Page 132 - Advanced Linear Algebra

116 Advanced Linear Algebra

shows that ann²4³ £ ¸ ¹ . On the other hand, this may fail if 9 is not an
integral domain. Also, there are torsion modules whose annihilators are trivial.
(We leave verification of these statements as an exercise. )
Free Modules
The definition of a basis for a module parallels that of a basis for a vector space.

Definition Let 4 be an -module. A subset of 4 is a basis if is linearly
8
9
8
9
independent and spans 4 . An -module 4 is said to be free if 4 ~ ¸ ¹ or if
8
4 4 has a basis. If is a basis for , we say that 48 is free on .
We have the following analog of part of Theorem 1.7.

Theorem 4.3 A subset of a module 4 is a basis if and only if every nonzero
8
8
#4 is an essentially unique linear combination of the vectors in .
In a vector space, a set of vectors is a basis if and only if it is a minimal
spanning set, or equivalently, a maximal linearly independent set. For modules,
the following is the best we can do in general. We leave proof to the reader.

Theorem 4.4 Let be a basis for an -module 4 . Then
9
8
)
1 8 is a minimal spanning set.
)
2 8 is a maximal linearly independent set.
{
The -module { has no basis since it has no linearly independent sets. But
since the entire module is a spanning set, we deduce that a minimal spanning set
need not be a basis. In the exercises, the reader is asked to give an example of a
module 4 that has a finite basis, but with the property that not every spanning
set in 4 contains a basis and not every linearly independent set in 4 is
contained in a basis. It follows in this case that a maximal linearly independent
set need not be a basis.
The next example shows that even free modules are not very much like vector
spaces. It is an example of a free module that has a submodule that is not free.

Example 4.5 The set d { is a free module over itself, using componentwise
{
scalar multiplication
² Á ³² Á ³ ~ ² Á ³

with basis ¸² Á ³¹ . But the submodule { d ¸ ¹ is not free since it has no
linearly independent elements and hence no basis.

Theorem 2.2 says that a linear transformation can be defined by specifying its
values arbitrarily on a basis. The same is true for free modules.