Why Does a Vector Space Have a Basis (Module Theory)

Module and vector space

First we recall some backgrounds. Suppose is a ring with multiplicative identity . A left module of is an additive abelian group , together with an ring operation such that for and . As a corollary, we see , which shows for all . On the other hand, which implies . We can also define right -modules but we are not discussing them here.

Let be a subset of . We say is a basis of if generates and is linearly independent. That is, for all , we can pick and such that and, for any , we have Note this also shows that (what happens if ?). We say is free if it has a basis. The case when or is trivial is excluded.

If is a field, then is called a vector space, which has no difference from the one we learn in linear algebra and functional analysis. Mathematicians in functional analysis may be interested in the cardinality of a vector space, for example, when a vector space is of finite dimension, or when the basis is countable. But the basis does not come from nowhere. In fact we can prove that vector spaces have basis, but modules are not so lucky.

Examples of non-free modules

First of all let's consider the cyclic group for . If we define which is actually copies of an element, then we get a module, which will be denoted by . For any , we see . Therefore for any subset , if , we have which gives the fact that has no basis. In fact this can be generalized further. If is a ring but not a field, let be a nontrivial proper ideal, then is a module that has no basis.

Following we also have another example on finite order. Indeed, any finite abelian group is not free as a module over . More generally,

Let be a abelian group, and be its torsion subgroup. If is non-trival, then cannot be a free module over .

Next we shall take a look at infinite rings. Let be the polynomial ring over a field and be the polynomial sub-ring that have coefficient of equal to . Then is a -module. However it is not free.

Suppose we have a basis of , then we claim that . If , say , then cannot generate since if is constant then we cannot generate a polynomial contains with power ; If is not constant, then the constant polynomial cannot be generate. Hence contains at least two polynomials, say and . However, note and , which gives Hence cannot be a basis.

Why does a vector space have a basis

I hope those examples have convinced you that basis is not a universal thing. We are going to prove that every vector space has a basis. More precisely,

Let be a nontrivial vector space over a field . Let be a set of generators of over and is a subset which is linearly independent, then there exists a basis of such that .

Note we can always find such and . For the extreme condition, we can pick and be a set containing any single non-zero element of . Note this also gives that we can generate a basis by expanding any linearly independent set. The proof relies on a fact that every non-zero element in a field is invertible, and also, Zorn's lemma. In fact, axiom of choice is equivalent to the statement that every vector has a set of basis. The converse can be found here.

Proof. Define Then is not empty since it contains . If is a totally ordered chain in , then is again linearly independent and contains . To show that is linearly independent, note that if , we can find some such that for . If we pick , then But we already know that is linearly independent, so implies .

By Zorn's lemma, let be the maximal element of , then is also linearly independent since it is an element of . Next we show that generates . Suppose not, then we can pick some that is not generated by . Define , we see is linearly independent as well, because if we pick , and if then if we have contradicting the assumption that is not generated by . Hence . However, we have proved that is a linearly independent set containing and contained in , contradicting the maximality of in . Hence generates .