This question says the following: Let $R$ and $S$ be local rings with the maximal ideals $M$ and $N$, respectively. Assume that $R\subset S$ and that $S$ is a finitely generated $R$-module. If there exists a proper ideal $I$ of $R$ such that $I=IS \cap R$ and the canonical image of $R/I$ in $S/IS$ equals $S/IS$, then prove that $R=S$.
Now, I am confused about the above result, since it seems that the following is a counterexample: $R=\mathbb{C}[x(x-1)]_{\langle x(x-1) \rangle}$, $S=\mathbb{C}[x]_{\langle x \rangle}$, $I=M=Rx(x-1)$. Notice that $IS=N=Sx$ and then $I=IS \cap R$. If I am not wrong, since $R \subseteq S$ is algebraic (actually, $S$ is finitely generared as an $R$-module by $1,x$), we obtain that $R/I \subseteq S/IS$ is algebraic. Moreover, this extension is actually $R/M \subseteq S/N$ and it is a field extension of degree two (it is enough for us that it is algebraic field extension). Also, $R/M = \mathbb{C}$ and $S/N = \mathbb{C}$. Since there are no proper algebraic extensions of $\mathbb{C}$, we get that $R/M = S/N$. Therefore, all the conditions of the theorem are satisfied, hence the conclusion $R=S$, but here clearly $R \neq S$.
Question: Where is my error?
Any hints and comments are welcome! Thank you.
This is based on @user237522's comment.
The extension $R \subset S$ is not finite. The extension $A:=C[x^2-x] \subset B:=C[x]$ is finite (of degree two). In $B$, $(x)$ and $(x-1)$ are the maximal ideals of $B$ containing $x^2-x$. With $W = A- (x^2-x)A$, we have the induced inclusion of locaizations $W^{-1}A \subset W^{-1}B$. By definition $W^{-1}A = R$. However, since $W^{-1}B$ has two maximal ideals (extensions of $(x)$ and $(x-1)$), it is not equal to $S$. Indeed, $S$ is the localization of $W^{-1}B$ at the extension of $(x)$. Note that $W^{-1}B \to S$ is not finite (this can be seen as there is no prime ideal in $S$ lying over the maximal ideal $(x-1)W^{-1}B$). Thus, $R \subset S$ is not finite.