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 $1\neq 0$ in both $R_1$ and $R_2$ . Therefore $(1,0)\neq (1,1)$ in $R$ , but $$(1,1)\cdot e_{i_0}=\pi_1(1,1)e_{i_0}=1 e_{i_0}=\pi_1(1,0) e_{i_0}=(1,0)\cdot e_{i_0}.$$ This situation is impossible in free modules (linear combination is uniquely determined by scalars). Contradiction. Analogously we prove that $R_2$ is not free. $\square$