Exercise 7.4 in Atiyah-Macdonald asks which of the following rings are Noetherian (numbering to match the book where there are other parts of the problem):
- The ring of power series in $z$ with a positive radius of convergence.
- The ring of power series in $z$ with an infinite radius of convergence.
For convenience I will call the ring in 2 $A$, and the ring in 3 $B$.
I am able to follow the reasoning in these solutions showing that $A$ is Noetherian, and the solution by Alex Youcis here showing that $B$ is not Noetherian. My question is - how does the argument showing that $B$ is not Noetherian not apply to $A$ as well?
If "positive radius" includes infinite radius, then $\sin(z)$ is in $A$, but is a nonzero, non-unit that is not a finite product of irreducibles, so $A$ should not be Notherian. On the other hand, if "positive radius" does not include infinite radius, then where do we use the assumption that our radius of convergence is finite in showing that $A$ is Noetherian?