I have a doubt as to what algebraic implication it has that a ring is not noetherian.
I have considered the ring of germs at the origin \begin{equation} \varepsilon_{n}=\lbrace f\colon(\mathbb{R}^{n},0)\longrightarrow\mathbb{R}:\text{$f$ is a map germ }\rbrace \end{equation}
with only maximal ideal $\mathfrak{m}_n$ of germs at the origin that vanishes out at the origin. I have proven that $\mathfrak{m}_n$ is not finitely generated, therefore $\varepsilon_{n}$ is not Noetherian ring.
However, if I change the space of smooth functions, to the space of holomorphic functions and the set of germs at the origin is defined in an analogous way, we have that if it is a Noetherian ring.
I have concluded very intuitively that being Noetherian changes the topology of analysis. However, if there is a condition or consequence of this, I would like to know it?