$R$ is a ring and $R[[x]]$ is the ring of formal power series over $R$.
I'm trying to get a better understanding of how zero divisors in $R[[x]]$ behave. Here's a simple question to this effect.
Suppose that $a,b \in R$ and $ax + b$ is not a zero divisor in $R[[x]]$. Does it follow that $\sum_{i=0}^n a^{n-i}b^ix^i$ is also not a zero divisor for every $n \in \mathbb{N}$?
I see an easy argument when $n = 2^k - 1$. In that case note that $ax - b$ is still not a zero divisor in $R[[x]]$ and so $(ax - b)(ax + b) = ax^2 - b^2$ is not a zero divisor, and similarly $ax^{2^k} - bx^{2^k}$ is not a zero divisor. The fact follows $\sum_{i=0}^n a^{n-i}b^ix^i$ divides $ax^{2^k} - bx^{2^k}$ when $n = 2^k - 1$.
This observation aside, I don't even understand why or why not $ax^2 + abx + b^2$ is a zero divisor in $R[[x]]$.
Part of the difficulty for me is that for all the rings in my comfort zone, the non zero divisors of $R[x]$ remain non zero divisors of $R[[x]]$. In those cases, $ax + b$ being a nonzero divisor just means that $(a,b)$ is a dense ideal which would imply that $(a^n,b^n)$ is also a dense ideal, and consequently $\sum_{i=0}^n a^{n-i}b^ix^i$ is a regular element of $R[x]$ (and therefore $R[[x]]$). This class of rings includes all rings which have 'reasonable' primary decompositions of $(0)$.