The ring of convergent power series over $\mathbb C$ isn't noetherian How can one prove that the ring of convergent everywhere power series in $\mathbb C[[z]]$ isn't Noetherian?
 A: The ring of everywhere convergent power series is isomorphic to the ring $\mathcal{O}(\mathbb{C})$ of entire functions. More generally, the ring of power series with radius of convergence at least $R$ is isomorphic to the ring $\mathcal{O}(D_R)$ of holomorphic functions on the disk $D_R = \{ z \in\mathbb{C} : \lvert z\rvert < R\}$. The isomorphisms are given by the Taylor expansion about $0$.
A consequence of the Weierstraß product theorem is that no ring $\mathcal{O}(U)$, where $U\subset\mathbb{C}$ is a domain, is Noetherian.
Choose any infinite discrete subset $V$ of $U$, and let
$$\mathfrak{I} = \{ f \in \mathcal{O}(U) : f(z) = 0 \text{ for almost all } z \in V\}.$$
This ideal is not finitely generated.
A: Alternatively, in every Noetherian ring, any non-unit, non-zero element is, up to units, a finite product of irreducibles. The irreducibles in $\mathcal{O}(\mathbb{C})$ are easily seen to be the functions $z-a$. Thus, if $\mathcal{O}(\mathbb{C})$ was Noetherian, then every non-unit, non-zero element would need to be a polynomial. This is certainly not true :)
