Open basis for Quasi-Projective Varieties

by ayushkhaitan3437

Today we’re going to generate a basis for a quasi-projective variety in the Zariski topology. These open sets will be affine charts (affine charts are open sets in the Zariski topology as the zth affine chart, for instance, is the complement of the variety z=0). Hence, as every point lies in an affine chart, we can that every quasi-projective variety locally looks like an affine variety, which allows us to do all kinds of affine-type calculations and draw affine-like conclusions.

We first notice that if Q\subseteq \Bbb{A}^n is an affine variety, and f\in C[Q], then Q\setminus V(f) is also an affine variety. This is because it can be bijectively mapped to an affine variety in \Bbb{A}^{n+1}. So we’re not working with objects anymore. We’re working with isomorphism classes. What is the variety that Q maps to? It is the variety that is the image of the map x\to \left(x,\frac{1}{f(x)}\right). How do we know that the image is an affine variety? Let I(Q) be the corresponding ideal for Q in \Bbb{A}^n. Then the ideal for its image is the the one generated by I(V)\cup (zf(x)-1). How does this work? When we just consider I(Q) in $\Bbb{A}^{n+1}$, than all values of z satisfy these equations. With the inclusion of the last equation, what happens is that for each n-tuple, straight up imported from \Bbb{A}^n, there is only one value of z that is attached. Hence, each n-tuple gets a unique z coordinate.

Now we try and generate the basis. Any quasi projective variety Q can be embedded in \Bbb{P}^n for some n. For any affine chart U_i, consider Q\cap U_i. Q is the union \bigcup\limits_{i=0}^n (V\cap U_i). Now each V\cap U_i is a quasi-projective variety in \Bbb{A}^n. Why is this? Because you you can take the ideal corresponding to the variety in projective space, and make the z_i=1 substitution. You’ll get an ideal, and hence a corresponding algebraic set. Now remember that this algebraic set is a quasi-projective variety, and not a traditional affine variety. Hence, it is the intersection of a closed set and an open set. Say it is something of the form V(F_1,F_2,\dots,F_s)\setminus V(G_1,G_2,\dots,G_t). This set is covered by open sets of the form V(F_1,F_2,\dots,F_s)\setminus V(G_i), where i\in\{1,2,\dots,t\}. How? First of all, notice that V(F_1,F_2,\dots,F_s)\setminus V(G_i) is smaller than V(F_1,F_2,\dots,F_s)\setminus V(G_1,G_2,\dots,G_t). This is because we’re intersecting V(F_1,F_2,\dots,F_s) with the complement of a bigger set, and hence intersecting it with a smaller set. Now we have to prove that we indeed have a cover. Let p\in V(F_1,F_2,\dots,F_s)\setminus V(G_1,G_2,\dots,G_t). Then p\in V(F_1,F_2,\dots,F_s) and p\notin V(G_1,G_2,\dots,G_t). Hence, there has to exist a G_i such that p\notin V(G_i). This implies that p\in V(F_1,F_2,\dots,F_s)\setminus V(G_i). Hence, we have an open cover. Taking such open covers over all affine charts, we see that the quasi-projective variety has been covered by open sets.

Do the affine open sets count, when what we wanted was projective-open sets? Can open sets be “projectively completed” to create open sets in projective space? Yes they can be. How? The complement of open sets in affine space would be closed sets. Take the projective closure of those affine closed sets and then take their complement. The complement of the projective closure will contain the affine open sets. Hence we’re done.