Let $f: X\to B$ be a family of curves, i.e. $f$ is flat, surjective and of relative dimension 1. If each fiber is an affine curve, can we conclude that $f$ is an affine morphism? If it is not true, what additional conditions should we impose to make sure we get an affine morphism? Thanks.

Actually my real question is: given a family of proper curves $f:X\to B$, and given a collection of sections $p_1,p_2,⋯,p_s$. If we know $X_b\backslash \{ p_1(b),⋯,p_s(b)\}$ is affine for each $b\in B$, then is $f:X \backslash p_1(B)\cup \cdots \cup p_s(B)\to B$ an affine morphism?

  • 5
    $\begingroup$ The map $\mathbb A^2-\{0\} \to \mathbb A^1$ given by projecting on to the x axis seems to be a counterexample. $\endgroup$ – Sam Gunningham Apr 25 '18 at 15:59
  • 2
    $\begingroup$ Related (but not identical): mathoverflow.net/q/296764/82179 $\endgroup$ – R. van Dobben de Bruyn Apr 25 '18 at 16:04
  • $\begingroup$ Actually my real question is: given a family of proper curves $f: X\to B$, and given a collection of sections $p_1,p_2,\cdots, p_s$. If we know $X_b\backslash \{ p_1(b), \cdots, p_s(b)\}$ is affine, then is $f: X\backslash p_1(B)\cup \cdot \cup p_s(B) \to B $ also affine? $\endgroup$ – JJH Apr 25 '18 at 16:25
  • 4
    $\begingroup$ That is also false, and the example of @SamGunningham adapts to give a counterexample. Begin with $B=\mathbb{A}^1$ and with $\widetilde{X}=B\times \mathbb{P}^1$. Now define $\nu:\widetilde{X}\to X$ to be the morphism obtained by glueing the point $(0,\infty)$ to the point $(0,0)$. The projection $\text{pr}_1:B\times \mathbb{P}^1\to B$ factors through $\nu$ to give $f:X\to B.$ This morphism is proper and flat. The infinity section of $\widetilde{X}$ gives an infinity section $p_1:B\to X$. The complement of the image of $p_1$ is $\mathbb{A}^2-\{(0,0)\}$. $\endgroup$ – Jason Starr Apr 25 '18 at 17:37
  • $\begingroup$ @Jason Starr , in your counter-example, at the fiber over $o\in \mathbb{A}^1$ the marked point is not smooth. If we assume the section is a family of marked smooth points, can we conclude the affineness? $\endgroup$ – JJH Apr 25 '18 at 18:03

Lemma. Let $f \colon X \to B$ is a proper, flat family of relative dimension $1$ with geometrically connected fibres. Let $\sigma_1,\ldots,\sigma_r$ for $r \geq 1$ be sections landing in the locus where $f$ is smooth. If all fibres $X_b \setminus \{\sigma_1(b),\ldots,\sigma_r(b)\}$ are affine, then $X \setminus \sigma_1(B) \cup \ldots \cup \sigma_r(B) \to B$ is affine.

Proof. The question is local on $B$, so we may assume $B = \operatorname{Spec} A$ is affine. For every $b \in B$ and every $i \in \{1,\ldots,r\}$, the point $\sigma_i(b) \in X_b$ is a Cartier divisor, since it sits in the smooth locus of $X_b$. Hence, $\sigma_i(B)$ is a relative effective Cartier divisor [Stacks, Tag 062Y]; in particular it is a Cartier divisor. Write $\mathscr L_i = \mathcal O_X(\sigma_i(B))$, and $$\mathscr L = \bigotimes_{i=1}^r \mathscr L_i.$$ For any $b \in B$, we have $\deg((\mathscr L_i)_b) > 0$ on the component of $X_b$ containing $\sigma_i(b)$. Since $X_b \setminus\{\sigma_1(b),\ldots,\sigma_r(b)\}$ is affine, every component of $X_b$ contains at least one marked point. Hence, $\deg(\mathscr L_b) > 0$ on each component of $X_b$, so $\mathscr L_b$ is ample [Stacks, Tag 0B5Y].

Therefore, $\mathscr L$ is ample [Stacks, Tag 02DN], using that $B$ is affine (see [Stacks, Tag 01VK]). Then some multiple $\mathscr L^d$ for $d \gg 0$ is very ample and defines a closed immersion $X \to \mathbb P^N_B$ for some $N$. Then $$d \sum_{i=1}^r \sigma_i(B) \in |\mathscr L^d|$$ is the intersection of $X \subseteq \mathbb P^N_B$ with a hyperplane, hence the complement $X \cap \mathbb A^N_B$ is affine. $\square$.


[Stacks] A.J. de Jong et al, The stacks project.

  • $\begingroup$ Given a family of curves of genus zero $f:X\to B$, with a degeneration to $X_{b_0}$: two projective lines with one intersection point. I thought we may have a section $\sigma: B\to X$ with $\sigma(b_0)$ in one component of $X_{b_0}$. In this case, $X_{b_0}\backslash \{\sigma(b_0)\}$ is not affine. $\endgroup$ – JJH Apr 25 '18 at 19:43
  • 1
    $\begingroup$ @Hong. I am sure that RvDdB intended to add the hypothesis that every geometric fiber $X_b\setminus \{\sigma_1(b),\dots,\sigma_r(b)\}$ is affine (which was your hypothesis above). $\endgroup$ – Jason Starr Apr 25 '18 at 19:46
  • $\begingroup$ Yes, with such assumption, we don't need to move points from one component to the other, since sometimes it is not possible. $\endgroup$ – JJH Apr 25 '18 at 19:49
  • $\begingroup$ @Hong: of course you are absolutely right; this is a slight oversight. Let me fix that. $\endgroup$ – R. van Dobben de Bruyn Apr 25 '18 at 21:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.