When does $f,g \in R[x]$ relatively prime imply $f,g \in R[[x]]$ relatively prime. Recently in some research I came to the point where the strength of my conclusion bottlenecks at my ability to precisely address this question:
Let $R$ be a ring such that $R[[x]]$, the ring of formal power series with coefficients in $R$, is a GCD domain (whatever that entails for $R$).  What additional properties (if any) do $R$, $R[x]$, or $R[[x]]$ need to possess so that elements relatively prime in $R[x]$ are also still relatively prime in $R[[x]]$?
With a little effort we can show (by way of a weak Bezout-type identity that holds for any polynomial GCD domain) that it's sufficient for $R[[x]]$ to be atomic in addition to a $GCD$ domain (hence a $UFD$) — or equivalently it is sufficient for $R$ to satisfy the $ACCP$.  This is an OK result, but I'm hoping that there's some slack here. In particular, since $R[[x]]$ a $GCD$ already implies that $R$ is Archimedean, I wonder if there's any point in the ground between Archimedean and $ACCP$ where $R$ is still structured enough for coprimeness to lift from $R[x]$ to $R[[x]]$. 
And if there isn't, I'd love to better understand where the limitation is coming from!
Update.  If the answer below checks out, then it's indeed enough for $R$ to be Archimedean (along with being a GCD domain).
 A: Revisiting this a year later, I think this question has a clean answer and that the sweet spot is a completely integrally closed $GCD$ domain, which means that in my original phrasing of the question (with $R[[x]]$ a GCD domain), the answer was actually no additional properties. I think the proof below works, please comment if you spot an error.

proposition: let $R$ be a $GCD$ domain.  Then coprime polynomials in $R[x]$ remain coprime in $R[[x]]$ iff $R$ is completely integrally closed.

The forward direction is very easy.  It suffices to show that if $R$ is not archimedean, then there exist coprime polynomials in $R[x]$ which are not coprime in $R[[x]]$ (since for GCD domains, being CIC and archimedean are the same thing).
 Let $b$ be a bounded element, i.e. a nonunit such that $\bigcap_n b^nR \supseteq (c) \supsetneq (0)$.  Since $c$ is not a unit, clearly $b-x$ and $bc$ are coprime in $R$. But setting $G = \sum_n c b^n x^n \in R[[x]]$ we have $(b-x)G = bc$ so $(b-x)$ and $bc$ are not coprime in $R[[x]]$.  
Now the converse.
Suppose that $f,g$ are coprime polynomials.  Since $R[x]$ is a GCD domain, there exist $s,t \in R[x]$, and a nonzero $\lambda \in R$ such that $fs + gt = \lambda$.  Suppose that $f,g$ have a common factor $H$ in $R[[x]]$.  Thus $HU = f$ and $HV = g$ for some $U,V \in R[[x]]$.   We aim to show that $H$ is a unit. 
Set $Q = tV + sU$ so that $HQ = \lambda$.
This gives us the identities $\lambda U = fQ, \lambda V = gQ$.  For $F \in R[[x]]$, let $c(F)$ denote the ideal generated by the coefficients of $F$, and let $I_v = (I^{-1})^{-1}$ denote the divisorial closure of an ideal $I \subset R$.  From the Dedekind-Mertens content formula, we find that there exists an $n$ such that both $$c(f)c(Q)^n =\lambda c(Q)^{n-1}c(U)$$ $$c(g)c(Q)^n =\lambda c(Q)^{n-1}c(V)$$  From the coprimeness of $f$ and $g$ it follows, since the divisorial closure of an f.g. ideal in a GCD domain is the GCD of the generators, that $(c(f), c(g))^{-1} = (c(f), c(g))_v = R$.  Moreover since $\big(c(f), c(g)\big) \subseteq \big(c(U), c(V)\big)$ we have that $\big(c(U), c(V)\big)_v = R$.  Thus for any $q \in c(Q)$, 
$$\big[\big( c(f), c(g)\big) \frac{q}{\lambda} c(Q)^{n-1} \big]_v \subseteq \big[\big( c(U), c(V)\big)  c(Q)^{n-1} \big]_v $$
so that $$\frac{q}{\lambda} \big(c(Q)^{n-1}\big)_v \subseteq \big(c(Q)^{n-1}\big)_v$$ which implies that $\frac{q}{\lambda}$ is almost integral over $R$.  If $R$ is completely integrally closed, we have thus shown that $\lambda$ divides $Q$, and $H\frac{Q}{\lambda} = 1$.  Hence $H$ is a unit. 
A: Actually if $R[[x]]$ is a PID, you can use the Bézout lemma. If $f,g\in R[[x]]$ are coprime than exist $h,k\in R[[x]]$ such that $h·f+k·g=1$ this equality holds also in $R[x]$, in this direction you don't even need again Bézout lemma (a common divisor of $f$ and $g$ in $R[x]$ would also divide $1$). I don't see any mistake in this reasoning, is there?
