# 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 $R_{i}$.

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

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

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

Title example of a projective module  which is not free ExampleOfAProjectiveModuleWhichIsNotFree 2013-03-22 18:49:55 2013-03-22 18:49:55 joking (16130) joking (16130) 6 joking (16130) Example msc 16D40