# Why is Vakil's definition of "distinguished affine base" of a scheme not a base in the usual sense?

From Vakil's Foundations of Algebraic Geometry:

The open sets of the distinguished affine base are the affine open subsets of $$X$$. We have already observed that this forms a base. But forget that fact. We like distinguished open sets $$\operatorname{Spec} A_f \hookrightarrow \operatorname{Spec} A$$, and we don’t really understand open embeddings of one random affine open subset in another. So we just remember the "nice" inclusions.

13.3.1.Definition. The distinguished affine base of a scheme $$X$$ is the data of the affine open sets and the distinguished inclusions.

Vakil writes that this a "not a base in the usual sense."

Is this not a base in the usual sense?

If $$X$$ is a topological space, a collection of open subsets $$\mathcal B$$ forms a base if every open subset of $$X$$ is a union of elements of $$\mathcal B$$.

Let $$U$$ be an open subset of a scheme $$X$$. Let $$p \in U$$. Then $$p$$ is in some affine open subset of $$X$$, say $$\operatorname{Spec} A$$. Then $$p \in U \cap \operatorname{Spec} A$$, which is open in $$\operatorname{Spec} A$$, hence $$p \in \operatorname{Spec} A_f$$ for some $$f \in A$$. $$\operatorname{Spec} A_f$$ is an open subset of an open subset, hence open in $$X$$. So, $$p \in \operatorname{Spec} A_f \subset U$$.

• A base in the usual sense requires that the intersection of two basic open sets is a basic open set. Commented Nov 12, 2021 at 21:59
• @ZhenLin Can you point me to a reference? That's not the definition of basis I have. It seems you are describing the fact that if we start we a set $X$ and collection of subsets satisfying some conditions (one of them being the one you listed), then there is a topology on $X$ where the collection of subsets form a base. Commented Nov 12, 2021 at 23:25
• @ZhenLin You may describing Lemma 5.5.2 here: stacks.math.columbia.edu/tag/004O Commented Nov 12, 2021 at 23:26
• @ZhenLin That’s already false in the case of the canonical basis of $\mathbb{R}^2$ in the Euclidean topology Commented Nov 13, 2021 at 1:58
• I don't know which version you was reading when asking this question. I got a similear doubts when reading the context about it. But at the end I found I get it wrong. Vakil does not say it's not a base. In fact, his original word is (at least in the recent versions): "The distinguished affine base isn’t a topology in the usual sense — the union of two affine sets isn’t necessarily affine, for example." Hence he is saying it's not a topology, which does make sense. And I think that this sentence is really easy to read it in a wrong way and change the word "topology" to "base"(at lest for me). Commented Jan 26, 2023 at 15:49

As he says in the text, Vakil is describing a (non-full) subcategory of the category of open sets of the scheme $$X$$. Namely, he is considering the category whose objects are open affine subsets $$U$$ and where there is a single morphism $$U \to V$$ if and only if $$U$$ is a distinguished open affine of $$V$$
The reason this is not a base in the usual topological sense is because it could be that we have two affine opens $$U \subseteq V$$, with a distinguished affine open $$W$$ in $$U$$ that is not distinguished open affine in $$V$$. In other words, in the distinguished affine base we “forget” that $$W$$ is a subset of $$V$$. The upshot is that to construct a quasicoherent sheaf on a scheme, we don’t need to remember arbitrary inclusions of open sets.
• Thanks. I think I see what is happening. Vakil is defining a sheaf on the distinguished affine base which is different from defining a sheaf on the base. When defining a sheaf on a base, we define a restriction map whenever we have $V\subset U$ are basis elements; on the contrary, when defining a sheaf on the distinguished affine base, we only define restriction maps whenever $V$ is a distinguished affine open of the affine open $U$. Commented Nov 14, 2021 at 19:09