$\mathbb{Z}[[x]]$ and $R[[x,y]]$ are not PID Show $\mathbb{Z}[[x]]$ and $R[[x,y]]$ are not principal ideal domains.
I've been thinking that the same ideals I have used for proving that $\mathbb{Z}[x]$ and $R[x,y]$ are not PID (respectively $<2,x>$ and $<x,y>$) might as well be used to show the $\mathbb{Z}[[x]]$ and $R[[x,y]]$ are not PID.
I know $\mathbb{Z}[x] \subset \mathbb{Z}[[x]]$ and $R[x] \subset R[[x]]$, thought I've been struggling to see if it is of any use that they contain subrings that are not PID.
I can not longer use that the degree of a product of polynomials is their sum of the degrees, which is what I used when proving such thing for $\mathbb{Z}[x]$.
I'm not so sure how to proceed for formal power series rings. Any lead would be appreciated.
 A: You never explained what the notation $R$ means. Probably you meant the real numbers, and if so then write it as $\mathbf R$ or $\mathbb R$.
It's true that merely because one ring contains another, the ideal structure of the smaller ring doesn't have to tell you very much about the ideal structure of the larger ring. But your intuition is correct that $(2,x)$ for $\mathbf Z[x]$ should still help you out for $\mathbf Z[[x]]$, of course with the understanding that the notation $(2,x)$ is a much bigger set as an ideal in $\mathbf Z[[x]]$ than as an ideal in $\mathbf Z[x]$.
First, show each element of $\mathbf Z[[x]]$ with constant term $\pm 1$ is a unit.  (That is definitely not true in $\mathbf Z[x]$, where the only units are $\pm 1$.) Assuming next that $(2,x)$ is principal, say $(2,x) = (f)$, you have $fg = 2$ for some formal power series $g$. Show $f$ is a unit or $g$ is a unit, so $(2,x) = (f) = (1)$ or $(2,x) = (f) = (2)$.  Then show each of those two cases leads to a contradiction.
For the case of $(x,y)$ in $\mathbf R[[x,y]]$, think of it as $\mathbf R[[y]][[x]]$ and try to carry out some analogue of the previous argument with $\mathbf Z$ replaced by $\mathbf R[[y]]$.
