Download Algebra II — exercise sheet 9

Algebra II — exercise sheet 9
41. Show that varieties are compact in the Zariski topology: any open cover has
a finite subcover.
Solution: This holds true for any noetherian topological space, i.e. a space X where descending
chains of closed
S
subsets stabilise or, equivalently, ascending chains
of
open
subsets
stabilise.
If
X
=
U
is
an open cover,
i
i∈I
S
then choose a well-ordering on I and define Vj := i≤j Ui . Then the Vj form an ascending chain of open subsets.
[In fact, a topological space is noetherian if and only if every open subset is compact.]
Affine varieties are noetherian topological spaces because their affine coordinate rings are noetherian, so that
asscending chains of ideals stabilise or, equivalently, descending chains of Zariski-closed subsets stabilise. The
same reasoning works for projective varieties. For quasi-affine or quasi-projective varieties, we observe that a
subspace of a noetherian topological space is still noetherian.
42. Show that a locally compact Hausdorff space X is compact if and only if the
projection X × Y → Y is a closed map for all topological spaces Y .
Solution: Let X be compact and take a closed subset Z ⊆ X × Y . We have to show that π(Z) ⊆ Y is closed,
where π : X × Y → Y is the projection. Let y ∈ Y \ π(Z). Then (x, y) ∈
/ Z for all x ∈ X. By the definition
of the product topology, for each x ∈ X, there
exist
open
neighbourhoods
x ∈ Ux ⊆ X and y ∈ Vx ⊆ Y such
S
that Ux × Vx ⊆ X × Y \ Z. In particular, x∈X Ux = X. By compactness of X, there are finitely many points
x1 , . . . , xr such that X = Ux1 ∪ · · · ∪ Uxr . Let V := Vx1 ∩ · · · ∩ Vxr , then y ∈ V ⊆ Y open. By construction,
V ⊂ Y \ π(Z), so that Y \ π(Z) is open, hence π(Z) closed.
For the reverse implication, assume that X is not compact. Then we consider its one-point compactification
Y := X̂ = X ∪ {∞} — this is a compact space; Y is Hausdorff space since X is locally compact and Hausdorff.
Moreover, let Z := {(x, x) | x ∈ X} ⊂ X × X̂ be the diagonal. As X is Hausdorff, Z ⊂ X × X is closed; thus X is
also closed in X × X̂. Then π(Z) = X ⊂ X̂, a closed subset in a compact Hausdorff space, hence compact itself.
S
The statement actually holds for arbitrary topological spaces X: for the reverse implication, let X = i∈I Ui be
an open covering without any finite subcovering. Let now Y := X ∪ {ω}, topologised with the following open sets:
all subsets of X; all sets containing {ω} ∪ (X \ U ) where U is a finite union of the Ui . This is indeed closed under
arbitrary unions and finite intersections. Moreover, by our assumption on the covering Ui , the subset X ⊂ Y is
not closed, i.e. {ω} ⊂ Y is not open. Let Z be the closure of the diagonal ∆ := {(x, x) | x ∈ X} in X × Y . By
hypothesis, its closure π(Z) is closed in Y , hence π(Z) = Y . Hence (x, ω) ∈ Z for some
x ∈ X. By the definitions
of closure, of the topology on Y and of the product topology, V × ({ω} ∪ (X \ Ui )) ∩ ∆ 6= ∅ for all i ∈ I and
open sets x ∈ V ⊆ X. Hence V ∩ (X \ Ui ) 6= ∅, i.e. V 6⊆ Ui for all V and i. But this would imply x ∈
/ Ui for all
i ∈ I, contradicting that the Ui cover X.
43. Let X = V (y 2 − x3 ) ⊂ A2 be the cuspidal cubic. Show that the map
A1 → X, t 7→ (t2 , t3 ) is regular, birational and a homeomorphism, but not an
isomorphism of varieties. Extend this to an example of a morphism of projective
varieties with the same properties.
Solution: The map is given by polynomials, so it is regular. It is bijective, because for any point (x, y) ∈ X, we
have y 2 = x3 and (with K algebraically closed), there are two square roots of x, call them t and −t. Moreover,
t6 = x3 = y 2 , regardless of sign. But exactly one of the two roots satisfies t3 = y. Next, A1 and X are infinite
sets with the cofinite topology, hence every bijection between them is automatically a homeomorphism.
Finally, the two varieties are birational: K(A1 ) = Quot(K[t]) = K(t), i.e. the function field of the affine line is
the field of rational functions in one variable. But A(X) = K[x, y]/(y 2 − x3 ) = K[t2 , t3 ]) has the same quotient
field: Quot(K[t2 , t3 ]) = K(t), as t3 /t2 = t in Quot(A(X)).
The two curves are not isomorphic because their affine coordinate rings aren’t. For example, the tangent space
T0 X = (M0 /M02 )∗ is 2-dimensional, whereas all tangent spaces of A2 are 1-dimensional.
By homogenisation, we get an example of projective curves: P1 → V (x0 x22 − x31 ), (t0 : t1 ) 7→ (t30 : t0 t21 : t31 ).
44. Let A be a commutative ring with unit and N a finitely generated A-module.
Show that a surjective module endomorphism N → N is an isomorphism.
Solution: The theorem of Vasconcelos.
Write f : N → N and use it to consider N as an A[X]-module, with X · n := f (n). Then XN = N by the
assumption that f is surjective. Applying the Nakayama lemma (with I = (X) ⊂ A[X]) yields an element
Y ∈ A[X] such that Y N = 0 and Y ∈ 1 + (X), i.e. Y = 1 + XZ for some Z ∈ A[X]. Let n ∈ ker(f ), then X · n = 0
and hence 0 = (1 + XZ)n = n + ZXn = n, so that f is indeed injective.
45. Find subgroups of SL2 such that the quotient is respectively projective, affine
or strictly quasi-affine.
Solution: The quotient SL2 /(T2 ∩ SL2 ) ∼
= PSL2 is an affine variety;
= P1 is a projective variety; SL2 /(D2 ∩ SL2 ) ∼
SL2 /U2 ∼
= A2 \ {0} is quasi-affine but not affine.