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Algebraic Geometry Lecture 20 – Projective Recappery
Lee Butler
Projective Space
For an algebraically closed field k recall that affine n-space is An^ = {(a 1 ,... , an) | ai ∈ k, 1 6 i 6 n}. Now consider an equivalence relation ∼ on An+1^ \ { 0 } given by
(a 0 ,... , an) ∼ (b 0 ,... , bn) ⇔ there exists λ ∈ k×^ such that ai = λbi for 0 6 i 6 n.
The space (An+1^ \ { 0 })/ ∼ is called n-dimensional projective space, Pn. So Pn^ = {[a 0 ,... , an] | ai ∈ k, 0 6 i 6 n}, where the ai aren’t all zero, and each ‘point’ [a 0 ,... , an] is really an equivalence class under ∼.
Projective Varieties
We defined an affine algebraic set U as the set of points in An^ that vanished on an ideal J ⊂ k[X]. For example, over C, V ((x^2 − y)) = {(x, y) ∈ C^2 | x^2 − y = 0} = {(x, x^2 ) | x ∈ C} = {(a + bi, a^2 − b^2 + 2abi) | a, b ∈ R}.
In Pn^ this doesn’t necessarily make sense. For example, x^2 − y is zero for [x, y] = [1, 1] ∈ P^1. But in P^1 we have [1, 1] = [2, 2], and 2^2 − 2 = 2 6 = 0. So instead we consider homogeneous polynomials. These satisfy f (λx 0 ,... , λxn) = λdf (x 0 ,... , xn) for any λ ∈ k and some d ∈ Z, d > 0, where d is called the degree of the polynomial. Homogeneous polynomials are just those polynomials, all of whose monomials have the same total degree – which coincides with the degree just defined. For example X^4 − 7 X^2 Y 2 + XY 3 − 19 Y 4 is a homogeneous polynomial of degree 4.
So if we let S ⊂ k[X] be a set of homogeneous polynomials then, utterly analo- gous to the affine case, we define V (S) = {P ∈ Pn^ | f (P ) = 0 for all f ∈ S}. Then a projective algebraic set is a subset U ⊆ Pn^ that can be written U = V (S) for some set of homogeneous polynomials S ⊂ k[X]. We also define the ideal of a set U ⊂ Pn^ as I(U ) = {f ∈ k[X] | f homogeneous, f (P ) = 0 for all P ∈ U }. 1
2
The ideal is called a homogeneous ideal since it only contains homogeneous poly- nomials. A projective variety is then just an irreducible projective algebraic set. Equivalently it’s a projective algebraic set whose homogeneous ideal is prime.
Functions
Given a projective variety we want to know about interesting functions on it. Recall that a function f : U → k was called regular in the affine case if there was a polynomial F (x) ∈ k[x] such that f (x) = F (x) for every x ∈ U. This is rather pointless in the projective case. Suppose we have a regular function f on a projective variety U and that it isn’t everywhere zero. So
f (P ) = α 6 = 0
at some P ∈ U. Then
f (λP ) = λdf (P ) = λdα
for any λ ∈ k. Clearly, then, d = 0, and so f has degree zero, i.e. it’s a constant function. So the only ‘regular’ functions are constant ones, thus using last weeks notation, O(U ) = k.
We may still define the coordinate ring on U as k[U ] = k[X]/I(U )
where, as usual, k[X] is the set of homogeneous polynomials. The fact that O(U ) 6 ∼= k[U ] is a consequence of projective varieties having more intrinsic structure than their affine counterparts^1.
This is still an integral domain so we take the function field on U to be the field of fractions of k[U ], however we have to add the proviso that the numerator and denominator in our rational functions have the same degree to ensure that f (λP )/ g(λP ) = f (P )/g(P ). So
k(U ) = {f /g | f, g ∈ k[X], homogeneous of same degree, g ∈ I(U )} / ∼
where ∼ is the expected equivalence relation for a field of fractions, that is
f 1 g 1
f 2 g 2
⇔ f 1 g 2 − f 2 g 1 ∈ I(U ).
We then define the dimension of a projective variety U as dim U := trdegk k(U ).
E.g.. Consider U = {X − Y = 0} ⊂ P^1. Then
C[U ] = C[X, Y ]/(X − Y ) ∼= C[X].
Hence C(U ) ∼= C(X), which has transcendence degree 1 over C.
(^1) Apparently.
