## Projective and Affine varieties

I'm having a little trouble seeing something Harris says in his intro book on alg. geom. Say X is contained in P^n. Harris says that X is a projective variety iff X intersect U_i is an affine variety for each i=0,...,n, where U_i are the points [Z_0 , Z_1 , ... , Z_n] in P^n with Z_i =/ 0. I'm a little confused about how he claims this:

If X is a projective variety, say its the locus of the homogeneous polynomials F_α(Z_0, ..., Z_n). Say we define on A^n the polynomials f_α(z_1, ..., z_n) = F_α(1,z_1, ...,z_n) = F_α(Z_0, ..., Z_n) / Zd_0 where d is the degree of F_α and z_i are the local coords (z_i = Z_i/Z_0). Then he claims the zero locus of the f_α is X intersect U_0. Now since there's a bijection between U_0 and A^n, are we just identifying X intersect U_0 with its image via the local coordinates, meaning its an affine variety too?

For the other direction, I don't really see this. If for example X intersect U_0 is an affine variety, say its the locus of f_α(z_1, ..., z_n),then we can define homogeneous polynomials F_α(Z_0, ... Z_n) = Zd_0 f_α(Z_1/Z_0,..., Z_n/Z_0) where d = deg(f_α). But then is the zero locus of the F_α just X?

Any help would be appreciated!

 PhysOrg.com science news on PhysOrg.com >> Front-row seats to climate change>> Attacking MRSA with metals from antibacterial clays>> New formula invented for microscope viewing, substitutes for federally controlled drug

Blog Entries: 8
Recognitions:
Gold Member
No, of course not. The entire set $\mathbb{P}^n\setminus U_0$ would also be in the zero locus of the $F_\alpha$ since it is easy to see that if $Z_0=0$, then $F_\alpha(0,Z_1,...,Z_n)=0$.
So you obtain $X\cup (\mathbb{P}^n\setminus U_0)$. If you want to obtain just X, then you need to perform the same procedure on the $U_1,...,U_n$. This will give you extra polynomials. The zero locus of everything will then be X.