I've run into this question that should be easy somehow, but I can't get a hold of:

$\displaystyle R = \{ \frac{a}{3^k} \in \mathbb{Q} : a \in \mathbb{Z}, k \in \mathbb{Z}_{\geq 0} \}$

Is R a PID?

Naturally, $\displaystyle \mathbb{Q}$ is a PID since it's a field. But since R isn't an ideal in $\displaystyle \mathbb{Q}$, I can't use that in an argument in any way. (It's not, because 1/2*r with r in R isn't in it by default, and 1/2 is in $\displaystyle \mathbb{Q}$).

Not every element has a multiplicative inverse in R, since a can be negative, so it's also surely not a field (which would also imply it is a PID)...

So how do I go about it? R is isomorphic with $\displaystyle \mathbb{Z} \times \mathbb{Z}_{\geq 0}$. Would that form an argument somehow?

Also, can anyone confirm for me that R/2R is isomorphic to $\displaystyle \mathbb{Z}/2\mathbb{Z}$ and R/3R is isomorphic to $\displaystyle \mathbb{Z}/3\mathbb{Z}$ for me?

My argument is that you can make a hormomorphism f: $\displaystyle R \to \mathbb{Z}/2\mathbb{Z}$ by $\displaystyle r \mapsto 0$ if a is even, and $\displaystyle r \mapsto 1$ if a is odd. This way only all elements in 2R can be found in $\displaystyle Ker(f)$. So by the isomorphism theorem $\displaystyle R/2R \cong \mathbb{Z}/2\mathbb{Z}$.

And the same argument applies for R/3R, by sending any r with a that is p mod 3 to p.