202506211906

Tags : Module Theory

M is Noetherian if for every sub-module N, N and MN are Noetherian

---

If $M$ is Neotherian, Then so is $M/N$ and $N$ Since every sub-module of a noetherian ring is finitely generated.

For the other direction, if $N$, $M/N$ is noetherian, let $P$ be a submodule of $M$.

We have that $P\cap N$ is a sub-module of $N$ and is finitely generated, so by the second isomorphism theorem we have:

$$\frac{P}{P\cap N}\cong \frac{P+N}{N}$$

Thus $P/P\cap N$ is isomorphic to a sub-module of $M/N$ but since $M/N$ is noetherian so is $P/P\cap N$.

Now we show that since both $P\cap N$ and $P/P\cap N$ are finitely generated, we show that $P$ is finitely generated.

Let $P\cap N$ be generated by $A$ and $P/P\cap N$ be generated by $B$, we get that $P$ is generated by $A\bigsqcup B$. This satisfies the universal property of quotients so works.

---

References

• Noetherian Modules
• Universal Property (Riehl)