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I am currently learning Algebraic geometry and I came over something that is bothering me.

I have an algebraic variety and the Picard group over it is defined as the set of classes of line bundles quotiented by the the relation of being isomorphic.

First of all it seems to me that these classes are actually proper classes and not sets, since I can equip almost any set that has the right cardinal with a structure of line bundle over a given variety. How can I collect them then into a set if they are in fact proper classes?

And even if I do, how can I guarantee that I actually won't get a proper class?

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The first difficulty is handled by Scott's trick: given a class $K$, let $\alpha$ be the smallest ordinal such that there are elements of $K$ of rank $\alpha$ (that is, in $V_\alpha$); then we can think instead of the set $K\cap V_\alpha$. This gives a canonical way of describing an equivalence (proper) class, without having to pick representatives.

So Scott's trick lets us replace each equivalence (proper) class with a set. Now we just need to count those sets. Since the powerset of $V_\alpha$ is again a set, it's enough to show that every line bundle has an isomorphic bundle of low rank; specifically, that there is some fixed $\alpha$ such that every line bundle over $X$ is isomorphic to one in $V_\alpha$. But this isn't hard to do; the space $X$ and the field $k$ lie in some $V_\beta$, the underlying set of the total space of a line bundle can be thought of as $k\times X$ (namely: for any line bundle, there's an isomorphic one with total space $k\times X$), and the additional data consists of some maps and some sets, all of which will have rank at most $\beta+3$ (my counting might be off by $1$ but oh well).

In particular, the Picard group can be taken to have underlying set consisting of a bunch of subsets of $V_\alpha$, which are the intersections of equivalence (proper) classes with $V_\alpha$, for $\alpha$ of appropriately high rank.

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You can fix a set of a certain cardinal and only look for line bundles whose underlying set is a subset of that set. $\Bbb R$ is probably enough most of the time.

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This is a common issue you will have to get used to. It's similar to the observation that zero-dimensional $k$-vector spaces are a proper class, because for any set $A$ we can give $\lbrace A\rbrace$ the structure of a $k$-vector space. However up to isomorphisms you of course obtain an actual set.

Ususally this is resolved by taking Grothendieck universes. This has the disadvantage that it assumes the existence of certain cardinals, whose existence is supposed to be independent of ZFC. So in order to do this properly you have to add one axiom to your set theory.

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  • $\begingroup$ I was hoping to avoid having to use Groethendieck universe. My reasoning is, that line bundles probably can't have too large cardinal. Seeing line bundle as a variety with a map to our original variety there inverse images of points are isomorphic to a line, it probably can't have too large cardinal. The only issue I have is, whether I can embed the structure sheaf of these varieties into some set. $\endgroup$
    – Keen
    Jun 23 '17 at 13:50
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    $\begingroup$ There is absolutely no need for Grothendieck universes here. $\endgroup$ Jun 25 '17 at 19:45
  • $\begingroup$ @Keen Note that a sheaf on a space is always a set. A sheaf is a map from the opens of a space to some category, satisfying some properties; since there are only set-many open sets in the space, the sheaf only "uses" set-many objects in the category. So there's really no difficulty at all, here. Set-theoretic issues can start to crop up when discussing the class of all sheaves, or similar, but that's different, and again they can still be sidestepped almost always by using Scott's trick. $\endgroup$ Jun 25 '17 at 19:49

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