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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?

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    $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ 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
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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. $\square$.


References.

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

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  • $\begingroup$ 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
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    $\begingroup$ @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

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