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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, $p \in \operatorname{Spec} A_f \subset U$.

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    $\begingroup$ A base in the usual sense requires that the intersection of two basic open sets is a basic open set. $\endgroup$
    – Zhen Lin
    Commented Nov 12, 2021 at 21:59
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    $\begingroup$ @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. $\endgroup$
    – user5826
    Commented Nov 12, 2021 at 23:25
  • $\begingroup$ @ZhenLin You may describing Lemma 5.5.2 here: stacks.math.columbia.edu/tag/004O $\endgroup$
    – user5826
    Commented Nov 12, 2021 at 23:26
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    $\begingroup$ @ZhenLin That’s already false in the case of the canonical basis of $\mathbb{R}^2$ in the Euclidean topology $\endgroup$
    – Exit path
    Commented Nov 13, 2021 at 1:58
  • $\begingroup$ 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). $\endgroup$
    – onRiv
    Commented Jan 26, 2023 at 15:49

1 Answer 1

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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.

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  • $\begingroup$ I'm still a bit confused as to why this doesn't form a basis though. Even if we only remember the "distinguished inclusions", it looks like we still have that every open subset is covered by affine opens. No? $\endgroup$
    – user5826
    Commented Nov 13, 2021 at 16:02
  • $\begingroup$ The set of all affine subschemes certainly forms a base. I think probably my answer is not very clear, but by remembering when one affine is distinguished open in another we’re endowing that set with extra structure. $\endgroup$
    – Exit path
    Commented Nov 14, 2021 at 8:57
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    $\begingroup$ 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 $U$. $\endgroup$
    – user5826
    Commented Nov 14, 2021 at 19:09

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