# $\Bbb Q$-line bundle definition

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?

• 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.) Jan 25 '20 at 4:53
• I think I got it right. See on page 38 of books.google.com/… Jan 25 '20 at 5:16
• 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. Jan 25 '20 at 6:47

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 $$\operatorname{Pic}(X)\otimes\mathbb{Q}$$.