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Let $X$ be a variety over a field $k$. Let $L,L' \in Pic(X)$ be two line bundles and $a,a' \in \Bbb Z$. In what I am reading, it says that $(L,a) \simeq (L,a')$ if $L^{a'} \simeq L'^{a}$. The equivalence classes of this relation are called $\Bbb Q$-line bundles.

I am struggling to prove why this relation is transitive to begin with: if $(L,a) \simeq (L',a')$ and $(L,a) \simeq (L'',a'')$, then I can deduce that $L'^{aa''} \simeq L''^{aa'}$, but how does this imply $(L',a') \simeq (L'',a'')$?

It is also claimed that the group $Pic(X) \otimes \Bbb Q$ represents the equivalence classes of this relation. However I don't see how this correspondence is made.

Can someone help me out with these two basic questions? Am I simply misinterpreting things?

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  • $\begingroup$ Are you sure you've written this down correctly? Transitivity is clear if instead of asking that $(L,a)\simeq (M,b)$ mean $L^b\cong M^a$, we instead require $L^a\cong M^b$. (Also, what is "what you're reading"? Knowing the source could help diagnose this, maybe as a known typo or a misread or something else.) $\endgroup$
    – KReiser
    Jan 25 '20 at 4:53
  • $\begingroup$ I think I got it right. See on page 38 of books.google.com/… $\endgroup$
    – Hammerhead
    Jan 25 '20 at 5:16
  • $\begingroup$ You're right, my suggestion was incorrect (what a time to forget how to define localization, silly me). The missing part is explained in Eric Wofsey's answer below. $\endgroup$
    – KReiser
    Jan 25 '20 at 6:47
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Indeed, this can fail to be transitive. The correct definition is that $(L,a)\simeq(L',a')$ if there exists a nonzero integer $b$ such that $L^{a'b}\cong L'^{ab}$. With this correction, transitivity should be easy to check. (Note that we should also only be considering pairs $(L,a)$ where $a$ is a nonzero integer.)

Note that this is just the usual definition of the localization of $\operatorname{Pic}(X)$ as a $\mathbb{Z}$-module with respect to the set of nonzero integers: $(L,a)$ here represents the fraction $\frac{L}{a}$ and we just have the usual equivalence on fractions $\frac{L}{a}=\frac{L'}{a'}$ iff there exists $b\in \mathbb{Z}\setminus\{0\}$ such that $ba'L=baL'$ (here I write the tensor product of line bundles in additive notation to match the more familiar context of localization). So, that's why this gives $\operatorname{Pic}(X)\otimes\mathbb{Q}$.

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  • $\begingroup$ I suspected this to be the correct route, but wanted some experienced eyes to confirm. $\endgroup$
    – Hammerhead
    Jan 26 '20 at 6:52

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