free modules over a ring which is not a PID
We will say that has the free submodule property if for any free module over and any submodule we have that is also free. It is well known, that if is a PID, then has the free submodule property. One can ask whether the converse is also true? We will try to answer this question.
Proof. Assume that is not a PID. Then there are two possibilities: either is not a domain or there is an ideal which is not principal. Assume that is not a domain and let be two zero divisors, i.e. , and . Let be an ideal generated by . Then obviously is a submodule of (regarded as a -module). Assume that is free. In particular there exists , such that if and only if . But is of the form and because is commutative we have
Contradiction, because . Thus is not free although is a submodule of a free module .
Assume now that there is an ideal which is not principal and assume that is free as a -module. Since is not principal, then there exist such that is linearly independent. On the other hand and is a free generator of . Thus is linearly dependent, so
for some nonzero (note that in this case both are nonzero, more precisely and ). Multiply the equation by . Thus we have
Note that here we used commutativity of . Since is linearly independend (in ), then the last equation implies that . Contradiction.
Corollary. Commutative ring is a PID if and only if it has the free submodule property.
|Title||free modules over a ring which is not a PID|
|Date of creation||2013-03-22 18:50:08|
|Last modified on||2013-03-22 18:50:08|
|Last modified by||joking (16130)|