proof of Nakayama’s lemma
Elements of can be written as linear combinations , where .
Suppose that . Since , we can express as a such a linear combination:
Moving the term involving to the left, we have
But this means that is redundant as a generator of , and so is generated by the subset . This contradicts the minimality of .
We conclude that and therefore .
|Title||proof of Nakayama’s lemma|
|Date of creation||2013-03-22 13:07:46|
|Last modified on||2013-03-22 13:07:46|
|Last modified by||mclase (549)|