If N and M/N are finitely generated as R-modules, then so is M

Let R be a ring with 1, and suppose M is a left R-module with submodule N. Suppose further that N and M/N are finitely generated. Prove that M is finitely generated.


Say M/N = (a_i + N \ |\ 1 \leq i \leq m) and N = (b_j \ |\ 1 \leq j \leq n). We claim that M = (a_i,b_j \ |\ 1 \leq i \leq m, 1 \leq j \leq n). Certainly (\supseteq) holds. Now suppose m \in M. Note that m + N = \sum_i r_i \cdot (a_i + N) = (\sum_i r_i \cdot a_i) + N. Now m - \sum_i r_i \cdot a_i \in N, so that m - \sum_i r_i \cdot a_i = \sum_j s_j b_j. Then m = \sum_i r_i \cdot a_i + \sum_j s_j \cdot b_j \in (a_i, b_j \ |\ 1 \leq i \leq m, 1 \leq j \leq n). In particular, M is finitely generated.

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