Dividend in Quotient Module Is Noetherian If Both The Quotient And Divisor Are

Theorem

Let \(M\) by an \(R\)-module with a submodule \(N\). Then if \(N\) and \(M/N\) are Noetherian, so is \(M\).

Proof

Let \(M\) be an \(R\)-module with submodule \(N\). Let \(L\) be an arbitrary submodule of \(M\). We will prove that \(L\) is finitely generated, and thus \(M\) is Noetherian.

By the third isomorphism theorem we have that

\[ \frac{L}{L \cap N} \cong \frac{L + N}{N}.\]

\((L + N)/N\) is a submodule of \(M/N\) (the one corresponding with \(L + N \leq M\) under the correspondence theorem) and thus finitely generated. As such, \(L/(L \cap N)\) must be finitely generated, and hence suppose it is generated by \(\{y_1 + L \cap N, \dots, y_n + L \cap N\}\). However \(L \cap N\) is also finitely generated since it is a submodule of the Noetherian module \(N\). As such, suppose \(L \cap N\) is generated by \(\{x_1, \dots, x_k\}\).

We then have that \(L\) is generated by \(\{y_1, \dots, y_n, x_1, \dots, x_k\}\).