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.

Any hints and comments are welcome!