Let $A\subset B$ be a ring extension and $x\in B$. Then if $A$ is noetherian the following two conditions are equivalent:
$i)$ $x$ is integral over $A$.
$ii)$ There exists a finitely generated $A$-module $M$ such that $A[x]\subset M \subset B$.
If we drop the condition of $A$ being noetherian the implication from i) to ii) stays true but I think that the other implication is now not true anymore. Using the determinant trick one can prove that the condition
$iii)$ There exists a faithful $A[x]$-module $M$ which is finitely generated as an $A$-module.
implies condition $i)$ but i dont see how we can make M in conditon $ii)$ into an $A[x]$-module. Unfortunately i cant find a counterexample which shows that $ii)$ does not imply $i)$ in general. Does anybody know of one? I would appreciate any suggestions or hints.
Let $A=k[y_1,y_2,\dots]$ be a polynomial ring in infinitely many generators over a field and let $B=A[x,z]/(x-y_1z,y_1x-y_2,y_1x^2-y_3,y_1x^3-y_4,\dots)$. Then $A[x]\subset B$ is contained in the $A$-submodule $M$ generated by $1$ and $z$, since $x^n=y_1zx^{n-1}=zy_n$ for each $n>0$. Suppose $x$ were integral over $A$. Then for some $n$, $x$ would be integral over $A_n=k[y_1,\dots,y_n]$ as an element of $B_n=A_n[x,z]/(x-y_1z,y_1x-y_2,\dots,y_1x^{n-1}-y_n)$. Letting $C_n=A_n[x]\subseteq B_n$, the induced map $\operatorname{Spec}(C_n)\to\operatorname{Spec}(A_n)=\mathbb{A}^n$ would have closed image. The intersection of this image with the hyperplane $y_n=1$ can be identified with the image of the map $\mathbb{A}^1\setminus\{0\}\to \mathbb{A}^{n-1}$ given by $x\mapsto(1/x^{n-1},1/x^{n-2},\dots,1/x)$ (since when you set $y_n=1$, $B_n=C_n=k[x,x^{-1}]$, with $y_k$ corresponding to $1/x^{n-k}$). But this image is not closed, because $(0,0,\dots,0)$ is in its closure. Thus $x$ cannot be integral over $A$.
(Probably you can translate this geometric argument into a more direct algebraic argument that $x$ isn't integral, but I find it easier to think about this way.)