Let $A \subseteq B$ be two GCD domains with $1$. Assume that the extension $A \subseteq B$ is algebraic, not necessarily integral.
Notation: For $w \in B$ denote its minimal polynmial over $A$ by $m_w$ and denote the leading coefficient of $m_w$ by $c_w$.
Question: Let $u,v \in B$. Assume we have $c_u$ and $c_v$. What can we say about $c_{u+v}$ in terms of $c_u$ and $c_v$? Especially, when $\gcd(c_u,c_{u+v})=1$ and $\gcd(c_v,c_{u+v})=1$? Or more strongly, when $(c_u,c_{u+v})=A$ and $(c_v,c_{u+v})=A$ (unit ideals).
Remarks:
(1) In the special case when the extension is integral, clearly all coefficients $c_u,c_v,c_{u+v}$ are units, so trivially $(c_u,c_{u+v})=A$ and $(c_v,c_{u+v})=A$.
(2) I do not mind to restrict to the case where $A$ and $B$ are Noetherian UFD's, if this simplifies my question.
(3) Please see this related question, which hints that considering resultants may help.
Edit: Perhaps it is better to change my question to the following one: Let $k$ be a characteristic zero base field $k \subsetneq A \subseteq B$. Instead of considering $u$, $v$ and $u+v$, let us consider $u$, $v$ and $u+\lambda v$, where $\lambda \in k$.
Therefore, I would like to ask:
New Question: Let $u,v \in B$. Assume we have $c_u$ and $c_v$. Can we fine $\lambda \in k$ such that $c_{u+\lambda v}$ satisfies $\gcd(c_u,c_{u+\lambda v})=1$ and $\gcd(c_v,c_{u+\lambda v})=1$? Or more strongly, $(c_u,c_{u+\lambda v})=A$ and $(c_v,c_{u+\lambda v})=A$ (unit ideals).
Thank you!
