I can't find the proof of this corollary: if $R$ is a PID, then every finitely generated projective $R$-module is free.
Please help me.
Projective modules over PID
0
$\begingroup$
abstract-algebra
modules
-
0What does "find the proof" mean? – 2017-02-28
1 Answers
2
A finitely generated projective $R$-module is a direct summand of a finitely generated free module. Furthermore, a submodule of a finitely generated free module over a P.I.D. is free.
-
1Yes. One of my many typos, sorry. Thank you for pointing it! – 2017-03-01