thus $1\eq 0$ in both $R_{1}$ and $R_{2}$ . Therefore $(1,0)\eq(1,1)$ in $R$ , but

(1,1)\\cdot e_{i_{0}}=\\pi_{1}(1,1)e_{i_{0}}=1e_{i_{0}}=\\pi_{1}(1,0)e_{i_{0}}=(1%,0)\\cdot e_{i_{0}}.

This situation is impossible in free modules (linear combination is uniquely determined by scalars). Contradiction. Analogously we prove that $R_{2}$ is not free. $\\square$