I am quite confused the definition given by Shafaverich:
Definition: A regular map $f:X \rightarrow \mathbb P^n$ of an irreducible quasiproj. variety $X$ to projective space $\Bbb P^n$ is given by an $(m+1)$ truple of form
$$(F_0 : \cdots : F_m)$$
of same degree in hom. coord. of $x \in \Bbb P^n$. We require that for every $x \in X$ there exists such an expression for $f$ such that $F_i(x)\not=0$ for at least one $i$.
The definition of regular maps between quasiprojective varieties allows us to define isomorphism.
Definition: We say a quasiproj. variety is isomoprhic to a closed subspace of an affine space , an affine variety
In this case, to apply the first definition, we must regard a closed subspace of an affine space as one in $\Bbb P^n$? What are we doing here?
I suppose we for "embed" $\Bbb A^n$ into $\Bbb P^n$, by one of the choice $\Bbb A_i^n$ (where $i$th coordinate is nonzero), and show it is independent? Also, are there not more ways to regard $\Bbb A^n$ as a subset in $\Bbb P^n$?
Best Answer
There's not much deep going on here, it's just a definition. In spirit, we want closed subsets of affine space $\Bbb A^n$ to be our "affine varieties", but sometimes you have a situation where $X$ is a quasiprojective variety that just happens to be also isomorphic to a closed subset of $\Bbb A^n$. Since we believe that "up to isomorphism" is the proper equivalence to put on varieties, we will call these things affine varieties as well.
Nothing about "quasiprojective" was important here; if we define more general types of varieties (I don't know in how much generality Shafarevich does this), then we will always call such varieties affine varieties when they are isomorphic to a closed subset of $\Bbb A^n$.
So a good example is the one given right here after the line you mention. If you take $X:=\Bbb A^1\smallsetminus\{0\}$, then $X$ is quasiprojective because it is open in $\Bbb A^1$, which is open in $\Bbb P^1$ under the identification $$\Bbb A^1\simeq\{[x:y]\mid x\neq0\}\subset\Bbb P^1,$$ but on the other hand it is isomorphic to the closed subset $\{(x,y)\mid xy=1\}\subset\Bbb A^2$. Since it "looks like" a closed subset of $\Bbb A^n$, we will call $X$ an affine variety.
I want to emphasize further that Shafarevich is not saying that every quasiprojective variety can be embedded as a closed subset of affine space. For instance, $\Bbb P^1$ is a quasiprojective variety, which is not an affine variety. If you know stuff about regular functions then you know that the only globally defined regular functions on $\Bbb P^1$ are constants (this is true for any $\Bbb P^m$), but this is never true for closed subsets $Z\subset\Bbb A^n$ unless $Z$ is a single point.