A module theoretical equivalence of integral dependence states as follows:
Let $S$ be a ring and $R \subseteq S$ be a subring, and $1\in R$. Then $x \in S$ is integral over $R$ if and only if $R[x]$ is finitely generated as an $R$-module, where $R[x]$ is the subring of $S$ generated by $x$.
I believe that this means elements of $R[x]$ are also integral over $R$. Since $x\in R$ is integral over $S$, $x$ satisfies an equation of the form $$x^n+a_1 x^{n-1}+\ldots+a_{n-1}x+a_n=0,$$ so $R[x]$ is generated by $\{1,x,\ldots,x^{n-1}$}. Now suppose $y \in R[x]$, then $y = b_0+b_1x+\ldots +b_{n-1}x^{n-1}$, where $b_i$ are elements of $R$, but this means $R[y]\subseteq R[x]$ is finitely generated by a subset of the generating set of $R[x]$, so the module theoretical equivalence theorem says $y\in R[x]$ is integral over $R$ as $R[y]$ is finitely generated as an $R$-module.
Is my reasoning correct?
Edit: On a second thought, I looked at the other two module theoretical equivalence of integral dependence, the third statement which is equivalent to the statement above says $R[x]$ is contained in a subring $C$ of $S$ such that $C$ is finitely generated $R$-module, now if we take $C=R[x]$, then $R[y]$ is contained in $R[x]$, the latter is a finitely generated $R$-module, then $y\in R[x]$ is also integral over $R$.
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