0

We work over a fixed algebraically closed field. The product of affine varieties is reasonably easy to describe; the product of projective varieties is already much more involved, using the Segre embedding. How about arbitrary varieties? By "variety" I mean a separated, locally ringed space admitting a finite open cover by affine varieties (so not necessarily quasi-projective). More in detail, my question splits down into the following two questions:

  • Varieties are schemes, and the (fibered) product of schemes exists. But is it again a variety?
  • How can we describe this product? If we are given affine open covers $(U_i)_i$ of $X$ and $(V_j)_j$ of $Y$, is $(U_i \times V_j)_{i,j}$ an affine open cover of $X \times Y$?
57Jimmy
  • 6,266
  • 1
    The answers to both of these questions depend on what your definition of a variety is. Please add this information in your post. – KReiser May 06 '20 at 04:00
  • @KReiser Thanks, I've made the question more precise – 57Jimmy May 06 '20 at 08:38
  • Note: Typically one calls something "variety" only after additionally imposing the reducedness (or even worse, integrality) condition. Anyway, the conclusion still holds in both these versions, since tensor product of two reduced algebras (integral domains, resp.) over an alg. closed field is again reduced (integral domain, resp.). – Pavel Čoupek May 07 '20 at 04:33

1 Answers1

1

Yes, both of these are true. One quibble before we start: the finiteness assumption you have for your definition could use a little work - presumably what you really want here is finite type over a field.

To see that the fiber product of varieties is again a variety, it is enough to know that if $X\to Z$ and $Y\to Z$ are both finite type and separated, then $X\times_Z Y\to Z$ is again finite type and separated. This is the fact that these notions are stable under base change and the composition of two morphisms which both have these properties also has those properties. StacksProject has references for the separated case at tag 01KU and the finite type case at tags 01T4 and 01T3; Vakil's text also has relevant material, as do typical books like Hartshorne. You may want to attempt to prove these at some point on your own, but when to do that is a matter of taste and mathematical maturity combined with how much algebraic geometry you end up wanting to do.

For your second question, this sort of thing is actually covered by the construction of the fiber product. If we have $X\to Z$ and $Y\to Z$, the construction of the fiber product $X\times_ZY$ happens by producing an open cover consisting of schemes of the form $X_i\times_{Z_k} Y_j$ where $X_i,Y_j,Z_k$ are affine open subschemes of $X,Y,Z$ respectively with $X_i$ and $Y_j$ mapping in to $Z_k$. This is easy in our case, because $Z=\operatorname{Spec} k$ is affine, so any open affines of $X$ and $Y$ map in to the affine scheme $Z$, and we have $X_i\times_Z Y_j$ is again an open affine, and the union of all of these covers $X\times_ZY$. Again, I'd encourage you to consult a textbook or other reference source on the fiber product if you've never seen the construction worked out all the way. (Here's the StacksProject section.)

KReiser
  • 65,137
  • Thanks a lot. Just for the sake of clarity, I agree that saying finite-type might be more precise, but why isn't it enough to say that there is a finite cover by open affines (over the base field)? Isn't it an equivalent condition, as long as we then see our variety as a locally ringed space over the base field? Or do you mean that there could be incompatible covers that give the same structure of locally ringed space but not the same structure of locally ringed space over the base field? – 57Jimmy May 06 '20 at 10:25
  • Mostly I said finite type because you didn't include your definition of an affine variety. What you really want to say is "there's a finite open cover by the spectra of finitely generated k-algebras" and all reasonable definitions are equivalent to this one. I do not mean to mention anything like the pathology you came up with in the last sentence: it is entirely possible to live life without thinking about stuff like that, and I find I am generally happier without thinking of things like that. – KReiser May 06 '20 at 11:13
  • Hahaha, ok, fair :) – 57Jimmy May 06 '20 at 12:02
  • I don't know if it's so pathological though. At the bottom, I think the point is really just that, for $k$-algebras $A$ and $B$, $\mathrm{Hom}\mathrm{Rings}(A,B) \neq \mathrm{Hom}{k\mathrm{-Alg}}(A,B)$. But I don't know whether the given situation I mentioned could actually happen. – 57Jimmy May 06 '20 at 12:07
  • It can, see this answer of Eric Wofsey. Frequently, when one works with $k$-varieties, one assumes all the maps involved are $k$-linear and thus there's no "funny business". – KReiser May 06 '20 at 19:09