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Exercise II.3.5 (c) in Hartshorne, Algebraic Geometry, asks to find an example of a surjective, finite-type and quasi-finite morphism of schemes which is not finite.

I need to find a finitely generated $A$-algebra $B$ which is not finite generated as an $A$-module. The only examples, I could find, of such a kind of $B$ give rise to a morphism which is not quasi-finite. Basically I was trying to use some modification of the classic $B=\mathbb{C}[x]$. I have also thought to find a morphism which is not closed, since we know that a finite morphism is always closed, but even this way didn't lead me anywhere.

Do you have any suggestion?

P.S.: $f$ quasi-finite means that $f^{-1}(y)$ is a finite set for every point $y\in Y$.

MOREOVER: while thinking at this example, I asked another question to myself. Which is a quasi-finite morphism which is not of finite-type?

Thank you very much!

A. Prufrock
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  • Hint: have you looked at the previous exercise? Is there some way you could mess with this to get something which satisfies your goals? – KReiser Sep 24 '15 at 16:09
  • @KReiser, thank you for your reply. Do you mean the exercise characterizing finite morphisms or the one stating that a finite morphism is closed? In the first case I cannot see any way to mess anything and in the second case I have actually tried, but the only example I could found was not surjective. – A. Prufrock Sep 24 '15 at 16:41
  • I mean exercise II.3.4, which asks you to prove that $f:X\to Y$ is finite iff for every open affine $\mathrm{Spec} B = V\subset Y$, $f^{-1}(V)$ is affine and Spec of some finite $B$-module. What conditions can you violate about this statement? – KReiser Sep 24 '15 at 16:53
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    Just a cultural note: often finite type is part of the definition of quasifinite, in order to make the concept stable under base change. With these definitions stuff like $k \subset k(x)$ won't count. – Hoot Sep 24 '15 at 17:05
  • For the problem: could you do it without the surjectivity requirement? We can probably cheat to get around that. The target being the affine line could certainly work. – Hoot Sep 24 '15 at 17:07
  • @Hoot: Simplest example I know has the affine line as a target and does satisfy the surjectivity requirement. I'm sure there's other good examples, but it might be a good idea to focus on the affine requirement of II.3.4. Are there any non-affine schemes which map nicely to the affine line? – KReiser Sep 24 '15 at 17:12
  • @KReiser I'd be interested to see your example (of course, we should let the OP think a bit). For what I have in mind, what I mean by "cheating" is that the source could be a disjoint union. – Hoot Sep 24 '15 at 17:16
  • Without the surjectivity requirement I have thought to the projection from the hyperbola to the affine line. – A. Prufrock Sep 24 '15 at 17:24
  • @KReiser I am still thinking about your hint. – A. Prufrock Sep 24 '15 at 17:26
  • @user273503 Okay, good. The problem with that example is that you miss the origin, but why not just add some random point to the hyperbola? Does that cause any problems? – Hoot Sep 24 '15 at 17:47
  • @KReiser what do you think about the projection $\mathbb{P}^1\to\mathbb{A^1}$? It is quasi-finite, surjective and of finite-type since $\mathbb{P}^1$ can be covered by two copies of $\mathbb{A}^1$. – A. Prufrock Sep 24 '15 at 17:50
  • @user273503 Could you say what that map is? A map to $\mathbf A^1$ is the same as a regular function, and $\mathbf{P}^n$ does not have many of those. – Hoot Sep 24 '15 at 17:51
  • @Hoot, sorry I have just noted your last comment. Yeah, it is easy, why I couldn't think about that..? – A. Prufrock Sep 24 '15 at 17:52
  • @Hoot, I was thinking to map $[x:y]$ to $x$. – A. Prufrock Sep 24 '15 at 17:54
  • @user273503 that is not even well defined. $[1:0]$ is the same as $[2:0]$ but they map to different places. – KReiser Sep 24 '15 at 17:58
  • @Hoot, of course it isn't. Forget it, I am stupid. But thinking at $\mathbb{P}^1$ as $\mathbb{A}^1$ plus the point at the infinity, is not possible to find a morphism mapping the $\mathbb{A}^1$ isomorphically on itself and the point at the infinity to some other point? – A. Prufrock Sep 24 '15 at 18:02
  • @Hoot, anyway coming back to your hint, I cannot see any problem caused by adding an extra point to the hyperbola. I create the scheme $X$ as disjoint union of the hyperbola and the point, in particular it has two connected components and I may define a map from each component to the affine line. Am I missing something? – A. Prufrock Sep 24 '15 at 18:18
  • That's what I had in mind. Hopefully we're not both missing something. – Hoot Sep 24 '15 at 18:21
  • @Hoot, thank you very much for the help. I will try to produce an example by following KReisen too, but now I am happy because I have at least one solution. All the best! – A. Prufrock Sep 24 '15 at 18:24

4 Answers4

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The example given, $Y=\mathbf A^1_k$ for $k$ any field, $X=A\sqcup B$ for $A=\operatorname{Spec} k[x,y]/(xy-1)$, $B=\{\ast\}=\operatorname{Spec} k$, is valid. Here the map $f:X\rightarrow Y$ is induced by the map $\varphi:k[x]\rightarrow\Gamma(X,\mathscr O_X)=\Gamma(A\sqcup B,\mathscr O_A\times\mathscr O_B)\cong k[x,y]/(xy-1)\times k$, which is in turn induced by the universal property of the product by the maps $k[x]\rightarrow k[x,y]/(xy-1)$, $k[x]\rightarrow k$. It is obviously quasi-finite; it is of finite type since $X=A\sqcup B$ is a cover of $X$ by open affines and $Y$ is itself affine. Surjectivity of closed points is obvious. Finally, if $\eta$ is the generic point of $A$, and $\xi$ is the generic point of $Y$, then we ask what is $f(\eta)$. If $\mathfrak m\subset\mathscr O_{X,\eta}$ is the maximal ideal of the stalk, and $p_\eta:\Gamma(X,\mathscr O_X)$ is the map into the stalk, then $f(\eta)$ is the point in $Y$ corresponding to $\varphi^{-1}p_\eta^{-1}(\mathfrak m)$. But now $$\mathscr O_{X,\eta}\cong (k[x,y]/(xy-1)\times k)_{(0,k)}\cong (k[x,y]/(xy-1))_{(0)}=k(x)$$ with $\mathfrak m=(0)$ in that ring, so $f(\eta)$ corresponds to $\varphi^{-1}((0,k))=(0)$.

Tomo
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  • Regarding the second paragraph: Do you argue that there is no affine example $\operatorname{Spec}(A) \to \operatorname{Spec}(B)$? But is your first example $X = A \sqcup B$ not just $\operatorname{Spec}(k[x,y]/(xy-1) \times k)$, hence affine? – red_trumpet Jan 28 '21 at 15:55
  • Yeah, thanks, I don’t know what that was about – Tomo Jan 28 '21 at 19:09
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Here is a different example.

Take $X=\operatorname{Spec}\mathbb Z[x]/(3x^2+2x+1)$ and $Y=\operatorname{Spec}\mathbb Z$. Then $X\to Y$ is clearly finite-type, but not finite since it isn't integral. Furthermore, the fibers over $(p)$ and $(0)$ are $\operatorname{Spec}\mathbb F_p[x]/(3x^2+2x+1)$ and $\operatorname{Spec}\mathbb Q[x]/(3x^2+2x+1)$, respectively, which are finite but nonempty, as desired.

Jingdan Hua
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One can also find examples over the complex numbers in which the source is irreducible:

Let $f \colon \mathbb{A}^{1} \to \mathbb{A}^{1}$ be the ramified cover given by $z \mapsto z^{2}$ at the level of coordinate rings. Let $i \colon \mathbb{A}^{1} \setminus \{ 1 \} \to \mathbb{A}^{1}$ be the open immersion of the complement of a closed point which is not the origin. Then $f_{0} := f \circ i$ is quasi-finite and surjective, but not proper, hence not finite.

One can check that $f_{0}$ is not proper with the valuative criterion for properness, but the intuititon for this example came from the analytic topology. In fact, when I googled "a finite morphism is closed in the analytic topology" I found this answer, which is exactly the kind of example that I had in mind and seems to confirm the suspicion I had about finite morphisms being closed in the analytic topology (doesn't really confirm it, but it seems that the author was thinking along similar lines at least).

Pedro
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Here's another example of a slightly different flavor that might be interesting to some: let $X$ be the line with doubled origin, $Y$ the line, $f:X\to Y$ the obvious map (i.e., the identity on each chart.) Then $f$ is surjective and quasi-finite, since the pre-image of any point has either one or two points. Furthermore, $f$ is finite-type, since $f^{-1}(Y)$ can be covered by affine opens such that the induced maps on rings are finite-type (in fact, they are even finite.) But $f$ is not finite, since finite maps are required to be affine, and this map is not affine.

This highlights a subtle but key difference between the way that finite-type and finite morphisms are defined: finite-type only requires that you can cover the pre-images of affine opens with affine opens that are finite-type, while finite requires that the pre-image of an affine itself be affine and finite. The asymmetry in the definition might seem strange at first, but we want finite-type to be a fairly weak assumption, while finite should be extremely strong.

cnunn
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