Let $r, R>0$ and $D_{R}=\{z\in\mathbb{C}: \|z\|<R\}$ and $D_{r,R}=\{z\in\mathbb{C}: r<\|z\|<R\}$. Consider the following
- $\mathcal{R}_{1}:=\mathcal{O}_{R}=\{f:D_{R}\to\mathbb{C}: f\hspace{0.1cm} \mbox{is analytic in}\hspace{0.1cm} D_{R}\}$
- $\mathcal{R}_{2}:=\mathcal{O}_{r,R}=\{f:D_{r, R}\to\mathbb{C}: f\hspace{0.1cm} \mbox{is analytic in}\hspace{0.1cm} D_{r, R}\}$.
Note that $\mathcal{R}_{1}$ and $\mathcal{R}_{2}$ are commutative rings with identity element, with the usual operations $+$ and $\cdot$.
My question is the following: Are $\mathcal{R}_{1}$ and $\mathcal{R}_{2}$ local and noetherian rings?
Thanks
"A ring $R$ is local" is equivalent to "for all $p\in R$, if $p$ is not a unit then $1-p$ is a unit".
But in either ring, obviously you can find some function which attains both $1$ and $0$. And if $f(z)$ vanishes for some value $z=c$, you won't find an element $g(z)$ of that ring such that $f(z)g(z)\equiv 1$.
In either ring, some functions vanish at an infinite sequence of points tending towards the boundary of the domain. So, for example, suppose the radius of convergence of $\mathcal{R}_1$ is $1$. Now, $f(z):=\sin\bigg(\dfrac{\pi}{z-1}\bigg)$ is an element of $\mathcal{R}_1$. We have that $f(z)=0$ whenever $z=1-\dfrac{1}{n}$ for any positive integer $n$. Therefore, $f(z)$ belongs to the ideal $\mathcal{I}$ in $\mathcal{R}_1$ defined as all those functions which vanish at every point $1-\dfrac{1}{n}$ for all $n\in\mathbb{Z}^+$.
Think about it, though. How about $g(z)=\sin\bigg(\dfrac{\pi/2}{z-1}\bigg)$, which is also in $\mathcal{R}_1$? It vanishes at every point of the form $1-\dfrac{1}{2n}$ for all $n\in\mathbb{Z}^+$. That's strictly fewer points than for $f(z)$.
How about $\sin\bigg(\dfrac{\pi/4}{(z-1)}\bigg)$? Or $\sin\bigg(\dfrac{\pi/8}{(z-1)}\bigg)$?
Do you see how this gives a chain of ascending ideals without a largest element?