Algebraic Geometry
Prof. Pandharipande
D-MATH, FS 2016
Exercise Sheet 11 - Solutions
1. Define an affine scheme (Spec(A), OSpec(A) ) and show that
a) the stalk at p ∈ Spec(A) is Ap , the localization of A along p.
b) for f be an element of A,
OSpec(A) (D(f )) = Af .
Here D(f ) is the open set of primes not containing f and Af is the localization of A at the element f . In particular OSpec(A) (Spec(A)) = A.
Solution Hartshorne, Chp II., Proposition 2.2.
2. Prove that the morphisms from Spec(B) to Spec(A) as locally ringed spaces are
in bijective correspondence to the ring homomorphisms A→B.
Solution Hartshorne, Chp II., Proposition 2.3.
3. Let F be a presheaf on a topological space X. For every x ∈ X, let Fx be
the stalk of F at x. For x ∈ U , U open, let ρU,x : F(U )→Fx be the induced
restriction map. Define a sheaf F sh , the sheafification of F or the sheaf associated
to F, by
F sh (U ) = {(sx ∈ Fx )x∈U | for all x ∈ U there exists an open set V with
x ∈ V ⊂ U and s ∈ F(V ), such that ρV,y (s) = sy for all y ∈ V }
(i) Show that F sh is a sheaf.
(ii) Prove that (F sh )x ∼
= Fx for all x ∈ X.
(iii) Let f : F→F sh be the natural map given by s ∈ F(U ) 7→ (ρU,x (s))x∈U ∈
F sh (U ). Show that it satisfies the following universal property: For any
sheaf G and any map of presheaves g : F→G, there exists a unique map
ḡ : F sh →G such that ḡ ◦ f = g.
Solution
(i) For V ⊂ U two open sets, we have a restriction map
ρV U : F sh (U ) → F sh (V ), (sx ∈ Fx )x∈U 7→ (sx ∈ Fx )x∈V .
As the condition on the families (sx ) is local, this map is well-defined.
It is clear that ρU U is the identity and that for W ⊂ V ⊂ U we have
ρW V ◦ ρV U = ρW U . This shows that F is a presheaf.
Let U be an open set in X and assume we have a cover U =
U by open sets Vα . Then giving sections
S
α∈A
Vα of
sα = (sαx ∈ Fx )x∈Vα ∈ F sh (Vα )
such that sα , sβ restrict to the same section on Vα ∩Vβ means simply sαx = sβx
for all x ∈ Vα ∩ Vβ . But then for every x ∈ U there is a unique sx = sαx for
any α ∈ A with x ∈ Vα and we obtain a section s = (sx ∈ Fx )x∈U on all of
U . This shows that F sh is a sheaf.
(ii) Given [(U, s)] ∈ Fx , where U is a neighbourhood of x and s ∈ F(U ), we
obtain an element (ρU,y (s))y∈U of F sh (U ) and thus [(U, (ρU,y (s))y∈U )] ∈ Fxsh .
Conversely, let [(U, (sy )y∈U )] ∈ Fxsh , then by assumption there exists an
open neighbourhood V of x and a section s ∈ F(V ) with ρV,y (s) = sy for
y ∈ V . Thus we have an element [(V, s)] ∈ Fx . These maps are well-defined
and give inverse isomorphisms Fx ∼
= Fxsh .
(iii) Assume we are given a sheaf G and a map g : F → G and we want ḡ with
ḡ◦f = g. Given x ∈ X, this equation implies that we have ḡx ◦fx = gx as the
corresponding maps of the stalks (F sh )x → Gx . But from (ii) we know that
fx is an isomorphism, so ḡx = gx ◦ (fx )−1 is uniquely determined. We define
the map ḡ(U ) : F sh (U )→G(U ) as follows: given (sx ∈ Fx )x∈U ∈ F sh (U )
we obtain a family (tx = (gx ◦ (fx )−1 )(sx ) ∈ Gx )x∈U . We need to show that
there is a section t ∈ G(U ) with stalk tx at x ∈ U , which will be the image
of (sx )x∈U under ḡ(U ).
By definition of F sh , for every x ∈ U there exists an open neighbourhood
Vx of x in U and a section Sx ∈ F(Vx ) with sy = ρVx ,y (Sx ). But then the
section Tx = g(Vx )(SSx ) ∈ G(Vx ) has stalk ty at all y ∈ Vx . Thus we have
the open cover U = x∈U Vx and the elements Tx ∈ G(Vx ) and they agree
on intersections Vx0 ∩ Vx00 (as their stalks ty , y ∈ Vx0 ∩ Vx00 agree there). As
G is a sheaf, these sections indeed glue to a unique section T ∈ G(U ).
The maps ḡ(U ) described above are compatible with restrictions and thus
define a sheaf map ḡ. From the construction, we see that as desired ḡx ◦fx =
gx . As a map into a sheaf is uniquely determined by its induced map on
stalks, we have ḡ ◦ f = g and we also have that ḡ is unique with this
property.
Due May 27.