Under what kind of assumptions on $X$ and $k$ (here $X$ is a scheme over $k$ and $\bar{k}$ the algebraic closure) is the map $\mathcal{L}\mapsto \mathcal{L}\otimes_k\bar{k}$ from $\mathrm{Pic}(X)$ to $\mathrm{Pic}(X\otimes_k\bar{k})$ injective?
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For example, if $X$ is proper, integral and normal, then this is true. – Cantlog May 28 '14 at 17:38
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Why do one needs normality? Maybe proper and integral suffices... – user109227 May 28 '14 at 20:03
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In my proof, the normality is to insure that a rational function without zero nor pole divisor is a unit. But properness is useless. – Cantlog May 29 '14 at 16:48
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Probably this holds for any scheme X and any field $k$. – Cantlog May 29 '14 at 22:12