Let $A$ be a domain, $f \in A[x]$ irreducible and $b$ a zero of $f$, so we have $A[x]/(f) \cong A[b]$. I want to show, under certain circumstances, that $b$ integral over $A$ implies that $f$ has an invertible leading coefficient.
I consider the following: $A = K[Y]$, the coordinate ring of an affine variety $Y$. If $A$ would be integrally closed, then we would know that $b$ had a monic minimal polynomial $p$ over $A$ and would be done ($f = u \cdot p$, $u$ unit in $A[x]$, i.e. constant).
But here, as a coordinate ring, $A$ in general will not be integrally closed. How do we proceed here?