Minimal polynomial of an algebraic element over any domain

Question

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).

Answer 1

I think you answered (almost correctly) all your questions. In fact, $b$ is algebraic over $A$ iff $b$ is algebraic over $Q(A)$, the field of fractions of $A$. For more insight let's consider the field extension $Q(A)\subset Q(B)$, and $f$ the minimal polynomial of $b$ over $Q(A)$. If $p\in A[X]$, $p\ne0$ is such that $p(b)=0$, then $f\mid p$ in $Q(A)[X]$, so $\deg p\ge\deg f$. But we can multiply $f$ by an appropriate $\alpha\in A$, $\alpha\ne0$ in order to have $\alpha f\in A[X]$. This shows that $\alpha f$ is a minimal polynomial of $b$ over $A$.

Now few remarks:

(1) I don't think a minimal polynomial is irreducible: for instance, $X^2-2$ is a minimal polynomial of $\sqrt 2$ over $\mathbb Z$ and it is irreducible. Instead $2(X^2-2)$ is also a minimal polynomial, but is not irreducible. (If $A$ is a GCD domain, then you can choose from all the possible minimal polynomials one (and only one!) having the gcd of coefficients equal to $1$.)

(2) In your example a minimal polynomial seems to be $X^2T-X^3$.

(3) Yes, when $A$ is a GCD domain one can choose a minimal polynomial having the usual properties.