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Projective and Affine varieties 
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#1
Dec1812, 08:05 AM

P: 2

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) / Z^{d}_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) = Z^{d}_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! 


#2
Dec1912, 11:59 AM

Mentor
P: 18,080

Could you give the relevant page number in the book?
So you obtain [itex]X\cup (\mathbb{P}^n\setminus U_0)[/itex]. If you want to obtain just X, then you need to perform the same procedure on the [itex]U_1,...,U_n[/itex]. This will give you extra polynomials. The zero locus of everything will then be X. 


#3
Dec1912, 02:19 PM

P: 2

Ooh, I see now! So the roots of the polynomial F_α are just (X intersect U_0) union (P^n \ U_0) = X union (P^n \U_0). Constructing similar polynomials for U_1, ...., U_n, we have that the zero locus for these polynomials is just the intersection of all the X union (P^n \U_i), which is just X. Cool! Thanks, that was helpful. (This was on page 5 in his book).



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