Let $R$ be an integral domain and $A$ it's integral closure. Futhermore $A$ is a finitely generated $R$-module.
I want to show that if $A$ is flat over $R$ then already $R = A$ holds.
My attempts: I tried (without success) following: Assume $R \neq A$. Let $a \in A \backslash R$. Since $A$ integral closure there exist a minimal $n$ with
(*) $$a^n +r_{n-1}a^{n-1} +... +r_0 =0$$
Since $R$ integral then $R \subset Frac(R)$ and the multiplication map $m: R \to A, r \mapsto ar$ is injective Since $A$ flat it should be stay injective after tensoring with $-\otimes A$.
Now I can multipy $a$ also with all elements from $A$. My hope was to deduce the contradiction using (*) by multiplying with a appropiate polynomial in $a$. The problem is the "constant" term r_0.
I think that I'm very close with this proof to the solution but I don't see the final step. Can anybody help. BTW: Up to now I don't used the finitely generated property.
Source: Liu's "Algebraic Geometry". Futhermore: Could the statement be true without the finiteness assumtion. Liu's says yes but how to show?