This is possibly a very silly question. I am reading Hartshorne's Algebraic Geometry, and in Chapter 1.4 (Varieties — Rational Maps) one of the propositions is as follows:
On any variety, there is a base for the topology consisting of open affine subsets.
I'm simply confused about what an "affine subset" is. (Is it just any subset of $\mathbb A^n$? But then the open affine subsets, which are the open subsets, obviously form a base.. Does he mean "algebraic subset"? Then no open set is affine, besides $\mathbb A^n$ itself..) In Chapter 1.1 Hartshorne defines affine varieties, quasi-affine varieties, affine curves, but not what an affine set is!
I have a feeling that I have just badly misunderstood something. Some help in clearing up this misunderstanding would be greatly appreciated!
Best Answer
In my edition of Hartshorne this question is answered near the beginning of 1.4:
So "open affine subset" means "open subset which is isomorphic, as a variety, to an affine variety." Somewhat confusingly, it should be read as "affine (open subset)"; that is, "affine" modifies "open subset." "Affine subset" is meaningless because Hartshorne has not defined what a morphism from an arbitrary subset of a variety to a variety is, which means he hasn't defined what it means for an arbitrary subset of a variety to be affine. (It's possible to give such a definition using the language of scheme theory.)
You ask in the comments:
This is a subtle point which is often not properly addressed. The answer is that Hartshorne defines a variety to be any of an affine, quasi-affine, projective, or quasi-projective variety. An open subset of an affine variety is a quasi-affine variety, and in 1.3 Hartshorne defines morphisms between varieties, which lets you write down a definition of a morphism from a quasi-affine variety to an affine variety, and hence lets you define what it means for two such things to be isomorphic.
The simplest example to keep in mind is the punctured affine line $\mathbb{A}^1 \setminus \{ 0 \}$, which is not an algebraic subset of $\mathbb{A}^1$. But it is open and so it's a quasi-affine variety, and as a quasi-affine variety it's isomorphic to an affine variety, namely the hyperbola $\{ xy = 1 \}$ in the affine plane $\mathbb{A}^2$. So it "is" affine. This is all cleared up by learning more about localization.
(On the other hand, the punctured affine plane $\mathbb{A}^2 \setminus \{ 0 \}$ is not affine in the sense that, as a quasi-affine variety, it is not isomorphic to an affine variety.)