(10)Let R be a ring with 1 and let M be a left R-module. Let N be a submodule of M. Prove that if both M/N and N are finitely generated then so is M.
From the condition, M and N are R-modules.
Since N is finitely generated, then N=<n1,n2,...,ns>. So each element in N has the
form r1n1+r2n2+...+rsns, where r1,r2,...,rs ...
R-Modules and Submodules are investigated.