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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 $Y$?
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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 $Y$.

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  • $\begingroup$ The map on the left may not be an injection, though it it is irrelevant for this discussion. $\endgroup$
    – Mohan
    Commented 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
    Commented 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
    Commented Sep 12, 2016 at 15:41
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    $\begingroup$ @abx What if you just remove like $33$ points from $\mathbb{P}^1$? $\endgroup$
    – SomeGuy
    Commented Sep 12, 2016 at 15:49
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    $\begingroup$ Oops! You (and Mohan) are right. I forgot that there are nontrivial invertible functions on $U$. I edit. $\endgroup$
    – abx
    Commented Sep 12, 2016 at 16:40

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