Let $A \subseteq B$ be integral domains ($A$ is not necessarily a UFD). Assume $b \in B$ is algebraic over $A$, namely: $a_nb^n+\cdots+a_1b+a_0=0$ for some $a_i \in A$ not all zero.
My (trivial) question: Does a "minimal polynomial for $b$" exist?
(By a minimal polynomial I mean a polynomial $f(T) \in A[T]$ with smallest degree such that $f(b)=0$.)
My answer: A minimal degree exists, but maybe there exist two or more polynomials with the same minimal degree (non is a multiple by scalar of the other, so there is no uniqueness up to scalar).
My (less trivial) questions:
(1) Is $m(T) \in A[T]$ irreducible? I think yes (just write $m=fg$, then $0=m(b)=u(b)v(b)$, so since $B$ is a domain, $u(b)=0$ or $v(b)=0$, a contradiction to the minimality of $m$). The difference is that now an irreducible element of $A[T]$ need not be prime?
(2) The property that $m$ divides any other polynomial $\in A[T]$ which $b$ vanishes is also not valid, since Euclid's algorithm is not valid? For example: $K[x^2,x^3] \subset K[x]$, $x$ is algebraic, its minimal polynomial is of degree $2$, $T^2-x^2$, and it does not divide $T^3-x^3$.
(3) If $A$ is a GCD-domain, is it true that $m$ behaves as over a field (namely: $m$ is prime and divides any other polynomial over $A$ which $b$ vanishes).