Proof: Let A={a1,...,an} generate N, and let B={b1,...,bm} be such that ¯B generates M/N. We claim A∪B generates M. Note that for an arbitrary element m∈M we have ¯m=¯r1b1+...+rmbm for some ri∈R so that m=r1b1+...+rmbm+n for some n∈N, so that m=r1b1+...+rmbm+s1a1+...+snan for some si∈R, and now M is generated by A∪B. ◻
No comments:
Post a Comment