example of free module with bases of diffrent cardinality
Let be a field and be an infinite dimensional vector space over . Let be its basis. Denote by the ring of endomorphisms of with standard addition and composition as a multiplication.
Let be any set such that .
Proposition. and are isomorphic as a -modules.
Proof. Let be a bijection (it exists since and is infinite) and denote by and the projections. Moreover let and .
Recall that is injective if and only if . So assume that for . Note that if and only if for all and this is if and only if for all and . So take any . Then (since is bijective) there exists such that . It follows that and . Thus we have
Since and were arbitrary, then which completes this part.
We wish to show that is onto, so take any . Define by the following formula:
It is easy to see that .
Corollary. For any two numbers there exists a ring and a free module such that has two bases with cardinality respectively.
Proof. It follows from the proposition, that for we have
For finite set module is free with basis consisting elements (product is the same as direct sum). Therefore (due to existence of previous isomorphisms) -module has two bases, one of cardinality and second of cardinality .
|Title||example of free module with bases of diffrent cardinality|
|Date of creation||2013-03-22 18:07:18|
|Last modified on||2013-03-22 18:07:18|
|Last modified by||joking (16130)|