An integral ring extension that $B/M$ is not separable over $A/m$ I am trying to find an example like this.
Let $A$ be integrally closed in its field of fraction $K$ and $L$ a finite Galois extension. Let $B$ be the integral closure of $A$ in $L$. Is there a constellation such that there  a maximal ideal $m$ of $A$ and a maximal ideal $M$ of $B$ lying above $m$, such that $B/M$ is not separable over $A/m$?
 A: Let $k$ be a field of characteristic $p$, and $x,t$ be two independent variables. We set $K=k(t,x)$, and $A=k(t)[x]_{(x)}$. In other words, $A$ is the local ring at the origin of the affine line over the field $k(t)$. Let $L$ be the finite extension of $K$ defined by the polynomial $f(y)=y^p+xy-t$, and $B=A[y]/(f(y))$.  It is easy to see that $f(y)$ is irreducible over $K$, and $f'(y)=x \neq 0$, so $L/K$ is separable.  By prop 15 on pp.18 of Local fields by J-P. Serre,  $B$ is a discrete valuation ring, and hence the integral closure of $A$ in $L$. Clearly the residue field extension $k(t^{1/p})/k(t)$ is not separable. 
A: Take $$A = \mathbb{Z}[X^{2}]_{(2,X^{2})}[T], \ B = \mathbb{Z}[X]_{(2,X)}[T],$$
where by $\mathbb{Z}[X]_{(2,X)}$ we mean localization of $\mathbb{Z}[X]$ in the prime ideal $(2,X)$.
Then $K = \mathbb{Q}(T, X^{2}), \ L = \mathbb{Q}(T,X)$, so $L/K$ is finite Galois as a quadratic extension in characteristic $\neq 2$. 
It is also clear that $A,B$ are integrally closed (since ring of polynomials over integrally closed domain is integrally closed and a localization of integrally closed domain is integrally closed; alternatively: observe that $A, B$ are UFDs), and $B$ is integral over $A$.
But if we take $$M = 2B + (X^{2}T-1)B,$$ then $$\mathfrak{m} = 2A + (X^{2}T-1)A$$
(for $\mathfrak{m}\subseteq M\cap A\subsetneq A$ and $\mathfrak{m}$ is maximal since $A/\mathfrak{m}\cong \mathbb{F}_{2}(X^{2})$) and the extension $B/M | A/\mathfrak{m}$ is isomorphic to $\mathbb{F}_{2}(X) | \mathbb{F}_{2}(X^{2})$ which is purely inseparable.
