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This problem came to my mind while working on a related from for a reading I am doing.

Let us say we have $X$ some scheme over a scheme $S$ (where most easily we could take $S$ to be affine). Given $T\rightarrow S$, we get $X_T=X\times_S T$ over $T$ (again we could take $T$ to be affine).

Let us say I knew all the line bundles over $X_T$, is there a good way to translate this into knowing the line bundles of $X$. Clearly, any line bundle of $X$ when base changed to $T$ must be one that we already know. Are there conditions we can give to say that there is only one line bundle (up to isomorphism) that base changes to each line bundle in $X_T$?

My question is vague, so a real answer may not be possible, but I am wondering what tools I should examine to look at this problem.

Thank you for any help or direction.

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    $\begingroup$ I think you may start browsing the key words relative Picard scheme, relative Picard functor. $\endgroup$
    – Brenin
    Commented Dec 4, 2014 at 19:58

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In the generality you envision the answer is a resounding no!
For example take a variety $X$ over a field and take $S=X$ with the identity $id: X\to S$ as structure morphism.
Then for any rational point $s\in S$ the corresponding morphism $T=Spec(k)\to S$ with image $s$ has as corresponding pull-back $X_T=Spec(k)$ and although you know the unique line bundle over $Spec(k)$ (duh!), that won't be a big help for determining the line bundles of a completely arbitrary $k$-variety $X$ (with a rational point) !

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  • $\begingroup$ Yeah. I figured that, but I was thinking about more general examples, like if the base extension is like some extension of base ring. Something there must be times where you can hope to recover the information $\endgroup$ Commented Dec 4, 2014 at 22:57
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    $\begingroup$ No extensions of base ring don't work. You could construct a similar example with $X=S=Spec(R)$ the spectrum of a domain $R$ and then extend to the spectrum $T=Spec(K)$ of the fraction field $K=Frac(R)$. Again $X_T$ is just the single point $Spec(K)$. $\endgroup$ Commented Dec 4, 2014 at 23:20

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