I am attempting to prove (or disprove) that $\mathbb{C}\{z\}$ is Noetherian, where $\mathbb{C}\{z\}$ is the subring of $\mathbb{C}[[z]]$ consisting of functions holomorphic in a neighborhood of $0 \in \mathbb{C}$.
Note : $\mathbb{C}[[z]]$ represents the ring of formal power series with coefficients in the field of complex numbers.
Firstly, I recognize that in the ring of smooth function $C^{\infty}(\mathbb{R}, \mathbb{R})$ if we consider the family of ideals $$I_n = \left\{f \in C^{\infty}(\mathbb{R}, \mathbb{R}) \>\Bigm|\> f \text{ vanishes in } \left]-\frac{1}{n}, \frac{1}{n}\right[\;\right\},$$ we obtain an increasing chain that does not stabilize.
Why does this work for $C^{\infty}(\mathbb{R}, \mathbb{R})$ and not for $\mathbb{C}\{z\}$?
Here's my explanation:
Using the principle of analytic extension, if $f$ is analytic and $f=0$ in an open set, then it is the null function.
Now For smooth functions, we can consider the ideal $I_n$ because it is not equal to the trivial ideal $\langle 0 \rangle$ since a smooth function is not necessarily analytic.
However, in $\mathbb{C}\{z\}$, the functions are holomorphic, which is equivalent to being analytic. Consequently, we would have $I_n = \langle 0 \rangle$.
So, is there another approach or is the one I mentioned sufficient to prove or disprove that $\mathbb{C}\{z\}$ is Noetherian?
Edit:
Additionally, one can consider the ring of smooth maps, denoted as $C^{\infty}(\mathbb{R}, \mathbb{R})$, instead of $\mathbb{C}\{z\}$, which is more general.