I'm trying to better understand polynomial division and the logic of common factors in domains where division isn't so well behaved, in particular in the absence of the pre-Schreier property. But my intuition goes out the window hand-in-hand with Schreierness.
Given polynomials $f,g \in R[x]$ I use the nonstandard notation $CD(f,g)$ to denote the set of common divisors of $f,g$. (Note that there's no assumption of greatness here!)
Suppose $R$ is a domain (with identity) but is not pre-Schreier (i.e. if $a | bc$ it's not necessarily the case that there exist $x,y$ with $a = xy$ and $x | b, y | c$).
Let $\lambda \in R$, and $f,g \in R[x]$ such that $CD(f,g) \subset R$.
I want to be able to say things like $$CD(f,g) \subset R \implies CD(f, \lambda g) \subset R \tag{1}$$ and $$CD(f,g) \subset R \implies CD(f, g^k) \subset R \ \ \ \forall k \tag{2} $$ and $$CD(f,g) \subset R \implies CD(f^k, g^k) \subset R \ \ \ \forall k \tag{3} $$ which are all immediately seen to be true in pre-Schreier domains.
I noticed that $(1) \implies (2) \implies (3)$ for commutative $R$ in general. Furthermore, immediately $(3) \implies (1)$ when $R$ is an almost-Schreier domain (i.e. $a | bc \implies a^k = xy$ with $x | b^k, y | c^k$).
It's nice to know that they're equivalent, but I want to know if/and when they are true! I have tried looking at almost Schreier domains such as $Z[\sqrt{-3}]$ for counterexamples to $(1)$, but I'm not seeing anything so far. I have a nagging suspicion that I am missing something obvious!
I'm hoping someone could shed some light on what the relevant mechanics are that cause $(1), (2), (3)$ to either hold or fail in non-Schreier domains. In particular, I'm curious if (complete) integral closure would have any bearing.
Along those lines I was thinking it might be easiest to prove (3) by an argument in the vein of:
Supposing $R$ is an integrally closed domain we consider polynomials in $R[x][y]$. Let $A_h$ denotes the content ideal of $h$, and consider $h^+ = fy + g, h^- = fy - g, h^* = h^+ h^- = f^2y^2 - g^2$. Then $$(A_{h^+}A_{h^-})_v = (A_{h^*})_v$$ Now if $R[x]$ were an $IP$ domain (i.e. integral v-ideals are the intersection of integral principal ideals) then the assumption that $CD(f,g) \subset R$ would imply that $A_h^*$ isn't contained in any principle ideal generated by a non-constant. Thus $CD(f^2, g^2) \subset R$, and a straightforward inductive argument can then extend this to $CD(f^k,g^k)$.
But that argument seriously cops out halfway through, because frankly I don't know when $R[x]$ is an $IP$ domain or how that interacts with the almost-Schreier property. But maybe we can do something along similar lines borrowing shallowly from v-ideal theory?