The title is the question. The parts in bold below are where I'm stuck. This is what I've tried (much of this is just me explaining a hint I received):
This comes from Eisenbud 3.17. The hint in the back of the book states:
First reduce to the case where $R$ is Noetherian by passing to a
subring finitely generated over $\mathbb{Z}$.
This part I'm completely lost at.
After that the hint goes on to say that
$f(x)\in S[x]$ is integral over $R[x]$, then $M:=R[x][f(x)]\subset S[x]$ is a finitely generated $R[x]$-module.
This follows from one of the theorems in the section.
Let $\text{coef}(M)$ be the submodule of $S$ generated by all coefficients of
elements of $M$. Show that $\text{coef}(M)$ is a finitely generated
$R$-module.
We know that $f$ is integral so satisfies some equation:
$$f(x)^n+b_{n-1}f(x)^{n-1}+\ldots+b_1f(x)+b_0=0\;\;\;\;\;b_i\in R[x]$$
This means that $M$ is finitely generated as an $R[x]$-module by $\{1,f(x),\ldots,f^n(x)\}$. From this, it's easy to see that $\text{coef}(M)$ is finitely generated.
The hint goes on
If $\alpha$ is the leading coefficient of $f$, show that $R[\alpha]\subset\text{coef}(M)$; it follows that $R[\alpha]$ is a finitely generated $R$-module so $\alpha$ is integral over $R$.
Shouldn't this just follow since $R\subset\text{coef}(M)$ and $\alpha\in\text{coef}(M)$…?
Finally,
Use induction on the degree and the fact that integral elements form a ring to show that $f\in R[x]$.
If $f(x)$ has degree $0$ we are done since the leading coefficient of $f$ is just a constant integral over $R$. And since $R$ is equal to its integral closure, $f\in R[x]$. Now assume $f$ has degree $n$ and the coefficients on lesser degree terms are integral in $R$. Well, the leading coefficient is integral over $R$ and all the lower coefficients are integral over $R$ by induction so all the coefficients of $f$ are integral over $R$. Since $R$ is integrally closed this implies $f(x)\in R[x]$.
Best Answer
No, it doesn't necessarily follow that $R[\alpha]$ is f.g. from being a submodule of $Coef(M)$, as $R$ should be Noetherian in order to deduce this.
First, let $g(T)\in (R[x])[T]$ be monic in the variable $T$ such that $g(f(x))=0$.
Consider the subring $R'$ of $R$ generated by the coefficients of the coefficients of $g(T)$, then $R'$ is a Noetherian ring as it is a direct image of $\Bbb Z[x_1,\ldots,x_n]$ for some $n$.
From this you can prove as you indicated that $R'[\alpha]$ is f.g. over $R'$, so it is integral over $R$ and thus $\alpha\in R$.