Chapter 1.2 Flashcards
Define: projective n-space, point, homogeneous coordinates. Discuss structure of P^n(k). Examples?
P^n = the set of equivalence classes of (n+1)-tuples (a0, … , an) of elements of k, not all zero, under the equivalence relation given by (a0, … , an) ~ (la0, … , lan) for all l in k* iei the quotient of the set A^n+1 - {(0, … ,0)} under the equivalence relation which identifies points lying on the same line through the origin. - moduli space of lines
An element of P^n is called a POINT. If p is a point, then any (n+1)-tuple in the equivalence class P is called a SET OF HOMOGENEOUS COORDINATES FOR P. Write these as (x_0 : x_1 : … : x_n)
If x_0 != 0, rescale (x_0, … , x_n) = (1 : y_1 : … : y_n) where y_i = x_i/x_0. This is a copy of A^n. When x_0 = 0 (0: x_1 : …. : x_n) is a copy of P^n-1(k). So we can view P^n(k) = disjoint union of A^n and P^n-1(k)
Define: graded ring, homogeneous element, homogeneous ideal
Facts about homogeneous ideals?
A GRADED RING is a ring S together with a decomposition S = (+) S_d of S into a direct sum of abelian groups S_d s.t. S_d * S_e < S_d+e.
An element of S_d is called a HOMOGENEOUS ELEMENT OF DEGREE d. Thus, any element of S can be written uniquely as a (finite) sum of homogeneous elements. An ideal a < S is a HOMOGENEOUS IDEAL if a = (+) (a intersect S_d)
An ideal is homogeneous <=> it can be generated by homogeneous elements.
The sum, product, intersection, and radical of homogeneous ideals are homogeneous.
To test whether a homogeneous ideal is prime, it is sufficient to show for any two homogeneous elements f,g that fg in a implies a f in a or g in a.
How can we talk about the zeros of a homogeneous polynomial?
A polynomial function clearly cannot define a function on P^n because of non-uniqueness of the homogeneous coordinates (i.e. if f(x) = x, then f(cx) = cx != x)
However, if f is a homogeneous polynomial of degree d, then f(ca0, … , an) = c^df(a0, … , an) so the property of f being zero or not depends only on the equivalence class of (a0, … , an).
Thus, we can talk about the zeros of a homogeneous polynomial, Z(f) = {p in P^n : f(p) = 0}. If T is any set of homogeneous element of S, we define the zero set of T to be Z(T) = { p in P^n : f(p) = 0 for all f in T}. Similarly for a homogeneous ideal…
What is an algebraic set of P^n? projective algebraic variety? quasi-projective algebraic variety? Homogeneous ideal of Y? Homogeneous coordinate ring?
A subset Y of P^n is ALGEBRAIC if there exists a set T of homogeneous elements of S such that Y = Z(T)
A PROJECTIVE VARIETY is an irreducible algebraic set in P^n with induced topology. An open subset of a projective variety is a QUASI-PROJECTIVE VARIETY
If Y is any subset of P^n, we define the HOMOGENEOUS IDEAL of Y in S, denoted I(Y), to be the ideal generated by {f in S : f is homogeneous and f(p) = 0 for all P in Y}. If Y is an algebraic set, we define the HOMOGENEOUS COORDINATE RING of Y to be S(Y) = S/I(Y).
Show that every projective/quasi-projective variety has an open covering by affine/quasi-affine varities. Why is this important?
Cover P^n just as in smooth manifold theory with n+1 open sets U_i = { (a0, … , an) : ai != 0}. Map (a0/ai … ) as before is an “isomorphism” of the structure involved – in this case a homeomorphism in the Zariski topology.
In general, projective varieties can be obtained by gluing together copies of affine varieties
pgs 10 -11
Discuss converting y = x^3 to a projective variety
B72
Discuss converting affine elliptic curve y^2 = x^3 + bx + c to projective variety
B74
Discuss converting twisted cubic to projective variey
B76
Discuss products of affine varieties
A^m x A^n = A^m+n just send (x1, ... , xm) x (y1, ... , yn) to (x1, ... , xm, y1, ... , yn) Polynomial ring is tensor product
Discuss products of projective spaces. Do example
P^m x P^n != P^m+n (x0 : … : xm : y0 : … yn) is not even well-defined. Multiplying xi’s by some lambda, we arrive at a different point
Instead define SEGRE EMBEDDING:
(x0 : … : xm) x (y0 : … : yn) –> (x0y0 :x1y0 … : xmy0 : x0y1 : … xmyn) from P^m x P^n to P^mn + m + n.
More abstractly: k^m+1 x k^m+1 –> k^m+1 (x) k^n+1 taking (v,w) –> v (x) w.
This map is clearly not onto. Let a point of this space be (w00 : w10 : …. : wmn). To see P^m x P^n is a projective variety, we need so show it’s image under this map is a closed subset (and map is injective). Claim: w_ijw_kl = w_ilw_kj are the relations satisfied by image.
Example P^1 x P^1 –> P^3
Try also to write things in terms of outer products of vectors
B80-83
Now show two more things:
- This is a MORPHISM of varieties
- Check that the universal property of products holds for SEGRE
Note: Segre embedding really doing two things:
- Constructing product P^m x P^n as abstract variety
- Embedding into P^mn+m+n
As with manifolds and other theories, we separate these two concepts now. First, deal with existence of abstract object without regard to embedding. Then look at problem of embedding in some space.
B125 -128
What is Veronese surface?
Map P^2 –> P^5 B84
Discuss the variety of all lines in P^3
equivalent to 2d subspaces of k^4 - special case of Grassmannian G(m,n) – G(2,2). Grassmannians are simplest case of Hilbert Schemes, G(2,2) simplest non-trivial example of a Grassmannian
PLUCKER RELATIONS
B85-88
Show Grassmannians are projective varieties
Applications of Grasmannians?
B89-93
Discuss Hirzebruch surfaces and scrolls
B95-96
Discuss Abstract Varieties
Used by Weil in proving Weil conjectures for curves over finite fields
Comparable to modern view of smooth manifolds…
B97-98
Discuss Toric Varieties. Examples?
Construct affine varieties from cones…B99-104
Exercise 2.1
Exercise 2.2
Exercise 2.3
Exercise 2.4
Exercise 2.5
Exercise 2.6
Exercise 2.7
Exercise 2.8
Exercise 2.9
Exercise 2.10
Exercise 2.11
Exercise 2.12
Exercise 2.13
Exercise 2.14
Exercise 2.15
Exercise 2.16
Exercise 2.17
