I will try to be as precise as possible, since my last post was probably too vague to explain well my problem, that now may be solved. I'll follow these conventions, that should be coherent with those used in Gathmann's Algebraic Geometry:
- when talking about ringed spaces, I'm always referring to the category of ringed spaces of $k$-valued maps, for a field $k=\bar k$.
- an affine variety by definition is a ringed space, and so any ringed space isomorphic to an affine variety will be called again affine variety. Then a prevariety is a ringed space $X$ whose underlying space admits an open covering $\{X_i\}_{i\in I}$, such that $X$ restricted to $X_i$ is an affine variety for all $i\in I$. The morphisms of prevarieties (and so also of affine varieties) are those of ringed spaces;
- the products, where not specified otherwise, will be meant in the category of prevarieties. In particular the product of two affine varieties is again an affine variety, and we always choose to construct it using as projections the set-theoretical ones.
Given prevarieties $X$ and $Y$, with affine open covering $\{X_i\}_{i\in I}$ and $\{Y_j\}_{j\in J}$, we want to construct $X\times Y$. The aim is to get a prevariety gluing all the affine varieties $\{X_i\times Y_j\}_{i,j\in I\times J}$.
(According to the conditions in Construction 5.6) such affine varieties will surely glue if the (set) identity between $U$ and $V$ is a ringed space morphism, where $U$, $V$ are the ringed spaces obtained restricting $X_{i_0}\times Y_{j_0}$, $X_{i_1}\times Y_{j_1}$ to the set $(X_{i_0}\cap X_{i_1})\times (Y_{j_0}\cap Y_{j_1})$, for fixed $i_0,i_1\in I$ and $j_0,j_1\in J$.
I thought this: if $Z_0$, $Z_1$ are affine varieties and $W_0\subset Z_0$, $W_1\subset Z_1$ are open subsets (so prevarieties), the restriction of $Z_0\times Z_1$ to the set $W_0\times W_1$ has the universal property of the product of prevarieties $W_0\times W_1$, with projections the set-theoretical ones. Hence $U$ and $V$ have both the universal property of $(X_{i_0}\cap X_{i_1})\times (Y_{j_0}\cap Y_{j_1})$, with the same (set-theoretical) projections, so the identity is the unique morphism commuting with them. (The ringed space structures on $X_{i_0}\cap X_{i_1}$ obtained by restricting $X_{i_0}$ and $X_{i_1}$ are equal, since they both come from $X$, so $X_{i_0}\cap X_{i_1}$ is a well-defined prevariety, and the same holds for $Y_{j_0}\cap Y_{j_1}$). Does it make sense to you now? Thanks
Best Answer
Yes, your argumentation seems correct to me.
I looked at some old notes of mine where I was trying to figure out the same thing: how products of prevarieties work, with the same conventions as in the question. I have added below an overview of what I ended up doing, as it may be helpful for comparison.
My argumentation went as follows (with most details omitted):
We hence understand products of affine varieties in the category of affine varieties.
We hence understand the product of affine varieties in the category of prevarieties.
We hence understand products of subprevarieties of affine varieties. We have also shown (2).
From now on let $X$ and $Y$ be arbitrary prevarieties.
Step 5. Consider the coverings $X = \bigcup \mathcal{U}$ and $Y = \bigcup \mathcal{V}$ by all affine open subsets. Then, $\{ U × V \mid U ∈ \mathcal{U}, V ∈ \mathcal{V} \}$ is a cover of $X × Y$, and we have on each $U × V$ the structure of an affine variety. To glue these structures together, we need to check that for all $U, U' ∈ \mathcal{U}$ and $V, V ∈ \mathcal{U}$:
Set $U'' ≔ U ∩ U'$ and $V'' ≔ V ∩ V'$, open prevarieties in $X$ and $Y$. The set $U'' × V''$ is open in the product topology of $U × V$ and thus also in the Zariski topology; similarly for $U' × V'$. As seen in step 3, restricting the structure of $U × V$ to $U'' × V''$ gives us the categorical product of the prevarieties $U''$ and $V''$ as described in (1). But the same also holds for $U' × V'$. By the uniqueness from step 4, both structures on $U'' × V''$ must agree.
We have thus made the set $X × Y$ into a prevariety. We also have (3) by construction.
Step 6. We check that $π_X$ and $π_Y$ are morphisms of prevarieties. This can be concluded from the fact that $π_U$ and $π_V$ are morphisms for all $U ∈ \mathcal{U}$, $V ∈ \mathcal{V}$.
Step 7. In needs to be shown that $X × Y$ together with $π_X$ and $π_Y$ satisfies the universal property of the product: given morphisms $φ \colon Z \to X$ and $ψ \colon Z \to Y$ we need to show that $⟨φ, ψ⟩ \colon Z \to X × Y$ is again a morphism. This can be checked on affine open subsets.
We have thus shown (1).