Let $A$ and $B$ ($A\subset B$) be commutative rings with unity. Let $a\in B$ be integral over $A$. Therefore, there exists a monic polynomial $p\in A[x]$, such that $p(a)=0$.
By well-ordering principle, there exists a monic polynomial $p_0 \in A[x]$, such that $p_0(a) = 0$ and $$\text{degree}(p_0) := \min\{\text{degree}(g);\ g\in A[x],\ g\ \text{is monic and }g(a)=0\}. $$
Question: Does anyone know an example of commutative rings $A$ and $B$ (with unity), such that there is a polynomial $g\in A[x]$ (not necessarily monic) such that $g(a)=0$ and $\text{degree}(g)<\text{degree}(p_0)?$
If $A$ is a field then it is obvious that it is impossible. But in the general case, I am not being able to find a counterexample. Can anyone help me?
One can choose $A =\mathbb Q[u]$ and $B =\mathbb Q[u,v]/(uv, v^2+1)$.
The equation of integrality of $v$ over $A$ is $T^2+1 \in A[T]$. If there exists a monic polynomial $f(T) \in A[T]$ of degree $1$ such that $f(v) = 0$, it must be of the form $T-v$. Since $v \not\in A$, this is impossible. So, $\deg p_0 = 2$.
For $g(T)$, one can take $g(T) = uT \in A[T]$ which is of degree $1$.