2
$\begingroup$

Let $f\colon X\to Y$ be a surjective morphism of smooth projective complex algebraic varieties. Assume $f$ has connected fibres. Let $\eta$ be the generic point of $Y$ and let $X_\eta$ be the generic fibre of $f$. Consider the pullback map $$\imath^*\colon {\rm Pic}(X)\to {\rm Pic}(X_\eta)$$ induced by the morphism $\imath\colon X_\eta\to X$. Questions:

  1. Under which conditions is $\imath^*$ surjective?
  2. Is the kernel of $\imath^*$ generated by line bundles corresponding to divisors whose support does not dominate $Y$?
$\endgroup$

1 Answer 1

6
$\begingroup$

If $U$ is any open subset of $X$, there is an exact sequence $$ \oplus\,\mathbb{Z} . D\rightarrow \mathrm{Pic}(X)\rightarrow \mathrm{Pic}(U)\rightarrow 0$$ where the first sum is over the irreducible divisors supported in $X\smallsetminus U$ -- this follows easily from the description of $\mathrm{Pic}$ as divisors modulo linear equivalence. In your situation, taking $U=f^{-1}(V)$ for $V$ open in $Y$ and passing to the limit, you get an exact sequence $$ \oplus\,\mathbb{Z}. D\rightarrow \mathrm{Pic}(X)\rightarrow \mathrm{Pic}(X_{\eta })\rightarrow 0$$ where the first sum is now over all irreducible divisors which do not dominate $Y$.

$\endgroup$
10
  • $\begingroup$ The map on the left may not be an injection, though it it is irrelevant for this discussion. $\endgroup$
    – Mohan
    Sep 12, 2016 at 15:21
  • $\begingroup$ I think it is injective. Just apply the snake lemma to the diagram of exact sequences where the first row is $0\rightarrow K_X^*/\mathbb{C}^*\rightarrow \oplus \,\mathbb{Z}.D\rightarrow \mathrm{Pic}(X)\rightarrow 0$, and the second row the same replacing $X$ by $U$. $\endgroup$
    – abx
    Sep 12, 2016 at 15:23
  • $\begingroup$ Many thanks. About passing to the limit, is there such an open subset $U\subseteq X$ with ${\rm Pic}(U)\simeq {\rm Pic}(X_\eta)$? $\endgroup$
    – pip
    Sep 12, 2016 at 15:41
  • 3
    $\begingroup$ @abx What if you just remove like $33$ points from $\mathbb{P}^1$? $\endgroup$
    – SomeGuy
    Sep 12, 2016 at 15:49
  • 1
    $\begingroup$ Oops! You (and Mohan) are right. I forgot that there are nontrivial invertible functions on $U$. I edit. $\endgroup$
    – abx
    Sep 12, 2016 at 16:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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