Let $n_1,\ldots,n_r$ generate $N$, and let the images of $m_1,\ldots,m_s$ generate $M/N$. For any $m\in M$, subtracting a suitable linear combination of the $m_j$ leaves an element of $N$, which is a combination of the $n_i$. Thus
$$
\boxed{M=Rn_1+\cdots+Rn_r+Rm_1+\cdots+Rm_s.}
$$


Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 101 / 1 / b / i / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-101/1/b/i/solution