Free Modules

That is we can define free module as follows: Definition Let F be an R-module. If F has a nonempty basis X, then F is a free R-module on the set X.

However, some authers consider the zero module a free module as illustrated in the follwing first example.

Thus Z is free as Z-module and Zn is free as Zn- module. Also Q is a free Q-module.

3. n Z= (n) is free as Z-module. Since X = {n} spans n Z as a Z-module. Also nm = 0 for m Z implies m = 0 as . Hence { n } is a basis for n Z. Thus if A is an infinite cyclic group then A is free as a Z-module.

6. Consider Q as a Z-module. Let p/q, r/t Q be two distinct elements. Then (q r) (p/q ) + (−p t) (r/t) = 0. Hence { p/q , r/t } is not linearly independent. Hence the only linearly independent subsets of Q are singletons. Thus if Q admits a basis, then Q = < p/q >, which is a contradiction as Q is not cyclic as an Abelian group under addition. Hence Q as a Z-module is not free.

7. Zm for m N is not free as Z-module. Because if a Zm, then ma = 0 , but m ≠ 0 in Z. Thus Zm has no basis elements and hence cannot be free. Thus if A is a finite cyclic group then A cannot be free as a Z-module.

Remark A submodule of a free R-module needs not be a free R-module. For example, Z6 is a free Z6-module, but 2Z6 ={ 0, 2, 4 } is not free as Z6- module since 2(3)=0, but 3 ≠ 0 in Z6 .

Remark: Note that the basis is not unique since the sets {(1,0), (0,1)} and {(1,1),(1,0)} are two different bases for R2 . One can ask about the cardinality of the different bases for a free module. In the case of free abelian groups (Z-modules) we know that any two bases of a free Z-module have the same cardinality. Unfortunately, this is not true for free modules over arbitrary rings with identity as we will see in the example below. First note that

However, the following result is satisfied Theorem. Let R be a ring with identity and F a free R-module with an infinite basis X . Then every basis of F is infinite and has the same cardinality as X. Proof. If Y is another basis of F, then we claim that Y is infinite. Suppose on the contrary that Y were finite. Since Y generates F and every element of Y is a linear combination of a finite number of elements of X, it follows that there is a finite subset{ x1, … , xm }of X, which generates F. Since X is infinite, there exists x X - { x1, … , xm }. Then for some ri R, x = r1 x1, … , rm xm, which contradicts the linear independence of X. Therefore, Y is infinite. ( complete the proof, see Hungerford Theorem 2.6. page 184 ).