Understanding the structure of $\mathbb{Z}[[x]]/(x-x^2)$

Question

I'm currently trying to understand the structure of quotients of power series rings, and found a particular example I'm confused about. Let $f = x-x^2$ be a polynomial in $\mathbb{Z}[[x]]$, and consider $R = \mathbb{Z}[[x]]/(f)$. What familiar ring is $R$ isomorphic to? I want to say that $R$ looks like $\mathbb{Z}^2$ (it has a constant part and a linear part), but I'm skeptical. I do know that $\mathbb{Z}^2$ is dense in $R$ since $\mathbb{Z}[x]$ is dense in $\mathbb{Z}[[x]]$, but I'm worried about examples like $h = 1 + x + x^2 + x^3 + \ldots$. In this case $h$ doesn't have a linear coefficient in $\mathbb{Z}$. What's happening here? Does the situation change if I consider $\mathbb{Z}_p[[x]]$ or $\mathbb{Q}_p[[x]]$ instead of $\mathbb{Z}[[x]]$?

Answer 1

You are using the letter $R$ in two different ways. I take the expression from the title, i.e. we look at the structure of $R[[x]]/(x - x^2)$.

For whatever coefficient ring $R$, we always have $x - x^2 = x(1 - x)$, and $1 - x$ is always a unit in $R[[x]]$, because we have $(1 - x)(1 + x + x^2 + \dots) = 1$.

Thus $R[[x]]/(x - x^2)$ is canonically isomorphic to $R[[x]]/(x)$, which is again isomorphic to $R$ via the map sending $a_0 + a_1 x + \dots$ to $a_0$.