Algebraic Geometry
Prof. Pandharipande
D-MATH, FS 2014
Solutions Sheet 4
1.
(i) Clearly every open and closed set is locally closed. Then the finite union
part is by definition and the finite intersection part by the usual distributivity law for ∩ and ∪.
(ii) If A = U ∩ V is a locally closed subset, then we have
Ac = U c ∪ V c = U c ∪˙ (U ∩ V c ).
This implies that the complement Ac is a disjoint union of locally closed
sets.
S
S
Now let A = ˙ i Ui ∩ Vi and B = ˙ j Uj ∩ Vj be two sets who are the
disjoint union of locally closed sets, then
A∩B =
[
˙
Ui ∩ Uj ∩ Vi ∩ Vj
is also a disjoint union of l.c. sets.
To prove that every union of l.c. sets can be written as a disjoint union of
such, we argue by induction on the number of unions. If C = ∪ni=1 Ci for
Ci l.c., then set A = ∪n−1
i=1 Ci and B = Cn . By induction we know that A
can be written as a disjoint union of l.c. sets. We have
A ∪ B = B ∪˙ (A \ B) = B ∪˙ (A ∩ B c )
By the above we know that A ∩ B c is the finite disjoint union of l.c. sets
and so we are done.
2. Denote the coordinates on CP2 by [z0 , z1 , z2 ] and let h = z13 − z02 z2 .
Claim: Im(f ) = V (h).
This can be seen as follows. On U = {z0 6= 0}, define the inverse map
g : U ∩ V (h)−→CP1 ,
[z0 , z1 , z2 ] 7→ [1, z1 /z0 ]
Its easy to check that (g ◦ f )|{x6=0} = id and f ◦ g = idU ∩V (h) . This shows that
U ∩ V (h) ∼
= {x 6= 0} ∼
= C ⊂ CP1 .
On the other hand, if P = [z0 , z1 , z2 ] ∈ V (h) with z0 = 0, then z1 = 0 so that
P = f ([0, 1]). This shows that f is a bijection of CP1 onto V (h) (but not a
biregular map as V (h) is singular) and hence we are done.
3. Set f = νn the Veronese embedding. The condition that the points f (p1 ), . . . , f (pn+1 )
are in general linear position means that they do not lie on a hyperplane. So
assume to the contrary that f (p1 ), . . . , f (pn+1 ) ∈ V (λ) for λ a non-zero linear
homogeneous polynomial. Then λ(f (x, y)) is a polynomial of degree n, with
n + 1 distinct roots. Hence λ(f (x, y)) = 0, and then so is λ. A contradiction.
4. There are two ways to prove this statement. Either use the Hint to construct
global non-constant functions f /Gd for f any homogeneous polynomial of degree
d. Otherwise, consider the d-th Veronese embedding νd : Pn −→PN with image
Y . There is a linear polynomial λ on PN such that Y ∩ V (λ) = V (Gd ). If V (Gd )
is disjoint from X, then X is a projective variety contained in PN \ V (λ) ∼
= CN .
But any projective variety that is affine is a finite collection of points.
5. Let X ⊂ CPn be a quasi-projective variety and let Z• = (Z1 ⊃ Z2 ⊃ Z3 ⊃ . . . )
be an decreasing sequence of closed subsets of X. Let X ⊂ CPn be the closure
of X and denote with Z • = (Z1 ⊃ Z2 ⊃ . . . ) the closure of the sequence in
CPn . If x0 , . . . , xn are the coordinates, then D(xi ) = {xi 6= 0} ∩ X is an affine
variety and therefore Z • ∩ {xi 6= 0} stabilizes (by the correspondence between
closed subsets and radicals ideals of C[z1 , . . . , zn ] and the fact that C[z1 , . . . , zn ]
is Noetherian). As D(xi ), i = 0, . . . , n cover CPn , Z • must stabilize and hence
so must Z• = Z • ∩ X.
6. E.g. Humphreys, Linear algebraic groups, Chapter 2.4.