Let $k$ be aa arbitrary field.
Let $R=k[x(x-1)]_{\langle x(x-1) \rangle}$ and let $S=k[x]_{\langle x \rangle}$, $m=x(x-1)R$, $n=xS$, $k(m)=R/m$, $k(n)=S/n$.
We have, $mS = n$ (since $x-1$ is invertible in $S$).
Now, $k(m) \cong k$ and $k(n) \cong k$, so $k(m) \cong k(n)$.
I am trying to figure out if the residue field extension $k(m) \to k(n)$ is finite-dimensional or not.
Question: Is $k(m) \to k(n)$ finite-dimensional or not? and why?
Remarks:
(1) I think that it is infinite dimensional; indeed, I know that $A=k[x(x-1)] \subseteq k[x]=\frac{k[x(x-1)][T]}{\langle T^2-T-x(x-1) \rangle}=B$ is not separable, since $(T^2-T-x(x-1))'=2T-1$ evaluated at $x$, $2x-1$, is not a unit of $k[x]$.
$B$ is Noetherian+finitely generated $A$-algebra, hence $B \otimes_A B$ is Noetherian, and so the kernel of $B \otimes_A B \to B$ is a finitely generated ideal. In the case separablity is equivalent to (formal) unramifiedness. Therefore, being non-separable impies that $A \subseteq B$ is not (formally) unramified, so there exists a maximal ideal $N$ of $B$, such that $A_{N \cap A} \subseteq B_N$ is not unramified.
Now I am not sure for which maximal ideals $N$ of $B$, such localizations are not unramified.
(2) $A \subseteq B$ is free with basis $\{1,x\}$, hence flat, hence every such localization $R \subseteq S$ is flat.
(3) Perhaps I am wrong, but it seems to me that $\{\frac{1}{x-1},1,x\}$ is a set of generators for $S$ as an $R$-module (I can explain), so if for example $k=\mathbb{C}$ and $\mathbb{C} \cong k(m) \subseteq k(n) \cong \mathbb{C}$ was finite-dimensional, then $k(m)=k(n)$ and then $R=S$, which is not true of course.
Any hints and comments are welcome!