Consider $R[x]$ and let $S$ be the subring generated by $rx$, where $r \in R$ is some non-invertible element. Then $x$ is not integral over $S$ 
Consider $R[x]$ and let $S$ be the subring generated by $rx$, where $r \in R$ is some non-invertible element. Then I want to show that $x$ is not integral over $S$

I'm not seeing why this is the case.
Note : Here R is a commutative ring.
 A: 
In the following I suppose that $R$ is a commutative unitary ring and $R[x]$ denotes the polynomial ring in one variable over $R$. 

Assume that $x$ is integral over $S=\Bbb Z[rx]$ (I guess this is what the OP means by "the subring generated by $rx$"). Then there exist $a_i\in S$ such that $x^n+\sum_{i=0}^{n-1}a_ix^i=0$. In fact $a_i=a_i(rx)$, that is, they are polynomials in $rx$. Since $x^n+\sum_{i=0}^{n-1}a_i(rx)x^i=0$ the monomial $x^n$ should be cancelled by other monomial of degree $n$ which appears in $\sum_{i=0}^{n-1}a_i(rx)x^i=0$. Unfortunately all the monomials of degree $n$ of this sum have coefficients multiples of $r$, so the cancellation is not possible.
A: Hint $\,\ 1\cdot x^n+f_{n-1}(rx)\,x^{n-1}+\cdots+f_{0}(rx) = 0\,  $ in $\,R[x],\ $ which, reduced mod $\,r$  
$\quad \Rightarrow\ \ \color{#c00}1\cdot x^n+\,f_{n-1}(0)\ x^{n-1}\,+\cdots+\,f_{0}(0)\, =\, 0\,$ in $\,(R/rR)[x]$
$\quad \Rightarrow\ \ \color{#c00}1 = 0\ $ in $\,R/rR,\,$ since, by definition, a polynomial $= 0$ iff all coefficients $= 0.$  
$\quad \Rightarrow\ \ \color{}1 \in rR,\ $ so $\,r\,$ is invertible in $\,R,\,$ contra hypothesis. 
A: Let $\bar{R} = R / rR$. Then $R[x] / rR[x] \cong \bar{R}[X]$ and the image $\bar{S}$ of $S$ in $\bar{R}[\bar{x}]$ is contained in $\bar{R}$ by assumption.
Suppose $x$ is integral over $S$. Then $X$ is integral over $\bar{S}$. Hence $X$ is integral over $\bar{R}$. The indeterminate $X$ is integral over $\bar{R}$ inside the polynomial ring $\bar{R}[X]$ only when $\bar{R}$ is the zero ring. So $rR = R$ and $r$ is a unit in $R$, a contradiction.
A: Assume that $x$ is integral over $S$. Then there is a polynomial $p(X)=x^n+a_1X^{n-1}+...a_n$, with $a_i$ in $S$ such that $p(x)=0$.
But since $a_i$ are in $S$, they are of the form $q_i(rX)$ for some polynomials $q_i$ with integer coefficients. 
We get $0=p(x)=x^n+q_1(rx)x^{n-1}+-...+q_{n}(rx)$. The latter is a polynomial in $x$. For it to be zero, the leading term must be zero. But the $x^n$ term is a multiple of $r$, say $rA$ plus $1$. But if $rA+1=0$, then $(-A)r=1$, which implies $r$ is invertible.
