# Cover for product of varieties

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 $$X \times Y$$?
• The answers to both of these questions depend on what your definition of a variety is. Please add this information in your post. Commented May 6, 2020 at 4:00
• @KReiser Thanks, I've made the question more precise Commented May 6, 2020 at 8:38
• 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.). Commented May 7, 2020 at 4:33

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.)
• 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. Commented May 6, 2020 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". Commented May 6, 2020 at 19:09