Download Hartshorne Ch. II, §3 First Properties of Schemes

Hartshorne Ch. II, §3 First Properties of Schemes
Takumi Murayama
January 20, 2014
Remark. There’s a lot of commutative algebra here; I’ve included as much of my own
solutions to problems in [AM69] that I’ve cited as I can.
Lemma 1. Let Ui = Spec(Ai ) for i = 1, 2 be two affine neighborhoods in a scheme X.
Then for any x ∈ U1 ∩U2 , there exists an open subscheme V such that x ∈ V ⊂ U1 ∩U2 ,
and V is a distinguished open set in both U1 and U2 .
Proof. Since the distinguished open subsets of U1 form a basis of U1 , we know that
D(f ) ⊂ U1 ∩ U2 for some f ∈ A1 . Since D(f ) = Spec(A1 )f , we can replace U1 with
D(f ) and assume that D(f ) = U1 ⊂ U2 is an open immersion, given by a ring map
ϕ : A2 → A1 . Let g ∈ A2 such that D(g) ⊂ D(f ); we can do this since again, the
distinguished open subsets of U2 form a basis for U2 , and D(f ) is open in U2 .
We claim we then have an isomorphism D(g) ∼
= D(ϕ(g)) induced by ϕ; by the
fact that open subsets of a scheme have unique subscheme structures (Problem 2.2),
it suffices to show that they are equal as sets, i.e., to show that ϕ∗−1 (Spec(A2 )g ) =
Spec(A1 )ϕ(g) , where ϕ∗ denotes the morphism of affine schemes induced by ϕ. Note
that D(g) = Spec(A2 )g and D(ϕ(g)) = Spec(A1 )ϕ(g) .
Let p ∈ ϕ∗−1 (D(g)) = ϕ∗−1 (Spec(A2 )g ). Then, g ∈
/ ϕ−1 (p) = q. If ϕ(g) ∈ p, then
g ∈ ϕ−1 (p), a contradiction. Thus, ϕ(g) ∈
/ p, and p ∈ Spec(A1 )ϕ(g)) .
Now let q ∈ D(ϕ(g)) = Spec(A1 )ϕ(g) . Then, ϕ(g) ∈
/ q. Thus, g ∈
/ ϕ−1 (p) = ϕ∗ (p),
∗
∗−1
∗
∗−1
so ϕ (p) ∈ Spec(A2 )g . This implies p ∈ ϕ (ϕ (p)) ⊂ ϕ (Spec(A2 )g ).
Problem 3.1. Show that a morphism f : X → Y is locally of finite type if and only
if for every open affine subset V = Spec B of Y , f −1 (V ) can be covered by open affine
subsets Uj = Spec Aj , where each Aj is a finitely generated B-algebra.
Proof. ⇐ Trivial. ⇒ Suppose that f is locally of finite type. V = Spec B is covered
by V ∩ Spec Bi ’s, and so is covered by distinguished opens Spec Bfk = Spec(Bi )gk
by Lemma 1. Now, the preimage of the open subsets Spec(Bi )gk is covered by
1
Spec(Aij )gk , where we consider gk as an element of Aij and the (Aij )gk are finitely
generated (Bi )gk -algebras, and so we are done since the Bfk ∼
= (Bi )gk are finitely
generated B-algebras.
Problem 3.2. A morphism f : X → Y of schemes is quasi-compact if there is a
cover of Y by open affines Vi such that f −1 (Vi ) is quasi-compact for each i. Show
that f is quasi-compact if and only if for every open affine subset V ⊆ Y , f −1 (V ) is
quasi-compact.
Proof. ⇐ Trivial. ⇒ Suppose that f is quasi-compact. V is covered by V ∩ Vi , and so
has an open cover by open subsets D(fk ) ⊂ Vi . Since V is affine it is quasi-compact
by Problem 2.13(b), and so we can assume this cover is finite. We can cover each
f −1 (Vi ) with open affines Spec Aij ; the preimage of the D(fk ) in each Spec Aij is
Spec(Aij )fk , so now we have a finite cover of f −1 (V ) by open affines Spec(Aij )fk .
f −1 (V ) is therefore quasi-compact, since if we had a cover of f −1 (V ), it would restrict
to finite subcovers of each Spec(Aij )fk , and so lift to a finite subcover of f −1 (V ) since
there are only finitely many Spec(Aij )fk .
Problem 3.3.
(a) Show that a morphism f : X → Y is of finite type if and only if it is locally of
finite type and quasi-compact.
(b) Conclude from this that f is of finite type if and only if for every open affine
subset V = Spec B of Y , f −1 (V ) can be covered by a finite number of open
affines Uj = Spec Aj , where each Aj is a finitely generated B-algebra.
(c) Show also if f is of finite type, then for every open affine subset V = Spec B ⊆
Y , and for every open affine subset U = Spec A ⊆ f −1 (V ), A is a finitely
generated B-algebra.
Proof of (a). ⇐ Follows by definition of locally of finite type and Problem 3.2. ⇒
It suffices to show f is quasi-compact. But if Spec Aij is our finite cover of f −1 (Vi ),
then f −1 (Vi ) is the finite union of quasi-compact spaces hence quasi-compact as in
Problem 3.2.
Proof of (b). By (a), f is of finite type if and only if it is locally of finite type and
quasi-compact. The claim then follows since the stated condition is exactly the
combination of the two conditions in Problems 3.1 and 3.2.
Proof of (c). Let Uj = Spec Aj be the cover of f −1 (V ) from (b). Let U = Spec A ⊆
f −1 (V ). Then, it has a finite cover U ∩ Uj , and each U ∩ Uj can be covered by
distinguished opens of Uj that are also distinguished opens of U by Lemma 1; let
these be Uk = Spec Afk = Spec(Aj )gk , and we note that Afk are finitely generated
2
B-algebras by (b). Since Spec A is affine hence quasi-compact by Problem 2.13(b),
we can pick a finite subcover of Uk ’s.
It remains to show A is a finitely generated B-algebra. Let a ∈ A, and choose
n
fP
k1 , . . . , fkm ∈ A that generate Afk over B. There then exists n such that fk a =
n
` gk` fk` for all k. Now (fk )k = A by [AM69, Ex. 1.13iv], and so we can represent a
as a linear combination of fk` ’s, i.e., A is a finitely generated B-algebra.
Problem 3.4. Show that a morphism f : X → Y is finite if and only if for every
open affine subset V = Spec B of Y , f −1 (V ) is affine, equal to Spec A, where A is a
finite B-module.
Proof. ⇐ Trivial. ⇒ Suppose that f is finite. Then, V is covered by V ∩ Spec Bi , and
so is covered by distinguished opens Spec Bfk = Spec(Bi )gk by Lemma 1; by affineness
of V , we can assume these are finite. Then, the fk generate the unit ideal in B. Now,
the f # (fk ) ∈ Γ(f −1 (V ), OX ) = A generate the unit ideal in A by the properties of a
ring morphism. Moreover, since the preimage of the Spec Bfk is exactly Spec Af # (fk ) ,
we see that f −1 (V ) is affine by Problem 2.17(b), and that by assumption, we see that
Af # (fk ) is a finitely generated Bfk -module.
It remains to show that A is a finite B-module. Suppose ak1 , . . . , akm are generators
of Af # (fk ) over Bfk ; after clearing denominators, we
P can assume that the generators
n
are in A. Then, any element a ∈ A satisfies fk a = ` b` ak` for some b` ∈ B, for every
k, after clearing denominators. Again by [AM69, Ex. 1.13iv], we then know that any
a can be represented as a B-linear combination of the ak `’s, and so the ak `’s generate
A over B.
Problem 3.5. A morphism f : X → Y is quasi-finite if for every point y ∈ Y , f −1 (y)
is a finite set.
(a) Show that a finite morphism is quasi-finite.
(b) Show that a finite morphism is closed, i.e., the image of any closed subset is
closed.
(c) Show by example that a surjective, finite-type, quasi-finite morphism need not
be finite.
Proof of (a). y ∈ V for some affine open subset V = Spec B. Since f is finite,
f −1 (V ) = Spec A for some A a finite B-module. Now, it suffices to show that
f −1 (y) ∼
= Spec A ×Spec B Spec k(y) is finite by Problem 3.10. But Spec A ×Spec B
Spec k(y) ∼
= Spec(A ⊗B k(y)) by Thm. 3.3. Since A is a finite B-module, we see that
A ⊗B k(y) is a finite k(y)-module, and so is a quotient of a polynomial ring over
k(y) by [AM69, Prop. 2.3]. Thus, A ⊗B k(y) has finite dimension as a ring, and f is
quasi-finite.
3
Lemma 2. If A → B is integral, then f ∗ : Spec B → Spec A is closed.
Proof of Lemma 2. Let q ∈ Spec(B); we claim f ∗ (V (q)) = V (f ∗ (q)). By [AM69,
Exc. 1.21iii], we see f ∗ (V (q)) ⊆ f ∗ (V (q)) = V (f −1 (q)) = V (f ∗ (q)). Conversely, if
p ∈ V (f ∗ (q)), then f ∗ (q) ⊆ p, and so f (f ∗ (q)) ⊆ f (p) is a chain of prime ideals in
f (A). Since f (f ∗ (q)) = f (f −1 (q)) = q∩f (A), we have that q lies over f (f ∗ (q)), and so
since B is integral over f (A), by Going Up [AM69, Thm. 5.11] there exists r ∈ Spec(B)
such that r ⊇ q and r∩f (A) = f (p). Thus, p = f −1 (f (p)) = f −1 (r∩f (A)) = f −1 (r) =
f ∗ (r), where r ∈ V (q). This implies p ∈ f ∗ (V (q)); thus, f ∗ (V (q)) = V (f ∗ (q)), i.e., f ∗
is closed.
Proof of (b). Since a closed subset is closed if and only if it is closed in every element
of an open cover, we can reduce to the case when X = Spec A and Y = Spec B,
where A is a finitely generated B-module by the fact that f is finite. Moreover, since
by Problem 3.11(b) every closed subset of X will then be of the form Spec A/a, it
suffices to show f (X) is closed. By Lemma 2, it suffices to show B → A is integral.
But this is true by [AM69, Prop. 5.1] since f is finite.
Proof of (c). Consider the map
Spec k[x]x ⊕ k[x](x−1) → Spec k[x]
given by the ring homomorphism ϕ such that 1 7→ (1, 1) and x 7→ (x, x). Now, for
any prime ideal p = (f ) ⊂ k[x], we see that ϕ(p) = (f, f ), which is a prime ideal
since it is a prime ideal of at least one of k[x]x or k[x]x−1 ; thus, ϕ∗ is surjective. It
is of finite type since adjoining x−1 , (x − 1)−1 to two copies of k[x] clearly makes
k[x]x ⊕ k[x]x−1 a finitely generated k[x]-algebra. Finally, it is a quasi-finite morphism
since f −1 (p) = Spec k[x]x ⊕k[x](x−1) ⊗k[x] k(p) = Spec(k[x]x ⊗k(p))⊕(k[x]x−1 ⊗k(p)) =
Spec k(p) ⊕ k(p), which is clearly finite. On the other hand, this is not a finite map
since k[x]x , k[x](x−1) are both not finite k[x]-modules, hence k[x]x ⊕ k[x](x−1) is not a
finite k[x]-module.
Problem 3.6. Let X be an integral scheme. Show that the local ring Oξ of the
generic point ξ of X is a field. It is called the function field of X, and is denoted by
K(X). Show also that if U = Spec A is any open affine subset of X, then K(X) is
isomorphic to the quotient field of A.
Proof. We have that
Oξ = lim OX (U ) = lim OSpec A (U ) = k(A).
−→
−→
U 3ξ
U 3ξ
4
where Spec A is an affine neighborhood of ξ. Note we know that the image of ξ in
Spec A is the generic point of Spec A since generic points are unique (Problem 2.9),
and {ξ} = Spec A since Spec A has the subspace topology. This is true for any affine
neighborhood, since every affine neighborhood contains the generic point.
Problem 3.7. A morphism f : X → Y , with Y irreducible, is generically finite if
f −1 (η) is a finite set, where η is the generic point of Y . A morphism f : X → Y is
dominant if f (X) is dense in Y . Now let f : X → Y be a dominant, generically finite
morphism of finite type of integral schemes. Show that there is an open dense subset
U ⊆ Y such that the induced morphism f −1 (U ) → U is finite.
Lemma 3 ([AM69, Exc. 5.10i]). A ring homomorphism f : A → B is said to have
the going-up property if the conclusion of the going-up theorem [AM69, Thm. 5.11]
holds for B and its subring f (A).
Let f ∗ : Spec(B) → Spec(A) be the mapping associated with f .
Consider the following three statements:
(a) f ∗ is a closed mapping.
(b) f has the going-up property.
(c) Let q be any prime ideal of B and let p = qc . Then f ∗ : Spec(B/q) → Spec(A/p)
is surjective.
Then, (a) ⇒ (b) ⇔ (c).
Proof of Lemma 3. (a) ⇒ (b). Suppose p1 ⊆ p2 a chain of prime ideals in f (A) and
q1 ⊆ B lies over p1 . Since f ∗ (q1 ) = f −1 (p1 ) ⊆ f −1 (p2 ), we have f −1 (p2 ) ∈ V (f ∗ (q1 )).
Note V (f ∗ (q1 )) = V (f −1 (q1 )) = f ∗ (V (q1 )) = f ∗ (V (q1 )) by the AM Exercise 1.21iii)
and the fact that f ∗ is closed. Thus, f −1 (p2 ) ∈ f ∗ (V (q1 )), and so there exists
q2 ∈ V (q1 ) such that f ∗ (q2 ) = f −1 (p2 ). Since f ∗ (q2 ) = f −1 (q2 ∩ f (A)), we then see
that p2 = q2 ∩ f (A), and so q2 lies over p2 . By induction as in the original proof of
Going Up [AM69, Thm. 5.11], we are done.
(b) ⇒ (c). Suppose q ∈ Spec(B) and let p = f −1 (q) = f −1 (q ∩ f (A)). Consider
1−1
p0 ∈ V (p); note that V (p) ←−→ Spec(A/p), and so showing that p0 ∈ im f ∗ suffices.
Then, p ⊆ p0 , and so f (p) ⊆ f (p0 ) is a chain of prime ideals in f (A) and q∩f (A) = f (p).
By (b), the Going Up property, there exists q0 ⊆ B such that q0 ∩ f (A) = f (p0 ).
Finally, f −1 (q0 ∩ f (A)) = f ∗ (q0 ) = f −1 (f (p0 )) = p0 , and so f ∗ is surjective.
(c) ⇒ (b). Suppose p1 ⊆ p2 is a chain of prime ideals in f (A) and q1 ⊆ B
lies over p1 . Note that f −1 (p1 ) is prime since it is a contraction of a prime ideal,
and also f −1 (p1 ) = f −1 (q1 ∩ f (A)) = (q1 )c . By (c), f ∗ : V (q1 ) → V (f −1 (p1 )) is
surjective, and so f −1 (p2 ) ∈ V (f −1 (p1 )) has preimage q2 = (f ∗ )−1 (f −1 (p2 )) ∈ V (q1 ).
q2 ∈ V (q1 ) =⇒ q1 ⊆ q2 , and f −1 (p2 ) = f ∗ (q2 ) = f −1 (q2 ∩f (A)) =⇒ p2 = q2 ∩f (A),
5
i.e., q2 lies over p2 . By induction as in the original proof of Going Up [AM69,
Thm. 5.11], we are done.
Proof. We first claim that there exists a finite field extension K(Y ) ,→ K(X). Let
Y ⊃ V = Spec B 3 η be an affine neighborhood of the generic point, and let
X ⊃ f −1 (V ) ⊃ U = Spec A be an affine neighborhood contained in the preimage,
where A is a finitely generated B-algebra by Problem 3.3(c). Then, since X is integral,
A is an integral domain.
So, A is finitely generated over B and thus, so is k(B) ⊗B A ∼
= S −1 A over k(B),
where S = B \ 0. Then, by the Noether normalization lemma [Mat70, 14.G], we have
k(B) ⊂ k(B)[y1 , . . . , yn ] ⊂ S −1 A
where the second extension is integral. This implies that we have a surjection
Spec S −1 A Spec k(B)[y1 , . . . , yn ]
by Lemma 3. But Spec S −1 (A) = Spec k(B) ⊗B A = Spec k(B) ×Spec B Spec A, which
is homeomorphic to f −1 (η)∩U by Problem 3.10, and so it is finite since f is generically
finite. But Spec k(B)[y1 , . . . , yn ] is infinite unless n = 0, and so we see k(B) ⊂ S −1 A
is an integral extension. But then, since an integral extension of a field is a field by
Zariski’s lemma [Rei95, Prop. 4.9], we see that S −1 A = k(A), and is a finite algebraic
field extension of k(A). This produces a finite field extension for K(Y ) ,→ K(X) by
Problem 3.6.
Now choose a set of generators for A over B; these then satisfy polynomial
equations with coefficients in k(B), and so by clearing denominators they satisfy
polynomial equations in B. Letting S be the multiplicative set generated by all the
leading terms of these polynomials, we have that S −1 A is integral over S −1 B, and so
Spec S −1 A → Spec S −1 B is surjective, and moreover finite.
Now for arbitrary Y , any affine neighborhood is dense since Y is irreducible. The
proposition now follows by taking U = Spec S −1 B with preimage Spec S −1 A as in
the previous paragraph.
Problem 3.8. Normalization. A scheme is normal if all of its local rings are integrally
closed domains. Let X be an integral scheme. For each open affine subset U = Spec A
of X, let à be the integral closure of A in its quotient field, and let Ũ = Spec Ã. Show
that one can glue the schemes Ũ to obtain a normal integral scheme X̃, called the
normalization of X. Show also that there is a morphism X̃ → X, having the following
universal property: for every normal integral scheme Z, and for every dominant
morphism f : Z → X, f factors uniquely through X̃. If X is of finite type over a field
k, then the morphism X̃ → X is a finite morphism. This generalizes (I, Ex. 3.17).
6
Proof. We first claim this is true for X affine. If X = Spec A is affine, then define
X̃ = Spec Ã; let our morphism ν : Spec à → Spec A be the one induced by the ring
morphism A ,→ Ã. Now we want to show the universal property
ϕ
Z
Spec Ã
ν
f
Spec A
By Problem 2.4, this corresponds to the commutative diagram
Γ(Z, OZ )
ϕ∗
Ã
ν∗
f∗
A
in Rings. Since any element of à is in the fraction field, it is of the form a/b; we claim
that defining the map ϕ by having a/b 7→ f ∗ (a)/f ∗ (b) works. We first show that
f ∗ (a)/f ∗ (b) ∈ Z̃; this follows since if g(x) is the polynomial a/b satisfies in A[x], then
applying f ∗ to g(x) on each coefficient gives a polynomial satisfied by f ∗ (a)/f ∗ (b) by
the fact that f ∗ is a ring morphism.
To show the map is unique, we first show f ∗ is injective. Since f is dominant, we
know f (Z) = Spec Ã. But then, f (Z) = V (f ∗−1 (0)) = V (ker f ∗ ) = Spec Ã, and so
ker f ∗ = N(Ã) = 0 by the fact that A is an integral domain.
Now we can show ϕ∗ is unique. If any other map ψ ∗ satisfied the commutative
diagram, ϕ∗ (a/1) = f ∗ (a) = ψ ∗ (a/1), and so ϕ∗ (b/1)ϕ∗ (a/b) = ϕ∗ (a/1) = f ∗ (a) =
ψ ∗ (b)ψ ∗ (a/b) implies ϕ∗ (a/b) = ψ ∗ (a/b) for all a/b ∈ Ã, i.e., ϕ∗ is unique. By the
bijection in Problem 2.4, we then see that ϕ exists and is unique as well. Note that
this implies that X̃ is unique up to unique isomorphism just by having Z = X̃ 0
another normalization, and then switching the roles of X̃ and X̃ 0 .
We would like to show now that if ν exists and U ⊂ X, then ν −1 (U ) ∼
= Ũ . ν −1 (U )
is normal since normality is a local property, and is integral since this is again a local
property. Now, if Z → U is a dominant map from a normal integral scheme Z, then
Z becomes a dominant map on X since X is irreducible hence any open subset of U is
open in X and hence dense in X. Thus, the map factors uniquely through X̃ by the
universal property for (affine) normalization proved above, and moreover the image of
X̃ is in ν −1 (U ), so ν −1 (U ) satisfies the universal property hence is isomorphic to Ũ .
7
Finally, we would like to extend this to the case when X is an arbitrary scheme.
Suppose {Xi } is an open cover of X; then, for each Xi , X̃i exists. Now given i =
6 j,
let Uij ⊂ X̃i be νi−1 (Xij ), where Xij = Xi ∩ Xj . Then, Uij = X̃ij by the above
paragraph, hence by the uniqueness of (affine) normalization above, we see that there
exist isomorphisms ϕij : Uij → Uji for all i 6= j that are compatible in the sense of
Problem 2.12, since uniqueness up to unique isomorphism ensures that ϕjk ◦ ϕij = ϕik .
Thus, by Problem 2.12 the normalization X̃ exists. We claim by glueing the νi on
each Xi together as in Thm. 3.3 Step 3, we can obtain a morphism ν : X̃ → X. But
this is possible since νi |Xij = νj |Xij on each Xij since normalization is unique.
Finally, we must show that this scheme ν : X̃ → X satisfies the universal property
for normalization. Let f : Z → X be dominant, and let Zi = f −1 (Xi ). Then, since
Zi → Xi is then dominant, by the universal property for normalization in the affine
case proven above, we get maps ϕi : Zi → X̃i = ν −1 (Xi ). It remains to show that we
can glue these ϕi together. But ϕi |Zij = ϕj |Zij , just by the uniqueness of these maps
ϕ on the subset f (Zij ) = Xij , and so ϕ exists. ϕ is moreover unique since it is unique
locally by the universal property in the affine case.
Problem 3.9. The Topological Space of a Product. Recall that in the category of
varieties, the Zariski topology on the product of two varieties is not equal to the product
topology (I, Ex. 1.4). Now we see that in the category of schemes, the underlying
point set of a product of schemes is not even the product set.
(a) Let k be a field, and let A1k = Spec k[x] be the affine line over k. Show that
A1k ×Spec k A1k ∼
= A2k , and show that the underlying point set of the product is not
the product of the underlying point sets of the factors (even if k is algebraically
closed).
(b) Let k be a field, let s and t be indeterminates over k. Then Spec k(s),
Spec k(t), and Spec k are all one-point spaces. Describe the product scheme
Spec k(s) ×Spec k Spec k(t).
Proof of (a). It suffices to show k[x, y] ∼
= k[x] ⊗k k[y] by the fact that A1k ×Spec k
A1k = Spec(k[x] ⊗k k[y]). We have the map g : k[x] × k[y] → k[x, y] defined by
(p(x), q(y)) 7→ k[x, y]; we claim that this satisfies the universal property for the tensor
product in [AM69, Prop. 2.12]. So let P be a k-module with a k-bilinear mapping
f : k[x] × k[y] → P . This is determined uniquely by defining f on (xi , y j ) for all
i, j by bilinearity. But then, we see that defining the map f 0 : xi y j 7→ f (xi , y j ) and
extending by linearity (uniquely) makes it such that f = f 0 ◦ g. Thus, by the universal
property of the tensor product, k[x, y] ∼
= k[x] ⊗k k[y] and so A1k ×Spec k A1k ∼
= A2k .
We now claim that their underlying point sets are not the product of the factors.
But recall that as varieties, V (x − y) ⊂ A2k is a closed subvariety, while the corre8
sponding set ∆ = {(x, x)} ⊂ A1k × A1k is not closed, since ∆ is closed if and only if A1k
is Hausdorff by [Mun00, Exc. 17.13], but A1k is not Hausdorff since any two open sets
intersect. This means that the point corresponding to V (x − y) in the scheme A2k
from Prop. 2.6 does not have a corresponding point in the product set A1k × A1k .
Proof of (b). By Thm. 3.3, we have that Spec k(s)×Spec k Spec k(t) = Spec k(s)⊗k k(t).
But then, letting S = k(s) \ {0} and T = k(t) \ {0}, k(s) ⊗ k(t) = S −1 k[s] ⊗ T −1 k[t].
We claim that S −1 k[s] ⊗ T −1 k[t] ∼
= U −1 k[s, t], where U is the multiplicative subset of
polynomials p(s)q(t). But defining g : k[s, t] → S −1 k[s] ⊗ T −1 k[t] as the composition
of the isomorphism k[s, t] → k[s] ⊗ k[t] and the injection into S −1 k[s] ⊗ T −1 k[t],
then we see that if p(s)q(t) ∈ U , g(p(s)q(t)) = p(s) ⊗ q(t) which is a unit; g(a) = 0
implies a = 0 since we have an injective map, and every element of B is of the form
g(a)g(s)−1 by construction of the localization S −1 k[s] ⊗ T −1 k[t]. Hence we have an
isomorphism S −1 k[s] ⊗ T −1 k[t] ∼
= U −1 k[s, t] by the universal property for localization
[AM69, Cor. 3.2].
So, it suffices to find the prime ideals in k[s, t] that are disjoint from U , i.e.,
which do not contain polynomials that can be factored as p(s)q(t). Every maximal
ideal m ⊂ k[s, t] meets U , since once of its generators is a polynomial p(s) ∈ k[s] by
[Rei95, Prop. 1.5]. So, the nonzero prime ideals in k[s, t] are exactly the height one
primes generated by irreducible elements f (s, t) ∈ k[s, t]. But there are an infinite
number of f (s, t) that cannot be factored as p(s)q(t) for p(s) ∈ k[s], q(t) ∈ k[t], and
so Spec k(s) ×Spec k Spec k(t) is infinite.
Problem 3.10. Fibres of a Morphism.
(a) If f : X → Y is a morphism, and y ∈ Y is a point, show that sp(Xy ) is
homeomorphic to f −1 (y) with the induced topology.
(b) Let X = Spec k[s, t]/(s − t2 ), let Y = Spec k[s], and let f : X → Y be the
morphism defined by sending s → s. If y ∈ Y is the point a ∈ k with a 6= 0,
show that the fibre Xy consists of two points, with residue field k. If y ∈ Y is
corresponds to 0 ∈ k, show that the fibre Xy is a nonreduced one-point scheme.
If η is the generic point of Y , show that Xη is a one-point scheme, whose
residue field is an extension of degree two of the residue field of η. (Assume k
algebraically closed.)
Lemma 4 ([AM69, Exc. 2.2]). Let A be a ring, a an ideal, M an A-module. Then
(A/a) ⊗A M is isomorphic to M/aM .
h
π
Proof of Lemma 4. Consider the exact sequence 0 → a →
− A→
− A/a → 0. Tensoring
with M over A yields the right exact sequence
h⊗1
π⊗1
a ⊗ M −−→ A ⊗ M −−→ (A/a) ⊗ M → 0
9
by [AM69, Prop. 2.18]. [AM69, Prop. 2.14] gives the unique isomorphism f : A ⊗
M → M . We then consider g = (π ⊗ 1) ◦ f −1 : M → A ⊗ M → (A/a) ⊗ M .
Im(g) = Im(π ⊗ 1) = (A/a) ⊗ M , and ker(g) = f (ker(π ⊗ 1)) = f (Im(h ⊗ 1)) = aM .
Thus, we have the isomorphism g̃ : M/aM → (A/a) ⊗ M .
Lemma 5 ([AM69, Exc. 3.21]).
(i) Let A be a ring, S a multiplicatively closed subset of A, and φ : A → S −1 A
the canonical homomorphism. Then, φ∗ : Spec(S −1 A) → Spec(A) is a homeomorphism of Spec(S −1 A) onto its image in X = Spec(A). Let this image be
denoted by S −1 X. In particular, if f ∈ A, the image of Spec(Af ) in X is the
basic open set Xf .
(ii) Let f : A → B be a ring homomorphism. Let X = Spec(A) and Y = Spec(B),
and let f ∗ : Y → X be the mapping associated with f . Identifying Spec(S −1 A)
with its canonical image S −1 X in X, and Spec(S −1 B)(= Spec(f (S)−1 B)) with
its canonical image S −1 Y in Y , we have S −1 f ∗ : Spec(S −1 B) → Spec(S −1 A)
is the restriction of f ∗ to S −1 Y , and that S −1 Y = f ∗−1 (S −1 X).
(iii) Let a be an ideal of A and let b = ae be its extension in B. Let f : A/a → B/b
be the homomorphism induced by f . If Spec(A/a) is identified with its canonical
∗
image V (a) in X, and Spec(B/b) with its image V (b) in Y , then f is the
restriction of f ∗ to V (b).
(iv) Let p be a prime ideal of A. Take S = A − p in ii) and then reduce mod
S −1 p as in iii). Then, the subspace f ∗−1 (p) of Y is naturally homeomorphic to
Spec(Bp /pBp ) = Spec(k(p) ⊗A B), where k(p) is the residue field of the local
ring Ap . Spec(k(p) ⊗A B) is called the fiber of f ∗ over p.
Proof of (i). φ∗ is continuous by [AM69, Exc. 1.21i]. Since every prime ideal of S −1 A
is an extended ideal by [AM69, Prop. 3.11i], we see that φ∗ is injective by [AM69,
Exc. 3.20ii], and therefore bijective onto its image. We consider S −1 X = im(φ∗ ); this
is the set of prime ideals that do not meet S by [AM69, Prop. 3.11iv]. It therefore
remains to show that φ∗ is closed. Since any arbitrary ideal of S −1 A is an extended
ideal by [AM69, Prop. 3.11i], we only have to consider basis elements of the form
V (ae ) ⊆ Spec(S −1 A) for a ∈ A. We claim φ∗ (V (ae )) = S −1 X ∩ V (aec ) (note the
latter is closed in S −1 X since V (aec ) is closed in X). If p ∈ φ∗ (V (ae )), then p ∩ S = ∅
and ae ⊆ pe =⇒ aec ⊆ pec = p =⇒ p ∈ S −1 X ∩ V (aec ). On the other hand
if p ∈ S −1 X ∩ V (aec ), then p ∩ S = ∅ and aec ⊆ p =⇒ ae ⊆ pe . Thus φ is a
homeomorphism onto its image.
In particular, if f ∈ A, φ∗ (Spec(Af )) = Xf by having S = hf i.
Proof of (ii). We first claim the diagram, with ϕA : A ,→ S −1 A, ϕB : B ,→ S −1 B the
10
natural embedding maps,
A
f
B
ϕA
ϕB
S −1 A
S −1 f
S −1 B
commutes. But this is clear since S −1 B ' f (S)−1 B by [AM69, Exc. 3.4], and since
S −1 f (a/s) = f (a)/f (s). Moreover, calculating explicitly, if a ∈ A, (ϕB ◦ f )(a) =
ϕB (f (a)) = f (a)/1, while (S −1 f ◦ ϕA )(a) = S −1 f (a/1) = f (a)/1. Since [AM69,
Exc. 1.21] implies f ∗ ◦ ϕ∗B = (ϕB ◦ f )∗ = (S −1 f ◦ ϕA )∗ = ϕ∗A ◦ (S −1 f )∗ , we have the
commutative diagram
Spec(S −1 B) = S −1 Y
(S −1 f )∗
Spec(S −1 A) = S −1 X
ϕ∗B
ϕ∗A
f∗
Spec(B) = Y
Spec(A) = X
where the identification is by i), which shows the compatibility of (S −1 f )∗ and f ∗ .
We now show S −1 Y = f ∗−1 (S −1 X). The diagram above shows that f ∗ (S −1 Y ) ⊆
−1
S X, and so S −1 Y ⊆ f ∗−1 (S −1 X). On the other hand, suppose p ∈ f ∗−1 (S −1 X).
Then, pc = f ∗ (p) ∈ S −1 X, so pc ∩ S = ∅ by [AM69, Prop. 3.11iv]. To show p ∈ S −1 Y ,
it suffices to show p ∩ f (S) = ∅. So, suppose x ∈ p ∩ f (S); then, x = f (s) for
some s ∈ S ∩ pc = ∅, which is a contradiction. Thus, S −1 Y ⊇ f ∗−1 (S −1 X), and
S −1 Y = f ∗−1 (S −1 X).
Proof of (iii). Letting πA : A A/a, πB : B B/b be the natural quotient maps,
we have the commutative diagram
A
f
πA
A/a
B
πB
f
B/b
and by the same argument as in iv), we get
Spec(B/b) = V (b)
f
∗
∗
πB
Spec(B) = Y
Spec(A/a) = V (a)
∗
πA
f∗
11
Spec(A) = X
The identification comes from [AM69, 1.21iv], which says πB∗ is a homeomorphism
Spec(B/b) → V (ker(πB )) = V (b) and πA∗ is a homeomorphism Spec(A/a) →
∗
V (ker(πA )) = V (a). Thus, f and f ∗ are compatible.
Proof of (iv). Following the steps given, we get the commutative diagram
Spec(Bp /pBp )
fp
∗
Spec(Ap /pp )
∗
πB
Spec(Bp )
∗
πA
fp∗
Spec(Ap )
ϕ∗B
Spec(B)
ϕ∗A
f∗
Spec(A)
By iii), Spec(Bp /pBp ) is homeomorphic to V (pBp ). By ii), V (pBp ) is homeomorphic
to ϕ∗B (V (pBp )). We now claim that ϕ∗B (V (pBp )) = f ∗−1 (p). Suppose q ∈ f ∗−1 (p),
which gives p ∈ im(ϕ∗A ) =⇒ q ∈ im(ϕ∗B ). Then, since p = f −1 (q), i.e., f (p) ⊆ q, i.e.,
pB ⊆ q. Thus, qp is prime in Bp and contains pBp , i.e., f ∗−1 (p) ⊆ ϕ∗B (V (pBp )). In
the other direction, suppose q ∈ ϕ∗B (V (pBp )). Then, pBp ⊆ qp so pB ⊆ qcp = q, i.e.,
f (p) ⊆ q. So, p ⊆ f −1 (q). On the other hand, f −1 (q) ⊆ p since q ∩ f (A \ p) = ∅ by
choice of q. Thus, p = f −1 (q), and so q ∈ f ∗−1 (p), i.e., ϕ∗B (V (pBp )) = f ∗−1 (p).
We now want to show the isomorphism given. We have Bp /pBp ' Bp /(pB)p '
(B/pB)p by [AM69, Prop. 3.11v]. Then, (B/pB)p ' Ap ⊗A B/pB by [AM69, Prop. 3.5].
Lemma 4 gives Ap ⊗A B/pB ' Ap ⊗A (A/p ⊗A B). The associativity of the tensor
product from [AM69, Prop. 2.14ii] then gives Ap ⊗A (A/p ⊗A B) ' (A/p ⊗A Ap ) ⊗A B.
Applying Lemma 4 again yields (A/p ⊗A Ap ) ⊗A B ' Ap /pAp ⊗A B. But then
Ap /pAp = Ap /pp = k(p), since Ap is local and therefore pp can only be the maximal
ideal, and so we get the isomorphism Bp /pBp ' k(p) ⊗A B. We note that this also
preserves ring structure since this is just the map b/f (x) + pBp ↔ (1/x + pp ) ⊗ b.
Proof of (a). We consider f −1 (U ) ×Spec A Spec k(y) where y ∈ U = Spec A ⊂ Y . We
first see Xy = X ×Y Spec k(y) ∼
= f −1 (U ) ×Spec A Spec k(y) as
Sin Thm. 3.3 Step 7; thus
−1
it suffices to consider when Y is affine. Now, if f (U ) = i Bi , we see that
!
[
[
Spec Bi ×Spec A Spec k(y) ∼
(Spec Bi ×Spec A Spec k(y))
=
i
i
12
by Thm. 3.3 Step 5. If we show that Spec Bi ×Spec A Spec k(y) ∼
= (f |Spec Bi )−1 (y), then
we are done, for then
[
[
(Spec Bi ×Spec A Spec k(y)) ∼
= (f |Spec Bi )−1 (y) = f −1 (y).
i
i
But this is just Lemma 5(iv).
Proof of (b). Xy = Spec k[s, t]/(s − t2 ) ×k[s] Spec k(y) = Spec k[s, t]/(s − t2 ) ⊗k[s] k(y).
If y corresponds to a ∈ k, then k(y) = k[s](s−a) /(s − a)k[s](s−a) ∼
= (k[s]/(s −
a))(s−a) = k[s]/(s − a), since everything in the complement of (s − a) is already in
k, hence a unit. Thus, we have k[s, t]/(s − t2 ) ⊗k[s] k(y) ∼
= k[s, t]/(s − t2 , s − a) by
Lemma 4.
If a 6= 0, then
k[s, t]/(s − t2 , s − a) ∼
= k[t]/(a − t2 )
√
√
∼
= k[t]/( a − t)( a + t)
√
√
∼
= k[t]/( a − t) ⊗k[t] k[t]/( a + t)
∼
= k ⊗k[t] k,
again using Lemma 4, and so Xy ∼
= Spec k ⊗k[t] k, which has two points corresponding
to the prime ideals (1, 0) and (0, 1).
If a = 0, then k[s, t]/(s − t2 , s − a) ∼
= Spec k[t]/t2 , which
= k[t]/t2 . Thus, Xy ∼
has one point corresponding to (t), which is nilpotent; hence, we have a nonreduced
one-point scheme.
Now if y = η, then we have k(y) = S −1 k[s] where S = k[s] \ {0}, and k[s, t]/(s −
2
t )⊗k[s] k(y) ∼
= S −1 k[s, t]/(s−t2 ) by [AM69, Prop. 3.5]. But then, S −1 k[s, t]/(s−t2 ) ∼
=
k(s)[t]/(s − t2 ), which is a field; hence, Xy ∼
= Spec k(s)[t]/(s − t2 ) has one point. The
residue field has degree 2 since t has degree two over k(s).
Problem 3.11. Closed Subschemes.
(a) Closed immersions are stable under base extension: if f : Y → X is a closed
immersion, and if X 0 → X is any morphism, then f 0 : Y ×X X 0 → X 0 is also
a closed immersion.
(b) If Y is a closed subscheme of an affine scheme X = Spec A, then Y is also
affine, and in fact Y is the closed subscheme determined by a suitable ideal
a ⊆ A as the image of the closed immersion Spec A/a → Spec A.
(c) Let Y be a closed subset of a scheme X, and give Y the reduced induced
subscheme structure. If Y 0 is any other closed subscheme of X with the same
13
underlying topological space, show that the closed immersion Y → X factors
through Y 0 . We express this property by saying that the reduced induced structure
is the smallest subscheme structure on a closed subset.
(d) Let f : Z → X be a morphism. Then there is a unique closed subscheme Y
of X with the following property: the morphism f factors through Y , and if
Y 0 is any other closed subscheme of X through which f factors, then Y → X
factors through Y 0 also. We call Y the scheme-theoretic image of f . If Z is a
reduced scheme, then Y is just the reduced induced structure on the closure of
the image f (Z).
Proof of (a). For the sheaf condition, it suffices to show it on stalks p ∈ Y ×X X 0 by
Caution 1.2.1, and so we can restrict to the affine case. There, it suffices to show that
in the following commutative diagram, f ∗ surjective implies f 0∗ surjective:
(B ⊗R A)p
Bp
Ap
Rp
But recall (B⊗R A)p ∼
= Bp ⊗Rp Ap by [AM69, Prop. 3.7]. So since we have the surjection
Rp → Bp by assumption, tensoring over Rp by Ap gives a surjection Ap → Bp ⊗Rp Ap
by the right exactness of the tensor product. Now by Problem 2.18(c), we see that
since f 0∗ is surjective, f 0 |U is a closed immersion when Y ×X X 0 and X 0 are affine.
Now we extend to the case when X 0 is arbitrary. We claim that f (Y ×X X 0 ) is closed
if and only if it is closed in each open cover Ui . It suffices to show U = X 0 \f (Y ×X X 0 )
is open if and only if U ∩ Ui is open in each Ui . U ∩ Ui is open in U if and only if it
is open in X 0 , and so we are done.
Now cover Y ×X X 0 with open affines Vj . By the argument above,
f (Vj ) is
S each S
0
closed in S
X , hence the f (Vj ) are closed in every Ui .SThen, f (Vj ) = Ui ∩ f (Vj ),
and Ui ∩ f (Vj ) is closed for each Ui since it equals j Ui ∩ f (Vj ), each one of which
are closed, and there are only finitely many j such that f (Vj ) intersects Ui since we
can choose a refinement of our cover Vj that results from the finite type property in
Problem 3.13(a). Thus the image of Y ×X X 0 is closed, and we have a bijection since
we have a bijection locally.
14
Proof of (b). Since Y is a closed subscheme of the quasi-compact space X, it can
be covered by finitely many open affine subsets of the form D(fi ) ∩ Y with fi ∈ A
since the D(fi ) ∩ Y form a basis for Y . By possibly adding more sets of the form
D(fi ), such that D(fi ) ∩ Y = ∅, we can assume the D(fi ) cover X, since the D(fi )
form a basis for X. Then, the fi generate the unit ideal of A as in Problem 2.13(b),
and therefore generate the unit ideal in Γ(Y, OY ) by Problem 2.4. Thus, by Problem
2.17(b) Y is affine, and by Problem 2.18(d) we see that ϕ : A → B = Γ(Y, OY ) is
surjective, and so B ∼
= A/ ker ϕ and Y is determined by the ideal ker ϕ.
Proof of (c). First consider the case when X is affine. Then, the map
√ Y → X induces
the map on rings A → A/a for some a ⊂ A. Now since the radical a of a is exactly
the intersection of all the prime ideals in Y by√[AM69, Prop. 1.14], and so the reduced
induced subscheme structure is given √
by V ( a). By the universal property for the
quotient, we see that the map A → A/ a uniquely factors through A/a; thus, taking
Spec gives us the desired property.
Now in the case when X is arbitrary, it can be covered by affines Spec Ai ; the
restriction to Y, Y 0 give affine covers Spec Bi and Spec Bi0 , and so we have unique
maps ϕi : Spec Bi → Spec Bi0 by the above. We now claim that we can glue these
maps together. Now since for any point p ∈ Spec Bi ∩ Spec Bj , we can find an affine
neighborhood of the form D(f ) ⊂ Spec Bi ∩ Spec Bj for f ∈ Bi , and the factoring is
unique for the map Spec(Bi )f → Spec(Ai )f as well, we see that ϕi |Spec Bij = ϕj |Spec Bij ,
and so we can glue by Thm. 3.3 Step 4 to have a unique factorization map.
Proof of (d). Suppose first that X = Spec A is affine. Then, the map f : Z → Spec A
corresponds to a morphism f ∗ : A → Γ(Z, OZ ) of rings by Problem 2.4. We claim
that letting a = ker f ∗ , Y = Spec A/a satisfies the universal property. For, in the
category Rings, a closed immersion Y 0 → X corresponds to a surjection A → A/a0
by Problem 2.18(c), where a0 is the ideal defining Y 0 from (b). This map then factors
uniquely through A/a by the universal property for the quotient since a0 ⊂ a by the
assumption that our maps factor through A/a, A/a0 , and so translating back to Sch
by Problem 2.4 we have the universal property desired. Note that then, the map
Y and the morphism Y → X are unique up to unique isomorphism since if we had
another Y 0 that satisfied the universal property, then we can substitute the roles of
Y, Y 0 applying the universal property each time to get a unique isomorphism between
them. We moreover claim that Y = f (Z) as topological spaces. But this is automatic
since the universal property shows that Y is the smallest closed subscheme of X that
our map f could factor through, and the fact that f factors means that f (Z) ⊂ Y .
We claim this universal property holds for arbitrary X. Giving X an affine cover
Xi , we see that the universal property holds locally in each Xi . Thus, we have
15
scheme-theoretic images Yi of f −1 (Xi ) in each open affine subset. Given i =
6 j, we
−1
can then consider the image of f (Xi ∩ Xj ) = Yi ∩ Yj ; by the uniqueness of the
scheme-theoretic image proven above, we see that then the Yi are compatible in the
sense of Problem 2.12, since uniqueness up to unique isomorphism implies the tricycle
condition. The morphisms on each Yi glue together since in each intersection Xij ,
by Lemma 1 we can find an open subset that is distinguished in both Xi and Xj
around any point, and so the morphisms agree on intersections since our morphism is
unique locally. The map in the universal property exists for this same reason, and is
unique since it is unique on any open affine subset. Moreover Y = f (Z) by the same
argument as above.
Finally, suppose Z is a reduced scheme. Then the scheme-theoretic image of f ,
Y = f (Z), satisfies the universal property for the reduced induced subscheme by
Problem 2(c), and so we are done.
Problem 3.12. Closed Subschemes of Proj S.
(a) Let ϕ : S → T be a surjective homomorphism of graded rings, preserving degrees.
Show that the open set U of (Ex. 2.14) is equal to Proj T , and the morphism
f : Proj T → Proj S is a closed immersion.
(b) If I ⊆ S is a homogeneous ideal, take T = S/I and let Y be the closed subscheme
of X = Proj S defined as image of the closed immersion Proj S/I → X. Show
that different homogeneous ideals can give rise to
Lthe same closed subscheme.0
0
For example, let d0 be an integer, and let I = d≥d0 Id . Show that I and I
determine the same closed subscheme.
Proof of (a). It suffices to show U c = {p ∈ Proj T | p ⊇ ϕ(S+ )} is empty. But
ϕ(S+ ) = T+ by surjectivity and since ϕ preserves degrees, and so U c = ∅ since Proj T
consists of all homogeneous primes that do not contain all of T+ .
Now recall that the map f is defined by f (p) = ϕ−1 (p). We first claim the sheaf
morphism is surjective; as before, we can check this on stalks. But by Prop. 2.5(a)
the stalks are equal to Sϕ−1 (p) , Tp for p ∈ Proj T , and ϕp : Sϕ−1 (p) → Tp is surjective if
ϕ is by [AM69, Prop. 3.9].
We first claim that f is injective. Suppose f (p) = f (q), i.e., ϕ−1 (p) = ϕ−1 (q).
Suppose p =
6 q, and let x ∈ p \ q, say. Since ϕ is surjective, we then see that
ϕ−1 (x) 6= ∅. But since ϕ−1 (p) = ϕ−1 (q), we have ϕ−1 (x) ⊂ ϕ−1 (q), which implies
x ⊂ ϕ(ϕ−1 (q)) ⊂ q by [AM69, Prop. 1.17i], a contradiction.
T
Now we claim that f (Proj T ) = V (a), where a = p∈Proj T ϕ−1 (p). First suppose
q ∈ V (a), i.e., a ⊂ q. We want to show q = ϕ−1 (p) for some p ∈ Proj T . So
let p = ϕ(q). We claim this is a homogeneous prime ideal. So let ab ∈ p for a, b
homogeneous. Since ϕ is surjective, we see that a, b have homogeneous preimages
16
ã, b̃ ∈ S such that ãb̃ ∈ q, and so one of ã, b̃ lie in q, hence one of a, b lies in p. Thus,
q ∈ f (Proj T ), and V (a) ⊂ f (Proj T ). In the other direction, if p ∈ Proj T , then
f (p) = ϕ−1 (p) clearly contains a by construction.
Finally, since f is a bijection Proj T to V (a) closed, and ϕ preserves inclusion of
ideals, we see f is a closed map, and so f is a closed immersion.
Proof (b). We first have the commutative diagram of graded rings
S/I 0
S
S/I
where each map is a surjection. By (a), this gives the commutative diagram
Proj S/I 0
Proj S
Proj S/I
where each map is a closed immersion. But (S/I 0 )d ∼
= (S/I)d for d ≥ d0 , and
0 ∼
so Proj S/I = Proj S/I by Problem 2.14(c), and so they define the same closed
subscheme structure by definition.
Problem 3.13. Properties of Morphisms of Finite Type.
(a) A closed immersion is a morphism of finite type.
(b) A quasi-compact open immersion (Ex. 3.2) is of finite type.
(c) A composition of two morphisms of finite type is of finite type.
(d) Morphisms of finite type are stable under base extension.
(e) If X and Y are schemes of finite type over S, then X ×S Y is of finite type
over S.
f
g
(f ) If X → Y → Z are two morphisms, and if f is quasi-compact, and g ◦ f is of
finite type, then f is of finite type.
(g) If f : X → Y is a morphism of finite type, and if Y is noetherian, then X is
noetherian.
Remark. Our morphisms will be denoted f : X → Y , g : Y → Z.
17
Proof of (a). Let f : X → Y be our closed immersion. Cover Y with Vi = Spec Bi .
Then, for each i, Vi ∩ f (X) is a closed subscheme of Vi , and so by Problem 3.11(b),
the map locally looks like f −1 (Vi ∩ f (X)) = Spec Bi /bi → Spec Bi = Vi for some ideal
bi ⊂ Bi , and since Bi /bi is a finitely generated Bi -algebra, f is of finite type.
Proof of (b). Let f : X → Y be our quasi-compact open immersion. Cover Y with
Vi = Spec Bi such that f −1 (Vi ) is quasi-compact. Then, f −1 (Vi ) = f −1 (Vi ∩ f (X)) is
an open set in X since f is a homeomorphism between X and f (X), and since it is
quasi-compact, it is covered by finitely many distinguished sets Spec(Bi )fj , but these
(Bi )fj are finitely generated Bi -algebras.
f
g
Proof of (c). Let X → Y → Z with f, g of finite type. Cover Z with Wi = Spec Ci .
Then, each g −1 (Wi ) can be covered by a finite number of affines Vij = Spec Bij , where
Bij are finitely generated Ci -algebras. Moreover, each f −1 (Vij ) can be covered by a
finite number of affines Uijk = Spec Aijk , where Aijk are finitely generated Bij -algebras.
Thus, (g ◦ f )−1 (Wi ) can be covered by a finite number of affines Uijk = Spec Aijk .
By definition, we have surjections Ci [t1 , . . . , tn ] → Bij and Bij [s1 , . . . , sm ] → Aijk ,
and so we have surjections Ci [t1 , . . . , tn , s1 , . . . , sm ] → Aijk , i.e., the Aijk are finitely
generated Ci -algebras.
Proof of (d). We want to show that if f : X → S is of finite type, and S 0 → S is any
morphism, then f 0 : X ×S S 0 → S 0 is also of finite type.
Cover S with open affines Vi = Spec Bi ; then g −1 (Vi ) is a cover for S 0 . Now
cover these g −1 (Vi ) with open affines Vij0 = Spec Bij0 ; these have preimage X ×S Vij0 ∼
=
f −1 (Vi ) ×Vi Vij0 by Thm. 3.3 Step 7. But the f −1 (Vi ) can be covered by finitely
many Uik = Spec Aik ⊂ X where Aik are finitely generated Bi -algebras, and so the
f −1 (Vi )×Vi Vij0 can be covered by finitely many Uik ×Vi Vij0 . But since the three schemes
in this fibre product are affine, we see Uik ×Vi Vij0 = Spec Aik ⊗Bi Bij0 . Since Aik are
finitely generated Bi -algebras, we have a surjection Bi [t1 , . . . , tn ] → Aik , and tensoring
with Bij0 over Bi gives a surjection Bij0 [t1 , . . . , tn ] → Aik ⊗Bi Bij0 by the right-exactness
of the tensor product, and so Aik ⊗Bi Bij0 is a finitely generated Bij0 -algebra.
Proof of (e). We can consider X ×S Y as a base extension Y → S of X → S; then,
X ×S Y → Y is of finite type by (d). Since Y → S is also of finite type, X ×S Y → S
is of finite type by (c).
Proof of (f ). Let Wi = Spec Ci be an affine cover of Z, which have preimage (g ◦
f )−1 (Wi ) that is covered by Uij = Spec Aij for Aij finitely generated Ci -algebras.
Let each g −1 (Wi ) be covered by Vik = Spec Bik . Then, f −1 (Vik ) ∩ Uij are covered
by distinguished opens Spec(Aij )f` , and the (Aij )f` are clearly finitely-generated
18
Ci -algebras. Since f is quasi-compact, we can choose finitely many Spec(Aij )f` , for
they form a cover of g −1 (Wi ) by Problem 3.2.
Proof of (g). Since Y is noetherian, it can be covered by finitely many open affines
Spec Ai where each Ai is noetherian. Since f is of finite type, f −1 (Spec Ai ) can be
covered by finitely many Spec Bij for Bij finitely generated Ai -algebras by Problem
3.3(b). The Bij are noetherian by [AM69, Cor. 7.7], and so we have a finite cover of
X with Spec Bij for Bij noetherian.
Problem 3.14. If X is a scheme of finite type over a field, show that the closed
points of X are dense. Give an example to show that this not true for arbitrary
schemes.
Proof. If X is of finite type over k, X can be covered by finitely many Spec Ai for Ai
finitely generated k-algebras.
We first claim that if x is a closed point in some open set U , it is closed in X.
It suffices to show x is closed in each Spec Ai . If Spec Ai 63 x, there is nothing to
check, so x is closed in some affine neighborhood Spec(Ai )f ⊂ U , hence corresponds
to a maximal ideal in (Ai )f . Now, the inclusion Spec(Ai )f → Spec Ai corresponds to
the localization map Ai → (Ai )f , and so since (Ai )f is a finitely generated k-algebra,
hence a Jacobson ring by [AM69, Ex. 5.24], the contraction of x in Ai is maximal by
[Eis95, Thm. 4.19]. Thus x is closed in Spec Ai , hence in X.
Now it suffices to show if p ∈ U ⊂ X, U contains some closed points. p ∈
Spec B ⊂ U for some affine B, and by [AM69, Cor. 1.4] there exists a maximal ideal
m ⊃ p. Then, m ∈ Spec B is closed, hence closed in X by the above.
Now suppose X is the spectrum for a DVR. Then, the closed point m is unique
hence equal to its own closure, which is not all of X.
Problem 3.15. Let X be a scheme of finite type over a field k (not necessarily
algebraically closed).
(a) Show that the following three conditions are equivalent (in which case we say
that X is geometrically irreducible).
(i) X ×k k is irreducible, where k denotes the algebraic closure of k. (By
abuse of notation, we write X ×k k to denote X ×Spec k Spec k.)
(ii) X ×k ks is irreducible, where ks denotes the separable closure of k.
(iii) X ×k K is irreducible for every extension field K of k.
(b) Show that the following three conditions are equivalent (in which case we say
X is geometrically reduced).
(i) X ×k k is reduced.
(ii) X ×k kp is reduced, where kp denotes the perfect closure of k.
19
(iii) X ×k K is reduced for all extension fields K of k.
(c) We say that X is geometrically integral if X ×k k is integral. Give examples of
integral schemes which are neither geometrically irreducible nor geometrically
reduced.
Lemma 6. X is irreducible if and only if there exists an open cover Ui where each
Ui is irreducible, and Ui ∩ Uj 6= ∅ for all i, j.
Proof of Lemma. X is reducible ⇔ X = A ∪ B ⇔ ∅ = (X \ A) ∩ (X \ B), and so X
is irreducible if and only if every open subset is dense.
⇒ Clearly Ui ∩ Uj =
6 ∅ for all i, j. If U ⊂ Ui , U is of the form A ∩ Ui for A closed
in X. But since U is open in X, A = X and so U = Ui .
⇐ Let U ⊂ X. Suppose U ∩ Ui 6= ∅ but U ∩ Uj = ∅; then, U ∩ (Ui ∩ Uj ) = ∅,
contradicting that U is dense in Ui . So U intersects all Ui , and since a set containing
U is closed if and only if it is closed in every element of the open cover Ui , and since
U is dense in each Ui , the closure must contain all Ui ’s, hence is equal to X.
Thus, it suffices to check irreducibility for affine schemes, since intersections are
stable under base change by Thm. 3.3, Step 5.
Lemma 7. If X = Spec A for A a finitely generated k-algebra, and K/k is an
algebraic extension, then for every closed subscheme W ⊂ X ×k K, there exists a
finite field extension K 0 , k ⊂ K 0 ⊂ K 0 , and a closed subvariety Z ⊂ X ×k K 0 such
that W = Z ×k K.
Proof of Lemma. W = V (I) for I ⊂ A ⊗k K, and I = (f1 , . . . , fm ). There then exists
K 0 a finite extension of k such that fi ∈ A ⊗k K 0 for all i. Let I 0 be the ideal generated
by the fi , and Z = V (I 0 ) ⊂ X ×k K 0 . Then, I 0 ⊗K 0 K = I, and so Z ×k K = W .
Lemma 8. If K/k is a purely inseparable extension, then p1 : X ×k K → X is a
homeomorphism for X = Spec A, A a finitely generated k-algebra.
Proof of Lemma. Suppose K is a simple extension of k, i.e., generated by one element.
n
Then, K ∼
= k[x]/(xp − α) by [Isa09, Thm. 19.10] for some n and p = char k; hence,
A → A ⊗k K is an integral extension. If p ⊂ A is prime, then by lying over [AM69,
n
Thm. 5.10], there exists q ⊂ A ⊗k K such that qc = p. Thus pe ⊂ q. But αp ∈ qc = p
√
for all α ∈ q by [Isa09, Thm. 19.10], and so q ⊂ pe . Thus, q = p, and p1 is injective.
Now show that p1 is closed. If I ⊂ A ⊗k K is an ideal, then letting J = I ∩ A,
n
I p ⊂ J, and so p1 (V (I)) = V (J). Since p1 is surjective by Lemma 3, we see that
p1 is a homeomorphism. By successive simple extensions, we see the claim holds for
K/k finite. The arbitrary case follows by Lemma 7.
20
Proof of (a). (iii) implies (i) and (ii), and (i) ⇔ (ii) by Lemma 8 by [Isa09,
Thm. 19.14], which says any field extension can be factored as a separable extension
and a purely inseparable extension.
It remains to show (i) ⇒ (iii). We claim we can assume K is algebraically closed.
K ,→ K is integral, hence A ⊗k K ,→ A ⊗k K is also by [AM69, Exc. 5.3], and
Spec A ⊗k K → Spec A ⊗k K is surjective by Lemma 3. If X̃ → X is a continuous
surjection, then X̃ irreducible implies X irreducible, since X = A ∪ B lifts to
X̃ = Ã ∪ B̃, and so X̃ = Ã, say, and X = A.
Thus, we can assume K is an extension of k. Let X = X ×k k; by transitivity of
base change, it suffices to show X irreducible implies X ×k K is irreducible. If X ×k K
is not irreducible, then we can find f, g to make a decomposition X ×k K = V (f )∪V (g).
Now we can write K = K(B) for some integral domain B which is a finitely generated
k-algebra, and f, g ∈ A⊗k B[1/b] for some 0 6= b ∈ B. Then, D(f ), D(g) are nonempty
open subsets of Spec A ⊗k B[1/b], whose image in Spec B[1/b] are nonempty opens
since B[1/b] → A ⊗k B[1/b] is flat since we can consider them as infinite dimensional
vector spaces over k, and so by [AM69, Exc. 7.25], Spec A ⊗k B[1/b] → Spec B[1/b]
is open. Thus, their intersection is nonempty and contains a closed point p. But then
∼
∼
p−1
2 (p) = Spec A ⊗k k(p) = Spec A by Problem 3.10, and so we have covered Spec A
with two proper closed sets, the images of V (f ) and V (y), a contradiction.
For (b), since we can check reducedness on stalks by Problem 2.3(a), it suffices to
consider the affine case.
Proof of (b). As before, (iii) implies (i) and (ii), and so it suffices to show (ii) ⇒
(i) ⇒ (iii).
(ii) ⇒ (i) We claim that if K/k is separable, then X reduced implies X ×k K is
reduced; this amounts
if A is. We can assume A is a
Q to saying A ×k K is reduced
Q
domain, since A ,→ A/pi , and so A ×k K ,→ A/pi ⊗k K since K/k is flat, where
the products are taken over the minimal primes of A, and also because a product of
rings is reduced if and only if each ring in the product is. Now since A ,→ Frac(A),
it suffices to show that F ⊗k K is reduced for every extension field F/k. Replace
K by a finite extension L/k since every element of F ⊗K K is contained in F ⊗K L
for some finite L. Suppose we have a simple extension; then, L ∼
= k[t]/(f ) with f
∼
separable, so F ⊗K L = F [t]/(f ), and f is still separable, so F [t]/(f ) is reduced. By
induction, this works for any finite extension L/k, hence for K in general. Now since
k/kp is separable, we are done.
(i) ⇒ (iii) We can reduce to the case when K is an extension of k, for if X ×k K
is reduced, we have an injection of rings A ⊗k K ,→ A ⊗k K by the fact that A is flat
as a k-module (it is an infinite dimensional vector space), and so N(A ⊗k K) = 0.
21
Now if A ⊗k K had nilpotent elements, then we would have some system of equations
in k with solutions in K. We claim we then have solutions in k, since if we suppose
not, the system generates the unit ideal and hence does in K, i.e., cannot have a
solution. Thus we have a contradiction.
Proof of (c). X = Spec R[x]/(x2 + 1) ∼
= Spec C is irreducible but not geometrically
irreducible since X ×R C = Spec C[x]/(x2 + 1) = Spec(C ⊕ C) = Spec C q Spec C.
k = Fp (T ) and K = Fp (T 1/p ); then, Spec K is reduced but not geometrically
reduced as Spec K ×k Spec K = Spec(K ⊗k K), and K ⊗k K has nonzero nilpotent
x = 1 ⊗ T 1/p − T 1/p ⊗ 1 with xp = 0. This is also an example of an integral but not
geometrically integral scheme.
Problem 3.16. Noetherian Induction. Let X be a noetherian topological space, and
let P be a property of closed subsets of X. Assume that for any closed subset Y of X,
if P holds for every proper closed subset of Y , then P holds for Y . (In particular,
P must hold for the empty set.) Then P holds for X.
Proof. Let Σ be the set of closed subsets of X not satisfying P; then, by the noetherian
property Σ has a minimal element Z. Since P holds for every proper closed subset
of Z by minimality, P holds for Z, a contradiction. Thus, Σ is empty.
Problem 3.17. Zariski Spaces. A topological space X is a Zariski space if it is
noetherian and every (nonempty) closed irreducible subset has a unique generic point
(Ex. 2.9).
For example, let R be a discrete valuation ring, and let T = sp(Spec R). Then T
consists of two points t0 = the maximal ideal, t1 = the zero ideal. The open subsets
are ∅, {t1 }, and T . This is an irreducible Zariski space with generic point t1 .
(a) Show that if X is a noetherian scheme, then sp(X) is a Zariski space.
(b) Show that any minimal nonempty closed subset of a Zariski space consists of
one point. We call these closed points.
(c) Show that a Zariski space X satisfies the axiom T0 : given any two distinct
points of X, there is an open set containing one but not the other.
(d) If X is an irreducible Zariski space, then its generic point is contained in every
nonempty open subset of X.
(e) If x0 , x1 are points of a topological space X, and if x0 ∈ {x1 }− , then we say
that x1 specializes to x0 , written x1
x0 . We also say x0 is a specialization of
x1 , or that x1 is a generization of x0 . Now let X be a Zariski space. Show that
the minimal points, for the partial ordering determined by x1 > x0 if x1
x0 ,
are the closed points, and the maximal points are the generic points of the
irreducible components of X. Show also that a closed subset contains every
22
specialization of any of its points. (We say closed subsets are stable under
specialization.) Similarly, open subsets are stable under generization.
(f ) Let t be the functor on topogical spaces introduced in the proof of (2.6). If
X is a noetherian topological space, show that t(X) is a Zariski space. Furthermore X itself is a Zariski space if and only if the map α : X → t(X) is a
homeomorphism.
Proof of (a). Note that since X is a scheme, every (nonempty) closed irreducible
subset has a unique generic point by Problem 2.9. It therefore suffices to show that
sp X is noetherian. So let U ⊂ X, and cover U by Ui . Since X is a noetherian
scheme, it can be covered by a finite number of open affines Spec Aj ; since each Aj is
noetherian, sp(Spec Aj ) is a noetherian topological space by Problem 2.13(c). Thus,
each U ∩ Spec Aj can be covered by a finite cover Ui ∩ Spec Aj , and so taking the
union of the collection of Ui ’s such that Ui ∩ Spec Aj cover each U ∩ Spec Aj gives us
a finite cover of U , and so by Problem 2.13(a) sp(X) is noetherian.
Proof of (b). If Z is a minimal nomepty closed subset, then it is irreducible, hence
has a unique generic point η. If x ∈ Z, then Z minimal so Z = {x}, hence x = η.
Proof of (c). Let x, y ∈ X, and let U = X \ {x}. If y ∈ U , we are done, so suppose
not. Then, y ∈ {x}. If x ∈ {y} also, then x, y would be generic points of the same
irreducible subset, and so x ∈
/ {y}, and X \ {y} contains x but not y.
Proof of (d). Suppose not, and η ∈
/ U 6= ∅, where η is the unique generic point of
c
X. Then, η ∈ U ( X is a closed set smaller than X containing η, contradicting the
definition of closure.
S
Proof of (e). Let X = Xi be a decomposition into maximal irreducible components.
Then, let the generic points be ηi ∈ Xi . Let x such that η ∈ {x}; then, Xi ⊂ {x},
and so since {x} is irreducible and generic points are unique, η = x and η is therefore
maximal.
Now since minimal closed subsets consist of one point by (b), and so every closed
set contains closed points, then we see closed points are minimal.
Proof of (f ). By construction in Prop. 2.6, we see the lattice of closed subsets in X
is the same as in t(X), and so t(X) is noetherian.
If A ⊂ t(X) is closed and irreducible, its preimage in X is also closed and
irreducible, hence there is a unique η ∈ t(X) that represents A. Moreover, this then
shows the image of η in X must be {η} = A. If η 0 is another generic point, then
{η} = {η 0 }, hence η = η 0 .
23
Now if X is itself Zariski, there there is a bijection between points and irreducible
closed subsets, so α : X → t(X) is a bijection. The inverse of α is clearly also
continuous.
Problem 3.18. Constructible sets. Let X be a Zariski topological space. A constructible subset of X is a subset which belongs to the smallest family F of subsets
such that (1) every every open subset is in F, (2) a finite intersection of elements of
F is in F, and (3) the complement of an element of F is in F.
(a) A subset of X is locally closed if it is the intersection of an open subset with a
closed subset. Show that a subset of X is constructible if and only if it can be
written as a finite disjoint union of locally closed subsets.
(b) Show that a constructible subset of an irreducible Zariski space X is dense if
and only if it contains the generic point. Furthermore, in that case it contains
a nonempty open subset.
(c) A subset S of X is closed if and only if it is constructible and stable under
specialization. Similarly, a subset T of X is open if and only if it is constructible
and stable under generization.
(d) If f : X → Y is a continuous map of Zariski spaces, then the inverse image of
any constructible subset of Y is a construcible subset of X.
Proof of (a). Let G denote the subsets that can be written as a finite disjoint union
of locally closed subsets. (1) and (3) imply all closed sets are in F, and (2) and (3)
imply all finite unions of elements of F are in F, as long as they are disjoint. Thus,
G ⊂ F.
Now every set of form (1) is in G by taking U = U ∩ X, and so we want to show
applying (2) and (3) don’t affect anything. For (2),
!
!
n
n
n
[
[
[
Ui ∩ Fi ∪
Vi ∩ Gi =
(Ui ∩ Vj ) ∩ (Fi ∩ Gj ),
i=1
i=1
i,j=1
which is in F. For (3), we proceed by induction on n. If Gn is the set of subsetsSof X
that can be written as a finite disjoint union of n locally closed sets, then G = Gn ,
and so if we show applying (3) to elements in Gn for all n result in elements of G, we
will be done. For n = 1, we have
(U ∩ F )c = U c ∪ F c = U c ∪ (F c ∩ U ) ∈ G.
For S ∈ Gn , we can write S = Sn−1 ∪ S1 for some Sn−1 ∈ Gn−1 , S1 ∈ F1 . S c =
c
Sn−1
∩ S1c . But the latter sets are both in G by inductive hypothesis and (2) above
proves their intersection S c is in G as well.
24
Proof of (b). Let Z denote our constructible set. If it contains the generic point, it is
dense, so suppose not. Then, by (a) we can write
Z=
m
[
(Ui ∩ Fi )
i=1
where Ui open in X, andSFi closed and irreducible. Then, Ui ∩ Fi = Fi by irreducibility
of the Fi , and so Z = i Fi . Now if Z is dense, then one of the Fi = X; since any
nonempty open set contains the generic point, we see that then Ui ∩ Fi contains the
generic point.
Proof of (c). By Problem 3.17(e), a closed set is stable under specialization, and
is clearly constructible. Conversely, suppose Z is constructible and is stable under
specialization. Z has a decomposition as in (b), and let xi be the generic point of
S Fi ;
this is also in Ui since if not, Ui ∩ Fi = ∅. Then, Z contains
S all of Fi , and so Z ⊃ Fi .
Now, since any x ∈ Z is contained in a Fi , then Z ⊂ Fi , and so Z is closed.
Now if V is open, then its complement V c is closed, i.e., constructible and stable
under specialization. Thus, V is constructible, and is stable under generization, for
it contains the generic points of the irreducible components of X it meets, and the
generic points are maximal by Problem 3.17(e).
Proof of (d). We have
f
−1
m
[
!
(Ui ∩ Fi )
i=1
=
n
[
f
−1
(Ui ∩ Fi ) =
i=1
n
[
f −1 (Ui ) ∩ f −1 (Fi ),
i=1
and the last set is constructible by continuity of f .
Problem 3.19. The real importance of the notion of constructible subsets derives
from the following theorem of Chevalley: let f : X → Y be a morphism of finite
type of noetherian schemes. Then the image of any constructible subset of X is a
constructible subset of Y . In particular, f (X), which need not be either open or closed,
is a constructible subset of Y . Prove this theorem in the following steps.
(a) Reduce to showing that f (X) itself is constructible, in the case where X and Y
are affine, integral noetherian schemes, and f is a dominant morphism.
(b) In that case, show that f (X) contains a nonempty open subset of Y by using
the following result from commutative algebra: let A ⊆ B be an inclusion of
noetherian integral domains, such that B is a finitely generated A-algebra. Then
given a nonzero element b ∈ B, there is a nonzero element a ∈ A with the
25
following property: if ϕ : A → K is any homomorphism of A to an algebraically
closed field K, such that ϕ(a) 6= 0, then ϕ extends to a homomorphism ϕ0 of B
into K, such that ϕ0 (b) 6= 0.
(c) Now use noetherian induction on Y to complete the proof.
(d) Give some examples of morphisms f : X → Y of varieties over an algebraically
closed field k, to show that f (X) need not be either open or closed.
Proof of (a). If Z ⊂ X is constructible, then since we can restrict f |Z , we only have
to show f (X) is constructible. Now if we have an affine cover Vi for Y and
S Uij covers
−1
for each f (Vi ), then showing f (Uij ) is constructible implies f (X) = i,j f (Uij ) is
constructible by finiteness of Uij given by the finite type condition and noetherianness,
and so we can assume X, Y to be affine. By similar argument, we can assume X, Y
are irreducible. Reducing a scheme doesn’t change the topology so we can assume
X, Y are reduced. By Prop. 3.1, this equates to X, Y being integral.
Now we claim it suffices to consider f dominant. For, suppose f (X) is constructible
for all dominant f . For arbitrary f 0 : X → Y , there is an induced morphism f 0 : X →
f (X); this map, then, is clearly dominant, and so if f 0 (X) is constructible in f (X),
then it is constructible in f (X) since f (X) has the subspace topology, and so
each Ui , Fi in the decomposition from Problem 3.17(b) lift to open, closed sets
respectively.
Proof of (b). B can be written as A[x1 , . . . , xr , xr+1 , . . . , xn ] =: A∗ [xr+1 , . . . , xn ] where
A∗ is a polynomial ring and xr+1 , . . . , xn are integral over A∗ by Noether normalization
[Mat70, 14.G]. Let gj (x1 , . . . , xr )(xj ) be the polynomial relations each xj satisfies for
r < j ≤ n. Taking the leading terms gj0 (x1 , . . . , xr ), let
n
Y
a0 (x1 , . . . , xr ) = b0
gj0
j=r+1
be the product of these leading terms multiplied by b0 some coefficient of a monomial
m in a representative of b. Then, the image of a0 in K[x1 , . . . , xr ] through ϕ extended
by having xi 7→ xi ∈ K[x1 , . . . , xr ] is
n
Y
0
ϕ(a ) = ϕ(b0 )
ϕ(gj0 ).
j=r+1
Now let a be some coefficient in A of this polynomial a0 (x1 , . . . , xr ). Then, we see
that
ϕ(a) = ϕ(b0 )ϕ(a00 )
26
for some a00 ∈ A, and so if ϕ(a) 6= 0, then ϕ(b0 ) 6= 0 since K is a field. Finally, we see
that if g(x1 , . . . , xn ) is a representative of b, then ϕ(g(x1 , . . . , xn )) is a polynomial
in K[x1 , . . . , xn ]; letting α be a non-root of this polynomial that is not zero, we see
that then, since ϕ(b0 )m(α) 6= 0 by the fact that ϕ(b0 ) 6= 0, that ϕ0 (b) 6= 0 where ϕ0
denotes the composition of the extension of ϕ to K[x1 , . . . , xn ] and evaluation at α.
Now that we’ve proved the algebraic result, we can prove the claim. Since Y is
integral it has a generic point η = (0), and f dominant implies η has preimage in X.
That is, there is letting X = Spec B and Y = Spec A, there exists p ⊆ B such that
ϕ−1 (p) = (0). ϕ is therefore injective since 0 ∈ p. Since f is of finite type, we know
B is a finitely-generated A-algebra and so we can apply the algebraic fact above.
Letting b = 1, we claim that D(a) ⊂ f (X). For, if p ∈ D(a), then a =
6 p and so the
image of a under the composition A → A/p → k(A/p) → k(A/p) = K is nonzero,
and so we can lift ϕ to a homomorphism ϕ0 : B → K in which 1 6= 0. Thus, ker ϕ0 is
a proper prime ideal q of B, and so A ∩ q = A ∩ ker ϕ0 = ker ϕ = p, and f (q) = p.
Hence D(a) ⊂ f (X).
Lemma 9. Z ⊂ Y is constructible if for each irreducible closed set Y0 ⊂ Y , either
Y0 ∩ Z is not dense in Y0 , or Y0 ∩ Z contains a non-empty open set of Y0 .
Proof of Lemma. Z ⊂ Y is noetherian by I, Problem 1.7(c). Since ∅ is constructible,
we suppose Z 6= ∅. We proceed by noetherian induction, and so assume for Z that
the lemma holds for all Z 0 ⊂ Z.
Let Z = F1 ∪ · · · ∪ Fr be a decomposition into irreducible components. Then,
F1 ∩ Z is dense in F1 , and so by the inductive hypothesis there exists a proper closed
subset F 0 ⊂ F1 such that F1 \ F 0 ⊂ Z. Then, putting F ∗ = F 0 ∪ F2 ∪ · · · ∪ Fr , we
have Z = (F1 \ F 0 ) ∪ (Z ∩ F ∗ ). The set (F1 \ F 0 ) is locally closed in X, and (Z ∩ F ∗ )
satisfies the inductive hypothesis since if Y0 is irreducible and if Z ∩ F ∗ ∩ Y0 = Y0 , the
closed set F ∗ must contain Y0 and so Z ∩ F ∗ ∩ Y0 = Z ∩ Y0 . Since Z ∩ F ∗ ⊂ F ∗ ⊂ Z,
the set Z ∩ F ∗ is constructible by inductive hypothesis. Thus Z is constructible.
Proof of (c). Consider f (X), and an irreducible closed set Y0 ⊂ Y . If the intersection
with f (X) is empty, then Y0 ∩ f (X) is not dense in Y0 , so suppose not. Then, the
preimage of Y0 in X is closed, and so restricting f to f −1 (Y0 ), f (f −1 (Y0 )) contains a
nonempty open subset of Y by (b), which is a nonempty open subset of Y0 , and so
the conditions for the lemma are met, and f (X) is constructible.
Proof of (d). Let A2 → A2 be defined by (x, y) 7→ (x, xy).
Problem 3.20. Dimension. Let X be an integral scheme of finite type over a field k
(not necessarily algebraically closed). Use appropriate results from (I, §1) to prove
the following.
27
(a) For any closed point P ∈ X, dim X = dim OP , where for rings, we always
mean the Krull dimension.
(b) Let K(X) be the function field of X (Ex. 3.6). Then dim X = tr. d. K(X)/k.
(c) If Y is a closed subset of X, then codim(Y, X) = inf{dim OP,X | P ∈ Y }.
(d) If Y is a closed subset of X, then dim Y + codim(Y, X) = dim X.
(e) If U is a nonempty open subset of X, then dim U = dim X.
(f ) If k ⊆ k 0 is a field extension, then every irreducible component of X 0 = X ×k k 0
has dimension = dim X.
Proof of (a). If P ∈ X, let Spec A be the affine scheme containing it. Then, we have,
by I, Prop. 1.10 and Thm. 1.8A(b),
dim X = dim Spec A = dim A = ht P + dim A/P = ht P = dim OP ,
since P is a maximal ideal.
Proof of (b). By (a) and I, Thm. 1.8A(b), we see dim X = dim A = tr. d. K(A)/k =
tr. d. K(X)/k since K(X) ∼
= K(A) by Problem 3.6 by the fact that Spec A contains
the generic point by Problem 3.17(d).
Proof of (c). codim(Y, X) = inf P ⊇I codim(Spec A/P, Spec A) after passing to an open
affine Spec A containing a closed point in Y by (a), where Spec A/I = Y ∩ Spec A.
But codim(Spec A/P, Spec A) = ht P = dim OP,X by (a), and so codim(Y, X) =
inf P ⊇I ht P = inf P ∈Y {dim OP,X | P ∈ Y }.
Proof of (d). We have
codim(Y, X) = inf codim(Spec A/P, Spec A)
P ⊇I
= inf ht P
P ⊇I
= inf dim A − dim A/P
P ⊇I
= dim A − dim A/P
= dim X − dim Y.
Proof of (e). This is I, Prop. 1.10.
Proof of (f ). By (e), it suffices to show this for affine schemes. Now by (b), dim X 0 =
tr. d. Frac(A) ⊗k k 0 = tr. d. Frac(A) + tr. d. k 0 = dim A + dim k 0 = dim X.
28
Problem 3.21. Let R be a discrete valuation ring containing its residue field k. Let
X = Spec R[t] be the affine line over Spec R. Show that statements (a), (d), (e) of
(Ex. 3.20) are false for X.
Proof. dim X = dim R[t] = 2, and so for (a) it suffices to find a maximal ideal of
height 1. Let I = (ut − 1) for (u) = m ⊂ R. Since any element in R can be written
αun for α a unit, we see that R[t]/I ∼
= R[1/u] = Frac(R), and so I is a height one
maximal ideal since R[t] is a UFD and by Prop. 1.12A, and so dim OI = 1.
For (d), then letting Y = V (I) ∼
= Spec(R[t]/I), we see that dim Y +codim(Y, X) =
1 + 0 6= 2 = dim X.
For (e), consider Spec R[t]u ; as above, dim Spec R[t]u = dim Frac(R)[t] = 1 6= 2 =
dim R[t].
Problem 3.22. Dimension of the Fibres of a Morphism. Let f : X → Y be a
dominant morphism of integral schemes of finite type over a field k.
(a) Let Y 0 be a closed irreducible subset of Y , whose generic point η 0 is contained
in f (X). Let Z be any irreducible component of f −1 (Y 0 ), such that η 0 ∈ f (Z),
and show that codim(Z, X) ≤ codim(Y 0 , Y ).
(b) Let e = dim X − dim Y be the relative dimension of X over Y . For any point
y ∈ f (X), show that every irreducible component of the fibre Xy has dimension
≥ e.
(c) Show that there is a dense open subset U ⊂ X, such that for any y ∈ f (U ),
dim Uy = e.
(d) Going back to our original morphism f : X → Y , for any integer h, let Eh
be the set of points x ∈ X such that, letting y = f (x), there is an irreducible
componenet Z of the fibre Xy , containing x, and having dim Z ≥ h. Show that
(1)Ee = X; (2) if h > e, then Eh is not dense in X; and (3)Eh is closed, for
all h.
(e) Prove the following theorem of Chevalley. For each integer h, let Ch be the set
of points y ∈ Y such that dim Xy = h. Then the subsets Ch are constructible,
and Ce contains an open dense subset of Y .
Proof of (a). By choosing an affine neighborhood Spec B ⊂ Y that contains η 0 , and
an affine neighborhood Spec A ⊂ X such that η 0 ∈ f (Spec A), dimension counts stay
the same by Problem 3.20(e), and so it suffices to show this in the affine case.
Let ϕ : B → A be the ring morphism associated to f . If Z = V (ξ) and Y 0 = V (η 0 ),
where Z, Y 0 irreducible imply ξ, η 0 are prime, it suffices to show ht ξ ≤ ht η 0 . So
let p0 ⊂ · · · pn ⊂ ξ be an ascending chain of primes to ξ in A; then, we see that
since contractions of primes are prime, we have a chain q0 ⊂ · · · ⊂ qn ⊂ ϕ−1 (ξ)
29
in B. Now V (ϕ−1 (ξ)) = f (V (ξ)) 3 η 0 , and so η 0 ⊇ ϕ−1 (ξ), i.e., and so our chain
q0 ⊂ · · · ⊂ qn ⊂ ϕ−1 (ξ) extends to the right to η 0 , and so ht ξ ≤ ht η 0 .
Proof of (b). Let Y 0 = {y}. Then, if Z is an irreducible component of Xy , then
codim(Z, X) ≤ codim(Y 0 , Y ) by (a). Then, dim X − dim Z ≤ dim Y − dim Y 0 by
Problem 3.20(d), and so e = dim X − dim Y ≤ dim Z − dim Y 0 ≤ dim Z.
Lemma 10. If f : Spec A → Spec B is dominant if and only if ker(ϕ : B → A) ⊂
N(B).
Proof of Lemma. V (ϕ−1 (I)) = f (V (I)), so V (ϕ−1 (0)) =Tf (X) = Spec B ⇔ f dominant. The former condition is equivalent to ϕ−1 (0) = p, where the intersection
runs over all primes in A, so ϕ−1 (0) = N(A).
Proof of (c). As in (a), reduce to the affine case. Then, A is a finitely generated
B-algebra, for both are finitely generated k-algebras, and so taking the basis for A
not in B gives a finite basis over B. Take t1 , . . . , te ∈ A that forms a transcendence
base of K(X) over K(Y ), and let X1 = Spec B[t1 , . . . , te ]. Then, X1 is isomorphic to
affine e-space over Y . We see the morphism g : X → X1 is generically finite, since
f −1 (η) = Spec A × k(B)(t1 , . . . , te ) = Spec A × K(A). Thus, by Problem 3.7, there
exists an open dense subset U ⊂ X1 such that g −1 (U ) → U is finite. We then see
that for this U , for any y ∈ h(U ), where h is the map that factors f = h ◦ g, we
have dim(X1 )y ≥ e by (b), but it must be at most e, and so dim(X1 )y = e. If we
show dim g −1 (U ) = e, then, we are done. But this is true since a chain of primes in
A maps down to a chain of primes in B[t1 , . . . , te ] by Problem 3.5(b), and moreover
these primes in B[t1 , . . . , te ] must be distinct by the fact that B[t1 , . . . , te ] → A is
injective on prime ideals by the Lemma.
Proof of (d). (1) This is clear by (b).
(2) By (b), Eh ⊂ X \ U where U is as constructed in (c), hence Eh is not dense in
X.
(3) We proceed by induction on X. If dim X = 0, then the claim is trivial by (1).
If h ≤ e, we are done by (1), and so suppose h > e. Let U ⊂ Y be the dense open
subset constructed in (c), then Eh does not meet the preimage of U . By replacing Y
with Y \ U , we are done by inductive hypothesis.
S
ProofSof (e). We see that Ch = f (Eh \ k>h Ek ), and so it suffices to show that
Eh \ k>h Ek is constructible by Problem 3.19. But
!c
[
[
Eh \
Ek = Eh ∩
Ek ,
k>h
k>h
30
which is constructible by the axioms in Problem 3.18.
Problem 3.23. If V, W are two varieties over an algebraically closed field k, and if
V × W is their product, as defined in (I, Ex. 3.15, 3.16), and if t is the functor of
(2.6), then t(V × W ) = t(V ) ×Spec k t(W ).
Proof. It suffices to show t(V ×W ) satisfies the universal property for t(V )×Spec k t(W ).
Since V × W is the product in the category of quasi-projective varieties by I, Problem
3.16(c), we have the commutative diagram
V ×W
Z
V
W
Applying t gives the commutative diagram
t(V × W )
t(Z)
t(V )
t(W )
Spec k
We claim we can complete the diagram with Spec k. But this is true since the maps
in the variety side all correspond to k-algebra homomorphisms locally by Prop. 3.5,
which correspond to morphisms of schemes over k by Problem 2.4, and by gluing.
References
[AM69]
M. F. Atiyah and I. G. Macdonald. Introduction to commutative algebra.
Reading, Mass.: Addison-Wesley Publishing Co., 1969.
[Eis95]
D. Eisenbud. Commutative algebra: With a view toward algebraic geometry.
Graduate Texts in Mathematics 150. New York: Springer-Verlag, 1995.
[Isa09]
I. M. Isaacs. Algebra: a graduate course. Graduate Studies in Mathematics
100. Reprint of the 1994 original. Providence, RI: American Mathematical
Society, 2009.
31
[Mat70]
H. Matsumura. Commutative algebra. W. A. Benjamin, Inc., New York,
1970.
[Mun00] J. R. Munkres. Topology. English. 2nd ed. Upper Saddle River, NJ: Prentice
Hall, 2000.
[Rei95]
M. Reid. Undergraduate commutative algebra. London Mathematical Society Student Texts 29. Cambridge: Cambridge University Press, 1995.
32