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Algebraic Geometry
Prof. Pandharipande
D-MATH, FS 2016
Exercise Sheet 8 - Solutions
1. Consider the category of
(*) affine algebraic varieties with algebraic maps,
(**) finitely generated reduced C-algebras with C-algebra homomorphism.
Prove the equivalence of the categories (*) and (**) via the contravariant functor
X → Γ(X)
Solution See Section 3 and in particular Corollary 3.8 in Hartshorne, Algebraic
Geometry Chapter I.
2. Let X be an affine variety with coordinate ring A.
a) Show that X is irreducible iff A is a domain.
b) Recall that by the Nullstellensatz, the operations Y 7→ I(Y ) and I 7→ V (I)
define inverse bijections
{closed subsets Y ⊂ X} ↔ {radical ideals I ⊂ A}.
Show that this bijection restricts to bijections
{irreducible closed subsets Y ⊂ X} ↔ {prime ideals I ⊂ A},
{points p ∈ X} ↔ {maximal ideals m ⊂ A}.
Solution
a) Assume first that X is irreducible and let f, g ∈ A with f g = 0. Then
X = V (f g) = V (f ) ∪ V (g) is the union of the closed sets V (f ), V (g). As
X is irreducible, this means that V (f ) = X or V (g) = X. Assume the
first, then as V (f ) = X = V (0), by the Nullstellensatz we have f n = 0 for
some n, hence f = 0.
Conversely, assume that A is a domain and let X = V (I) ∪ V (J) be the
union of two closed sets for ideals I, J ⊂ A. Then for f ∈ I, g ∈ J we have
V (I) ⊂ V (f ) and V (J) ⊂ V (g), thus X = V (f ) ∪ V (g) = V (f g). Again
by the Nullstellensatz we have (f g)n = 0, hence f g = 0. But as A is a
domain, we know either f = 0 or g = 0. As f, g were arbitrary in I, J, we
must have I = 0 or J = 0, so indeed V (I) = X or V (J) = X.
b) Recall that the coordinate ring of V (I) ⊂ X is given by A/I. Thus by
the previous exercise, V (I) is irreducible iff A/I is a domain. But this
is the case iff I is prime. On the other hand, if V (I) is a point, then
clearly Γ(V (I)) = C is a field, so I is a maximal ideal. On the other hand,
for a maximal ideal m look at V = V (m). Clearly V 6= ∅ as otherwise
V (m) = V (A), so there exists n with 1n = 1 ∈ m, a contradiction. But
for p ∈ V we clearly have m = I(V (m)) ⊂ I({p}), so as m is maximal we
have m = I({p}) and so the vanishing set V (m) = {p} is indeed a point.
3. Let X be a quasi-projective variety and p ∈ X. Define
OX,p = {(f, U ) : p ∈ U ⊂ X nonempty open, f : U → C algebraic}/ ∼,
where (f, U ) ∼ (g, V ) if there exists a nonempty open neighborhood W ⊂ U ∩V
of p with f |W = g|W . Show that OX,p has a natural C-algebra structure. It is
called the ring of germs of algebraic functions on X around p. If V ⊂ X is an
open neighborhood of p, show that there is a natural isomorphism OX,p ∼
= OV,p .
If V is affine with coordinate ring A and if p ∈ V corresponds to the maximal
ideal m ⊂ A, show that OV,p is isomorphic to the localization Am .
Solution The algebra structure on OX,p is given by
[(f, U )] + [(f 0 , U 0 )] = [(f |U ∩U 0 + f 0 |U ∩U 0 , U ∩ U 0 )],
[(f, U )] · [(f 0 , U 0 )] = [(f |U ∩U 0 · f 0 |U ∩U 0 , U ∩ U 0 )],
λ[(f, U )] = [(λf, U )] (λ ∈ C).
One checks that this is well-defined (i.e. does not depend on the choice of representatives (f, U ), (f 0 , U 0 )), because addition and multiplication commute with
restriction of functions.
If V ⊂ X is an open neighborhood of p, the map
OX,p → OV,p , [(f, U )] 7→ [(f |U ∩V , U ∩ V )]
defines an isomorphism with inverse [(f 0 , U 0 )] 7→ [(f 0 , U 0 )] (think about what
this means).
For an affine V with coordinate ring A, such that m = I({p}) ⊂ A is the
maximal ideal corresponding to p, we define a map Am → OV,p by noting that
an element f /g ∈ Am (f ∈ A, g ∈ A\m) defines a function on the neighborhood
D(g) = {q ∈ V : g 6= 0} ⊂ X of p. Indeed, by assumption g ∈
/ m, so g(p) 6= 0,
hence p ∈ D(g). Thus we define
Am → OV,p , f /g 7→ [(f /g, D(g))].
This is well-defined: if we choose a different representative f 0 /g 0 = f /g in Am ,
then by definition there exists u ∈ A \ m with u(f 0 g − f g 0 ) = 0 ∈ A. Then
restricted to the open neighborhood D(u) ∩ D(g) ∩ D(g 0 ) of p we have indeed
f 0 /g 0 = f /g.
To see injectivity note that f /g = 0 on D(g) iff f g = 0 as functions on V , thus
f g = 0 ∈ A. But as g ∈
/ m, this means f /g = 0 ∈ Am .
For surjectivity, assume we have a function h on a neighborhood U of p, then
as the sets D(g), g ∈ A, form a basis of the Zariski topology, we can find g
with p ∈ D(g) ⊂ U . But then [(h, U )] = [(h|D(g) , D(g))]. As we have seen,
the functions on D(g) are exactly the localization Ag , so h can be written as
h = f /g m and g ∈
/ m. This comes exactly from f /g m ∈ Am by the above map,
finishing the proof.
4. Let X be an affine algebraic variety and let A be the ring of algebraic functions
on X. Let p ∈ X be a point and let m ⊂ A be the associated maximal ideal. Let
Am be the localization of A at m. Let mAm be the maximal ideal of Am . Prove
that the natural map
m/m2 →mAm /m2 Am
is an isomorphism of A/m vector spaces.
Solution Let p ⊂ A be a prime ideal of a ring A. Let ϕ : A→Ap be the localization. For any ideal I ⊂ A, we write IAp for ϕ(I) · Ap , which is an ideal of Ap .
Let M be any A-module, then the localization of M along p is denoted Mp . It
is isomorphic to M ⊗A Ap .
Let
0→I→A→A/I→0
be an exact sequence for an ideal I. As localizing at a prime ideal p is exact,
we obtain the exact sequence,
0→Ip →Ap →(A/I)p →0.
Thus we get (A/I)p = Ap /Ip . Furthermore, it is easy to show that the kernel of
Ap →Ap /Ip is IAp , hence also IAp ∼
= Ip = I ⊗ Ap .
For our problem, consider the exact sequence
0→m2 →m→m/m2 →0
Localizing/Tensoring with Am and using IAp ∼
= I ⊗ Ap from above twice, we
obtain
0→m2 Am →mAm →(m/m2 ) ⊗ Am →0
which implies (m/m2 ) ⊗ Am ∼
= mAm /m2 Am . To conclude, note that
(m/m2 ) ⊗ Am = (m/m2 ) ⊗ Am /mAm = m/m2 ⊗ A/m = m/m2 ⊗ A ∼
= m/m2
Here the first and third equality use that if M is a A-module, and I ⊂ A an
ideal such that I · M = 0, then M ⊗A N = M ⊗ (N/IN ) for all A-modules N .
Also we use Am /mAm = (A/m)m = A/m, where we note that m ⊂ A/m is the
zero-ideal in the field A/m, so localizing here does not change A/m.
5. Let X and Y be irreducible quasi-projective varieties. Recall that X and Y are
birational if there are nonempty open sets U ⊂ X , W ⊂ Y , such that U is
isomorphic to W . Prove that X and Y are birational if and only if K(X) is
isomorphic to K(Y ).
Solution As the field of rational functions of a variety depends only on an
open set and two (irreducible) varieties are birational if and only if they are
birational on an open set, we can assume that X and Y are affine with coordinate rings A = k[x1 , . . . , xn ]/I and B = k[y1 , . . . , ym ]/J respectively. Clearly
birational implies isomorphic rings of rational functions so let us assume that
we have an isomorphism ϕ : Q(A)→Q(B) of the quotient fields with inverse
ψ : Q(B)→Q(A).
Let ϕ(xi ) =
maps
bi
,
gi
i = 1, . . . , n and ψ(yi ) =
ai
,
fi
i = 1, . . . , m. This defines regular
F : D(f1 ) ∩ · · · ∩ D(fm ) ⊂ X→Y, P 7→ (a1 (P )/f1 (P ), . . . , am (P )/fm (P ))
G : D(g1 ) ∩ · · · ∩ D(gn ) ⊂ Y →X, P →
7 (b1 (P )/g1 (P ), . . . , an (P )/gn (P ))
The ring homomorphism ψ|B : B→A[1/f1 , . . . , 1/fn ] is injective, hence the map
F has a dense image (if not, then there would be an open subset V ⊂ Y \ im(F )
and an element g ∈ B with D(g) ⊂ V . This would imply ψ(g) = g ◦ F = 0.)
As ϕ and ψ are inverse to each other, we have G◦F = id and F ◦G = id whenever
they are defined. Let U = D(f1 ) ∩ · · · ∩ D(fm ) and V = D(g1 ) ∩ · · · ∩ D(gn ).
Then G ◦ F (x) is defined when x ∈ U and F (x) ∈ V , or in other words G ◦ F
is defined on U ∩ F −1 (V ), and similarly F ◦ G is defined on V ∩ G−1 (U ). Let
e = U ∩ F −1 (V ),
U
Ve = V ∩ G−1 (U )
e →Y and G = G| e . By the
and consider the restricted maps F = F |Ue : U
V
e and Ve are non-empty.
argument above we have seen, that both U
e →Ve .
Claim: F : U
e . Then x ∈ U and F (x) ∈ V . If F (x) ∈ G−1 (U ), then
Proof of Claim: Let x ∈ U
F (x) ∈ Ve and we are done. But F (x) ∈ G−1 (U ) is equivalent to G(F (x)) ∈ U ,
e →Ve .
which is true since G(F (x)) = x. Hence F : U
e . But now G ◦ F is defined on U
e , and F ◦ G on
Analogously, we have G : Ve →U
∼
e
e
e
V , with G ◦ F = idUe , F ◦ G = idVe . Hence U = V and we are done.
Due May 06.