Let $R ⊆ S$ be an extension of rings, and let $u$ be a unit in $S$. Let $I$ be an ideal of $R$ with $I \ne R$. Show that $IR[u] \ne R[u]$ or $IR[u^{−1}] \ne R[u^{-1}]$.
Here is what I try: I have already proved that $u^{-1}$ is integral over $R$ if and only if $u^{-1}∈R[u]$. I suppose both $IR[u]=R[u]$ and $IR[u^{−1}]=R[u^{-1}]$. I want to should $u^{-1}\in R[u]$, so that I can conclude that $R[u^{-1}] $ is integral over $R$, and so $R[u^{-1}] $ is a finite generated $R$-module. so that $IR[u^{−1}] = R[u^{-1}]=<x_1,...,x_n>$, and so $Ix_1+Ix_2+...+Ix_n=Rx_1+Rx_2+...+Rx_n$. Thus we have $I=R$ is a contradiction.
my problem is I cannot succeed to show $u^{-1}\in R[u]$ after many tries, help please, or do I have the wrong idea somehow? thanks a lot!
Hint $\ $ If not then $\,IR[u] = (1) = IR[u^{-1}]\ $ which yields equations
$$\begin{align} a_0 + a_1 u + \cdots + a_n u^n &= 1,\ \ a_i \in I\\ b_0 + b_1 u^{-1} +\cdots + b_m u^{-m} &= 1,\ \ b_i \in I\end{align}$$
By symmetry, wlog we assume $\,n\ge m\,$ and, further, that $\,n\,$ is minimal.
Scale the 2nd equation by $\,u^n\,$ then use it to eliminate $\,u^n,\,$ obtaining an equation of the same type with smaller $\,n,\,$ contradiction.