202506211906

Tags : Module Theory

If $R$ is Noetherian and $M$ is finitely generated then $M$ is Noetherian

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This is a corollary of the theorem discussed here.

Corollary

If $R$ is Noetherian and $M$ is finitely generated then M is Noetherian.

We have an onto morphism $R^{\oplus n}\twoheadrightarrow M$. Hence $M$ is isomorphic to a quotient of $R^{\oplus n}$. By the mentioned theorem we need to show that $R^{\oplus n}$ is finitely Noethrian.

This can be done by induction. Since $R$ is Noetherian as a ring, it is Noethrian as a module. Now to show that $R^{\oplus n}$ is Noetherian, we show that $R^{\oplus (n-1)}$ is noetherian, which is again enough by the given theorem since

$$\frac{R^{\oplus n}}{R^{\oplus (n-1)}}\cong R$$

which we have by induction hypothesis. so we are done.

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References

• Noetherian Ring and PIDs
• Noetherian Modules
• M is noetherian if for every sub-module N, N and MN are noetherian