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ID: past-exam-of-the-rings-and-modules-course-of-the-university-of-oxford/2021/2/a/ii/solution

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Past exam of the rings and modules course of the University of Oxford / 2021 / 2 / a / ii / Solution by

Codex  0 2026-10-25

Let

n_{1}, \dots, n_{t}

generate the submodule

N

, and let

m_{1} + N, \dots, m_{s} + N

generate the quotient module

M / N

. Choose each

m_{i} \in M

as a representative of its class. For any

m \in M

, there are

r_{1}, \dots, r_{s} \in R

such that

m + N = \sum_{i = 1}^{s} r_{i} (m_{i} + N) .

Therefore

m - \sum_{i} r_{i} m_{i} \in N

. Expressing that difference using the generating set of a module

n_{1}, \dots, n_{t}

gives

m = \sum_{i = 1}^{s} r_{i} m_{i} + \sum_{j = 1}^{t} u_{j} n_{j} .

Hence

M = R m_{1} + \dots + R m_{s} + R n_{1} + \dots + R n_{t}

is a finitely generated module. This proves that finite generation is preserved by module extensions.

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