I had this very elementary question which baffles me. Most introductions to the topic define an affine variety as a subset of affine space that is the zero-locus of a set of polynomials.
Now, $X:=\mathbb{A}\setminus\{0\}$ does not satisfy this property. Yet I see in many places that it is treated as an affine variety (e.g. on Wikipedia) because it is isomorphic (not biregular, but birational, I guess) to the zero-locus of one variety that does satisfy the above definition (the zero-locus of $\langle xy-1\rangle$ on $\mathbb{A}^2$). Now, there's a circularity problem here, since for two objects to be isomorphic they must first be in the same category, and, according to our original definition, the first object $X$ is not an object in the category of affine varieties (over the given field, say $\mathbb{C}$). I assume there's some reference with a definition that captures this silly issue and gives a broader definition that pleases me, but I can't find it. The question then is: what should such a definition of an affine variety (in a classical sense, that is, avoiding schemes and the like) be? I mean a definition that includes the $X$ above (and similar other objects) as an affine variety.
Best Answer
I am also extremely annoyed when people say things like this. Until you describe what a non-affine variety is, statements like so-and-so is not an affine variety are meaningless: a priori being an affine variety is extra structure on a set, not a property. If being affine is a property if some larger class of objects, you first need to describe what that larger class of objects is!
Let me restrict my attention to $\mathbb{A}^1 \setminus \{ 0 \}$. What sort of object is this? One answer is that it can be usefully thought of as a "Zariski sheaf," namely a functor
$$\mathbb{G}_m : \text{CRing} \ni R \mapsto R^{\times} \in \text{Set}$$
from commutative rings to sets satisfying a sheaf condition. (Note that this does something a bit more general than just "remove the origin" on commutative rings that aren't fields: it removes all non-invertible elements.) The sense in which $\mathbb{G}_m$ is an affine variety, even though it is not a closed subvariety of $\mathbb{A}^1$, is that this sheaf is representable by a commutative ring, namely $\mathbb{Z}[x, x^{-1}]$.
(A more naive guess is $R \mapsto R \setminus \{ 0 \}$, but this isn't even a functor.)
Other answers are possible at various levels of sophistication; you can look up the terms "quasi-affine" and "quasi-projective" variety.