Alternative proof for a sufficient condition for $x$ to be integral over a ring $A$. I have a request for an alternative proof of a fact in ring theory. Given an extension of rings $R \supseteq A$, we say that $x\in R$ is integral over $A$ if there is a monic polynomial $p$ with coefficients from $A$ so that $p(x) = 0$.
In pg. 2 of Ash's "A Course in Algebraic Number Theory," Ash presents a theorem of the form "TFAE" for integrality over a ring. I am interested in an alternative proof of the following:
Theorem: If $x$ belongs to a subring $B \subseteq R$ which is a finitely-generated $A$-module so that $A \subseteq B$, then $x$ is integral over $A$.
Proof: Suppose that $B$ is $A$-generated by $\{\beta_1,\dots,\beta_n\}$. Then for each $i$:
$$x\beta_i = \sum_{i=1}^n c_{ij}\beta_j$$
for some $A$-coefficients $c_{ij}$. Writing $\overline{\beta}$ as the vector with elements $\beta_i$ and $C = [c_{ij}]$, it then follows that $(xI - C)\overline{\beta} = \overline{0}$. Recalling that $\text{adj}(A)A = \det(A)I$, it follows that $\det(xI-C)I\overline{\beta} = \overline{0}$, hence for every $b \in B$ it follows that $\det(xI - C)b = 0$. Setting $b=1$ gives the polynomial we sought. $\square$
Here is my question: is there a more "natural" proof of this fact? To avoid ambiguity, by "natural" I mean ring-theoretic. Since we have a foot in the world of modules, it's not unnatural to use some linear algebra, but when I tried to show this myself, it didn't even occur to me to use the determinant. So I suppose an alternative question I have is: "is there a nice proof here that doesn't use the determinant?"
 A: (I'll use $r$ instead of $x$ to avoid confusion with the polynomial algebra.)
We'll say is that an algebra $R$ over $A$ is finite over $A$ if it is finitely generated as an $A$ module.
It's easy (and "natural") to see that  $A[r]$ is finite over $A$ if and only if $r$ is integral over $A$, so we'll take that for granted.
The question now becomes: how can we conclude that $A[r]$ is finite over $A$ if $r$ belongs to a subring $B$ which is finite over $A$?
Well, the obvious observation is that $A[r]$ is a submodule of $B$, so if $A$ is Noetherian, we are done. From there, we can seek for a way to reduce the general case to the Noetherian case, which is indeed possible.
Suppose that $B$ is $A$-generated by $\beta_1, \dots, \beta_n$. Write
$$1=\sum_{i=1}^na_i \beta_i$$
$$\beta_i \beta_j= \sum_{k=1}^n c_{ijk} \beta_k$$
and $$r\beta_i=\sum_{j=1}^nd_{ij}\beta_j$$
For $a_i,c_{ijk},d_{ij} \in A$
Now let $A'=\Bbb Z[a_i,d_{ij},c_{ijk}] \subset A$ be the $\Bbb Z$-subalgebra generated by all these coefficients. Let $B'\subset B$ be the module $A'$-generated by $\beta_1, \dots, \beta_n$. The way we chose our coefficients implies that $B'$ is actually closed under multiplication, so it is a subalgebra and $r \in B'$. Thus we are in exactly the same situation as before except that $A$ was replaced by $A'$ and $B$ by $B'$. But $A'$ is Noetherian by the Hilbert basis theorem, so we win.
