Minimal generating set of a free module over local ring - MathOverflow most recent 30 from http://mathoverflow.net2013-06-20T04:53:52Zhttp://mathoverflow.net/feeds/question/83544http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/83544/minimal-generating-set-of-a-free-module-over-local-ringMinimal generating set of a free module over local ringIvo Jan2011-12-15T18:33:37Z2011-12-15T18:58:04Z
<p>Greetings,
in my studies I went into a statement "minimal generating set of a free module over a local ring is a free basis". The statement came without a proof, just with a reference to Kaplansky's theorem. I was unsuccessful trying to prove the statement myself, and I couldn't find the proof elsewhere either. I would be grateful for any hint.</p>
http://mathoverflow.net/questions/83544/minimal-generating-set-of-a-free-module-over-local-ring/83548#83548Answer by Francesco Polizzi for Minimal generating set of a free module over local ringFrancesco Polizzi2011-12-15T18:58:04Z2011-12-15T18:58:04Z<p>Any free module $M$ is projective. Then, when $M$ is finitely generated, you can use the argument given in <a href="http://books.google.it/books?id=yJwNrABugDEC&printsec=frontcover&dq=matsumura+commutative+ring+theory&hl=en&sa=X&ei=zEHqTtrqLcrS4QSFtdTsCA&redir_esc=y#v=onepage&q=kaplansky&f=false" rel="nofollow">Matsumura, Commutative Ring Theory</a>, proof of Theorem 2.5 page 10.</p>
<p>Matsumura also expains the argument used by Kaplanski ["Projective modules", Ann. Math. <strong>68</strong> (1958)] in the general case.</p>