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

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!

• I mean, the counterexample is going to have to be semi-contrived. For example, if $f$ is also locally quasi-finite then it's not hard to see that 1) implies the finiteness in 2). I feel like in most reasonable situations 1. should imply locally quasi-finite. Like, suppose that $f:X\to Y$ is surjective map of varieties (we can just replace $Y$ by its image), then $\dim X\geqslant \dim Y$. But, let $y$ be a closed point of $Y$ and let $x$ be a point of $X$ mapping to it. Necessarily $x$ is closed. 1. then implies that $\mathrm{codim}_x X=\dim X\leqslant \mathrm{codim}_y Y=\dim Y$. – Alex Youcis Oct 20 '15 at 13:14
• So, this implies that $\dim X=\dim Y$. so if $f$ satisfies mild conditions (e.g. flat) then the fibers will be zero-dimensional, and so $f$ will be quasi-finite and then, as I mentioned before, this should show that 1) implies the finiteness in 2). EDIT: when I say varieties I mean the usual list (separated, finite type, reduced, equidimensional,..) EDIT EDIT I made a boo-boo above. What I should have said is take $x$ a closed point, then $f(x)$ is a closed point and then proceeded. It's non-sense that any closed point has only closed point preimages. – Alex Youcis Oct 20 '15 at 13:23
• Hi @AlexYoucis ! Thanks for your comments. But I just realised something, how about the map induced by $\mathbb{C} \rightarrow \mathbb{C}(T)$? Isn't it an example? Its fibers are obviously finite, which is quasi-finiteness of the map by definition right? So why does your logic not apply in this case? – Joachim Oct 20 '15 at 13:42
• Those aren't varieties. And that map isn't locally of finite type :P A locally finite type map between fields is finite. – Alex Youcis Oct 20 '15 at 13:46
• @AlexYoucis Woops I messed up the definition of finite type maps. ^^ But that makes me think, since the map $\kappa(x) \rightarrow \kappa(y)$ is of finite type doesn't that already imply it is finite? Or am I being silly again? – Joachim Oct 20 '15 at 13:54

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.

• I feel like we've had this encounter recently :) The OP was curious about the non-finiteness in 2. not the separability, I believe. – Alex Youcis Oct 20 '15 at 14:02
• Thanks! But this extension is still finite right? I was actually looking for an infinite extension (not necessarily inseparable). – Joachim Oct 20 '15 at 14:03
• Dear @1Alex and Joachim: sorry, you are right: I skipped that part of the question ! – Georges Elencwajg Oct 20 '15 at 14:09
• I have now made another attempt in an edit... – Georges Elencwajg Oct 20 '15 at 14:38
• @GeorgesElencwajg Nice example! That is extremely easy. Do you agree that any examples with varieties are bound to be pathological? – Alex Youcis Oct 20 '15 at 14:43