Download 6.3. Definition of a scheme. Recall that a topological space X is

LECTURES ON ALGEBRAIC GEOMETRY
MATH 202A
57
6.3. Definition of a scheme. Recall that a topological space X is Hausdor↵ if the diagonal
(X) = {(x, x)} is a closed subset of X ⇥ X in the product topology.
Definition 6.3.1. A prevariety X over k is a variety if
(1) X is absolutely irreducible and
(2) (X) is a closed subset of X ⇥ X with the k-topology.
Proposition 6.3.2. An absolutely irreducible prevariety X is a variety if and only if, for
all prevarieties K and any two morphisms f, g : K ! X, the set
Y = {x 2 K : f (x) = g(x)}
is a closed subset of K.
Proof. By definition of product, f, g induce a map (f, g) : K ! X ⇥ X. Then, Y =
(f, g) 1 X. If X is a variety then X is a closed set. Since (f, g) is continuous, Y =
(f, g) 1 X is a closed subset of K. Conversely, if the condition holds for all f, g : K ! X
then take K = X ⇥ X and f, g = p1 , p2 . Then Y = X is closed in K = X ⇥ X. So, X is
a variety.
⇤
For schemes the condition f (x) = g(x) needs to be strengthened.
Definition 6.3.3. Suppose f, g : K ! X are morphisms of preschemes and x 2 K. We
write f (x) ⌘ g(x) if the compositions:
i
x
Spec(k(x)) !
K◆X
are equal. I.e., f (x) = g(x) = y and the induced ring maps f ⇤ , g ⇤ : k(y) ! k(x) are equal.
Thus, f, g are “equal at the scheme point x”.
Definition 6.3.4. A prescheme X is a scheme if, for all preschemes K and any two morphisms f, g : K ! X, the set
Y = {x 2 K : f (x) ⌘ g(x)}
is a closed subset of K.
Proposition 6.3.5. X is a scheme if and only if this condition is satisfied in the special
case K = X ⇥Spec(Z) X with f, g being the projection maps.
Proof. Every prescheme is a prescheme over Z since there is a unique ring homomorphism
X ! Spec(Z). (Why?). So, any two morphisms f, g : K ! X give a morphism (f, g) :
K ! X ⇥Spec(Z) X
f
Spec(k(x))
ix
/K
(f,g)
g
6 XO
p1
/ X ⇥Spec(Z) X
'
Spec(Z)
7
p2
( ✏
X
But f ix = g ix i↵ p1 (f, g)ix = p2 (f, g)ix . So, the set Y of all points x 2 K so that
f (x) ⌘ g(x) is equal to (f, g) 1 of the set Z of points y 2 X ⇥Spec(Z) X so that p1 (y) ⌘ p2 (y).
Since Z is closed and (f, g) is continuous, Y is closed. So, X is a scheme.
⇤