An example of a map that is not unramified in a specific way

Question

In Milne - Étale cohomology, the definition of an unramified morphism is given as follows (here I am using standard notation for sheaves, stalks, maximal ideals in the stalk and residue fields):

Let $f: Y \rightarrow X$ be a map of schemes, locally of finite type, and let $y \in Y$. Write $x = f(x)$. We say $f$ is unramified at $y$ if

1. The image of $m_x$ in $\mathcal{O}_{Y, y}$ generates $m_y$.

2. The resulting map of fields $\kappa(x) \rightarrow \mathcal{O}_{Y,y}/m_x \mathcal{O}_{Y,y} \overset{!}= \kappa(y)$ is a finite separable field extension.

My question now is: what is an example of such a map where condition 1. is satisfied, but the resulting field extension in 2. is infinite? Since this finiteness condition is part of the definition, such an example probably exists.. This is just out of curiosity and to build more intuition for the concept.

Some remarks: Since the question is local we can reduce to $X$ and $Y$ affine and $f$ of finite type. Then the map $\Gamma(X, \mathcal{O}_X) \rightarrow \Gamma(Y, \mathcal{O}_{Y})$ makes the latter into a finitely generated algebra of the former, which implies that in fact $\kappa(y)$ is a finitely generated $\kappa(x)$ algebra. In particular if this field extension would be algebraic, it would be finite, so an example like the one I am looking for must be a transcedental extension. (In fact I have seen that Hartshorne requires 2. only to be an algebraic extension, which by the above remark is equivalent.)

I have tried numerous examples but somehow satisfying condition 1. automatically implied the field extension to be finite. Maybe the problem is that I worked only with varieties (in my case separable and of finite type over $k = \overline{k}$. A proof that in that case 1. would imply that the extenson in 2. is finite would be very interesting in its own right!

Answer 1

Let $k$ be a non perfect field of characteristic $p\gt 0$, $a\in k$ an element with no $p$-th root in $k$ and let $$K=k[\sqrt[p] a]=\frac {k[T]}{\langle T^p-a\rangle}$$ We thus obtain a non-separable extension $k\to K$, of dimension $[K:k]=p$.
The associated finite morphism of one point affine schemes $$f: \operatorname {Spec}(K)\to \operatorname {Spec}(k)$$ satisfies 1. but not 2. at their unique point.

Edit
Here is (as was actually required!) an example with $\kappa(x)\to \kappa (y)$ infinite.
Let $\mathbb Z_p \to\mathbb Z_p[T] $ be the inclusion and let $$f:Y=\operatorname {Spec}(\mathbb Z_p[T])\to X=\operatorname {Spec}(\mathbb Z_p)$$ be the associated morphism of finite type between the corresponding affine schemes.
Taking $y= p\mathbb Z_p[T] \in Y$ and $x=p\mathbb Z_p \in X$, we get an example where 1. is satisfied at $y$ , but 2. is not since $\kappa (x)=\mathbb F_p\to \kappa (y)=\mathbb F_p(T)$ is not finite.