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Let $X$ be a smooth projective variety, and $D\subset X$ a smooth nef and big divisor. Assume that the restriction map $Pic(X)\rightarrow Pic(D)$ is an isomorphism over $\mathbb{Q}$.

Under which hypothesis may we conclude that there exists an isomorphism $\mathrm{Pic}(X)\rightarrow \mathrm{Pic}(D)$ over $\mathbb{Z}$ ?

I am particularly interested in the case $dim(X)=3$.

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  • $\begingroup$ Fix a linear subspace $\Lambda\subset \mathbb{P}^n$ of codimension $3$ (so $n\geq 3$). Denote by $\pi:\mathbb{P}^n\setminus \Lambda \to \Pi$ the linear projection to $\Pi\cong \mathbb{P}^2$. In the blowing up $\nu:X\to \mathbb{P}^n$ along $\Lambda$ with exceptional locus $E$, consider the linear system of $\nu^*\mathcal{O}(2)(-2\underline{E})$. It seems to me that this is a big and basepoint free linear system (I should double-check). Yet a general member $D$ of this linear system is a projective space bundle over a conic $C\subset \Pi$, so $\text{Pic}(X)$ has index $2$ in $\text{Pic}(D)$. $\endgroup$
    – Jason Starr
    Commented Mar 16, 2017 at 22:59
  • $\begingroup$ I just realized that the linear system above is basepoint free, but it is not big. $\endgroup$
    – Jason Starr
    Commented Mar 16, 2017 at 23:40
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    $\begingroup$ I think it is easy to arrange examples where this map is not an isomorphism. (Let me know if I messed up.) For example let $X$ be the blowup of $\mathbf P^n$ along a line $L$ and a point $p$ not in $L$. Let $Q$ be a smooth quadric hypersurface meeting $L$ in 2 distinct points but not containing $p$, and let $D$ be the proper transform of $Q$ on $X$. Then both $X$ and $D$ have Picard number 3, $D$ is nef and big, but the exceptional divisor over $p$ restricts to $0$ in $\mathrm{Pic} \, D$. $\endgroup$
    – Bertie
    Commented Mar 16, 2017 at 23:48
  • $\begingroup$ By the way, in light of Jason Starr's comment below, it would be good to clarify whether you are assuming just that $\rho(X)=\rho(D)$, or that the map on Picard groups tensored with $\mathbf Q$ is an isomorphism. $\endgroup$
    – Bertie
    Commented Mar 17, 2017 at 9:19
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    $\begingroup$ Indeed we can assume that the restriction map tensored by $\mathbb{Q}$ is an isomorphism. Thanks. $\endgroup$
    – user82886
    Commented Mar 17, 2017 at 12:03

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