# 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 $1\ne 0$ in both ${R}_{1}$ and ${R}_{2}$. Therefore $(1,0)\ne (1,1)$ 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. $\mathrm{\square}$

Title | example of a projective module^{} which is not free |
---|---|

Canonical name | ExampleOfAProjectiveModuleWhichIsNotFree |

Date of creation | 2013-03-22 18:49:55 |

Last modified on | 2013-03-22 18:49:55 |

Owner | joking (16130) |

Last modified by | joking (16130) |

Numerical id | 6 |

Author | joking (16130) |

Entry type | Example |

Classification | msc 16D40 |