zero of polynomial integral => polynomial has invertible leading coefficient 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?
So, if $A$ is a domain with $b$ integral over $A$, and if $f(x)\in A[x]$ is irreducible with $b$ as a root, then is the leading coefficient of $f$ invertible in $A$?

In general, no.  Let $A=\mathbb{Z}[\sqrt{-3}]=\{a+b\sqrt{-3}\,\vert\,a,b\in \mathbb{Z}\}$, and let $\omega=\frac{-1+\sqrt{-3}}{2}$ be a primitive third root of unity.  Note that $\omega\notin A$ and $\omega$ is a root of the monic polynomial $g(x)=x^2+x+1\in A[x]$, thus $\omega$ is integral over $A$.  Also, $g(x)$ is irreducible in $A[x]$, since the only roots of $g$ are $\omega$ and $\overline{\omega}:=\frac{-1-\sqrt{-3}}{2}$.

However, $f(x)=2x-2\omega = 2x+1-\sqrt{-3}\in A[x]$ is irreducible (and has $\omega$ as a root), but $2\notin U(A)=\{\pm 1\}$.
Isn't it by definition that a element is integral over a ring $A$ if it is the root of a monic polynomial with coefficients in $A$?