Exercise 3.6.18 (Finitely generated submodule and quotient module give a finitely generated parent module)

Question

Let M be an R-module and let N be a submodule of M. Prove that if N and M/N are both finitely generated, then M is finitely generated.

Shinkenjoe · Accepted Answer

We are given:
$$
N = \langle a_1, \dots, a_n \rangle \quad \text{and} \quad M/N = \langle b_1 + N, \dots, b_m + N \rangle.
$$

Take an arbitrary $ m \in M $. Then:
$$
m + N = \sum_{i=1}^m r_i(b_i + N),
$$
where $ r_i \in R $.
Using the properties of the module action in $ M/N $ and addition in $ M/N $ this simplifies to:
$$
m + N = \left( \sum_{i=1}^m r_i b_i \right) + N.
$$

This implies:
$$
m - \sum_{i=1}^m r_i b_i \in N.
$$

Thus, $ m - \sum_{i=1}^m r_i b_i = \sum_{j=1}^n s_j a_j $ for some $ r_i, s_j \in R $.

Finally:
$$
m = \sum_{i=1}^m r_i b_i + \sum_{j=1}^n s_j a_j.
$$

Since $ m $ was arbitrary, $ M $ is generated by $ \{a_1, \dots, a_n, b_1, \dots, b_m\} $.

Exercise 3.6.18 (Finitely generated submodule and quotient module give a finitely generated parent module)

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Exercise 3.6.18 (Finitely generated submodule and quotient module give a finitely generated parent module)

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