3. Algebraic Varieties

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An algebraic variety is a ringed space that is locally isomorphic to an affine algebraic

variety, just as a topological manifold is a ringed space that is locally isomorphic to

an open subset of Rn; both are required to satisfy a separation axiom.

Algebraic prevarieties. As motivation, recall the following definitions.

Definition 3.1. (a) A topological manifold is a ringed space (V,OV ) such that V

is Hausdorff and every point of V has an open neighbourhood U for which (U,OV |U)

is isomorphic to the ringed space of continuous functions on an open subset of Rn (cf.

(2.2a)).

(b) A differentiable manifold is a ringed space such that V is Hausdorff and every

point of V has an open neighbourhood U for which (U,OV |U) is isomorphic to a

ringed space as in (2.2b).

(c) A complex manifold is a ringed space such that V is Hausdorff and every point

of V has an open neighbourhood U for which (U,OV |U) is isomorphic to a ringed

space as in (2.2c).

The above definitions are easily seen to be equivalent to the more classical definitions

in terms of charts and atlases. Often one imposes additional conditions on V ,

for example, that it is second countable or connected.

Definition 3.2. An algebraic prevariety is a ringed space (V,OV ) such that V

is quasi-compact and every point of V has an open neighbourhood U such that

(V,OV |U) is an affine algebraic variety.

Equivalently, a ringed space (V,OV ) is an algebraic prevariety if there is a finite

open covering V = Vi such that (Vi,OV |Vi) is an affine algebraic variety for all i.

An algebraic variety will be defined to be an algebraic prevariety satisfying a certain

separation condition.

An open subset U of an algebraic prevariety V such that (U, OV |U) is an affine

algebraic variety is called an open affine (subvariety) in V .

Let (V,OV ) be an algebraic variety, and let U be an open subset of V . The functions

f : U k lying in Γ(U,OV ) are called regular. Note that if (Ui) is an open covering of

V by affine varieties, then f : U k is regular if and only if f|Ui U is regular for all

i (this is just a special case of condition (c) to be a sheaf, p12). Thus understanding

the regular functions on open subsets of V amounts to understanding the regular

functions on the open affine subvarieties and how these subvarieties fit together to

form V .

Example 3.3. Any open subset of an affine variety together with its induced

ringed structure is an algebraic prevariety (in fact variety). For example, A2 \{(0, 0)}

is an algebraic variety.

Example 3.4. (Projective space). Let

Pn = kn+1 \ {(0, . . . , 0)}/

where (a0, . . . , an) (b0, . . . , bn) if there is a c k× such that (a0, . . . , an) =

(cb0, . . . , cbn). Thus the equivalence classes are the lines through the origin in kn+1.

Algebraic Geometry: 3. Algebraic Varieties 45

Write (a0 : . . . : an) for the equivalence class containing (a0, . . . , an). For each i, let

Ui = {(a0 : . . . : ai : . . . : an) Pn | ai _= 0}.

Then Pn = Ui, and the map ui

(a1, . . . , an) _ (a0 : . . . : ai1 : 1 : ai+1, . . . : an) : kn Ui

is a bijection. We use this map to transfer the Zariski topology on kn to Ui, and we

endow Pn with the topology such that U Pn is open if and only if U Ui is open

in Ui for all i. Define a function f : U k on an open subset U of Pn to be regular

if f ui is a regular function on kn for all i. These definitions endow Pn with the

structure of a ringed space, and each map ui is an isomorphism of ringed spaces (An,

OAn) (Ui, OV |Ui). Thus Pn is an algebraic prevariety. Later (see Section 5), we

shall study Pn in detail.

Regular maps. In each of the examples (3.1a,b,c), a morphism of manifolds (continuous

map, differentiable map, analytic map respectively) is just a morphism of

ringed spaces. This motivates the following definition.

Let (V,OV) and (W,OW) be algebraic prevarieties. A map ϕ: V W is said to

be regular if it is a morphism of ringed spaces. A composite of regular maps is again

regular (this is a general fact about morphisms of ringed spaces).

Note that we have four categories:

(Affine varieties) (Alg. prevarieties) (ringed spaces).

Each subcategory is full (i.e., the morphisms Mor(V,W) are the same in the four

categories).

Proposition 3.5. Let (V,OV ) and (W,OW) be prevarieties, and let ϕ: V W

be a continuous map (of topological spaces). Let W = Wi be a covering of W by

open affines, and let ϕ1(Wj) = Vji be a covering of ϕ1(Wj) by open affines. Then

ϕ is regular if and only if its restrictions

ϕ|Vji : Vji Wj

are regular for all i, j.

Proof. We assume that ϕ satisfies this condition, and prove that it is regular. Let

f be a regular function on an open subset U of W. Then f|U Wj is regular for each

Wj (because the regular functions form a sheaf), and so f ϕ|ϕ1(U) Vji is regular

for each j, i (this is our assumption). It follows that f ϕ is regular on ϕ1(U) (sheaf

condition). Thus ϕ is regular. The converse is equally easy.

Aside 3.6. A differentiable manifold of dimension d is locally isomorphic to an

open subset of Rd. In particular, all manifolds of the same dimension are locally

isomorphic. This is not true for algebraic varieties, for two reasons:

(a) We are not assuming our varieties are nonsingular (see the next section).

(b) The inverse function theorem fails in our context. If P is a nonsingular point on

variety of dimension d, we shall see (in the next section) that there is a neighbourhood

U of P and a regular map ϕ: U Ad such that map (dϕ)P : TP Tϕ(P) on the

tangent spaces is an isomorphism. If the inverse function theorem were true in our

context, it would tell us that an open neighbourhood of P is isomorphic to an open

neighbourhood of ϕ(P).

46 Algebraic Geometry: 3. Algebraic Varieties

Algebraic varieties. In the study of topological manifolds, the Hausdorff condition

eliminates such bizarre possibilities as the line with the origin doubled, where a

sequence tending to the origin has two limits.

It is not immediately obvious how to impose a separation axiom on our algebraic

varieties, because even affine algebraic varieties are not Hausdorff. The key is to

restate the Hausdorff condition. Intuitively, the significance of this condition is that

it implies that a sequence in the space can have at most one limit. Thus a continuous

map into the space should be determined by its values on a dense subset, i.e., if ϕ

and ψ are continuous maps Z U that agree on a dense subset of Z then they

should agree on the whole of Z. Equivalently, the set where two continuous maps

ϕ, ψ: Z U agree should be closed. Surprisingly, affine varieties have this property,

provided ϕ and ψ are required to be regular maps.

Lemma 3.7. Let ϕ and ψ be regular maps of affine algebraic varieties Z V .

The subset of Z on which ϕ and ψ agree is closed.

Proof. There are regular functions xi on V such that P _ (x1(P), . . . , xn(P))

identifies V with a closed subset of An (take the xi to be any set of generators for

k[V ] as a k-algebra). Now xi ϕ and xi ϕ are regular functions on Z, and the set

where ϕ and ψ agree is

_n

i=1 V (xi ϕ xi ψ), which is closed.

Definition 3.8. An algebraic prevariety V is said to be separated, or to be an

algebraic variety, if it satisfies the following additional condition:

separation axiom: for every pair of regular maps ϕ, ψ: Z V with Z an algebraic

prevariety, the set {z Z | ϕ(z) = ψ(z)} is closed in Z.

The terminology not completely standardized:o ften one requires a variety to be

irreducible, and sometimes one calls a prevariety a variety.

Remark 3.9. In order to check that a prevariety V is separated, it suffices to

show that for every pair of regular maps ϕ, ψ: Z V with Z an affine algebraic

variety {z Z | ϕ(z) = ψ(z)} is closed in Z. To prove this remark, cover Z with

open affines. Thus (3.7) shows that affine varieties are separated.

Example 3.10. (The affine line with the origin doubled.) Let V1 and V2 be copies

of A1. Let V = V1 # V2 (disjoint union), and give it the obvious topology. Define an

equivalence relation on V by

x (in V1) y (in V2) ⇐⇒ x = y and x _= 0.

Let V be the quotient space V = V / with the quotient topology (a set is open if

and only if its inverse image in V is open). Then V1 and V2 are open subspaces of

V , V = V1 V2, and V1 V2 = A1 {0}. Define a function on an open subset to be

regular if its restriction to each Vi is regular. This makes V into a prevariety, but not

a variety:i t fails the separation axiom because the two maps

A1 = V1 8 V

, A1 = V2 8 V

agree exactly on A1 {0}, which is not closed in A1.

Algebraic Geometry: 3. Algebraic Varieties 47

Subvarieties. Let (V,OV ) be a prevariety. Then V is a finite union of open affines,

and in each open affine the open affines (in fact the basic open subsets) form a basis

for the topology. From this it follows the open affines form a basis for the topology on

V , i.e., every open subset U of V is a union of open affines (of V ). It follows that, for

any open subset U of V , (U,OV |U) is a prevariety. Obviously the inclusion U 8 V

is regular. A regular map ϕ: W V is an open immersion if ϕ(W) is open in V and

ϕ defines an isomorphism W ϕ(W) (of prevarieties).

Any closed subset Z in V has a canonical structure of an algebraic prevariety:

endow it with the induced topology, and say that a function f on an open subset of Z

is regular if each point P in the open subset has an open neighbourhood U in V such

that f extends to a regular function on U. To show that Z, with this ringed space

structure is a prevariety, check that for every open affine U V , the ringed space

(U Z,OZ|U Z) is isomorphic to U Z with its ringed space structure acquired as

a closed subset of U (see p45). A regular map ϕ: W V is a closed immersion if

ϕ(W) is closed in V and ϕ defines an isomorphism W ϕ(W) (of prevarieties).

A subset W of a topological space V is said to be locally closed if every point P in

W has an open neighbourhood U in V such that W U is closed in U; equivalently,

W is the intersection of an open and a closed subset of V . A locally closed subset

W of a prevariety V acquires a natural structure as a prevariety:wr ite it as the

intersection W = U Z of an open and a closed subset; Z is a prevariety, and W

(being open in Z) therefore acquires the structure of a prevariety. This structure

on W has the following characterization:the inclusion map W 8 V is regular, and

a map ϕ: V _ W with V _ a prevariety is regular if and only if it is regular when

regarded as a map into V . With this structure, W is called a sub(pre)variety of V .

A morphism ϕ: V _ V is called an immersion if it induces an isomorphism of V _

onto a subvariety of V . Every immersion is the composite of an open immersion with

a closed immersion (in both orders).

A subprevariety of a variety is automatically separated.

Proposition 3.11. A prevariety V is separated if and only if it has the following

property: if two regular maps ϕ, ψ: Z V agree on a dense subset of Z, then they

agree on the whole of Z.

Proof. If V is separated, then the set where ϕ and ψ agree is closed, and so must

be the whole of Z.

Conversely, consider a pair of maps ϕ, ψ: Z V , and let S be the subset of Z

on which they agree. We assume V has the property in the statement of the lemma,

and show that S is closed. Let S be the closure of S in Z. According to the above

discussion, S has the structure of a closed prevariety of Z, and the maps ϕ| S and

ψ| S are regular. Because they agree on a dense subset of S they agree on the whole

of S, and so S = S is closed.

Prevarieties obtained by patching. Let V = Vi (finite union), and suppose that

each Vi has the structure of an algebraic prevariety satisfying the following condition:

for all i, j, ViVj is open in both Vi and Vj and the structures of an algebraic prevariety

induced on it by Vi and Vj are equal. Then we can define the structure of a ringed

space on V as follows: U V is open if and only if U Vi is open for all i, and

48 Algebraic Geometry: 3. Algebraic Varieties

f : U k is regular if and only if f|U Vi is regular for all i. It is straightforward to

check that this does make V into a ringed space (V,OV ).

Proposition 3.12. The ringed space (V,OV ) is a prevariety, and the inclusions

Vi 8 V are regular maps.

Proof. One only has to check that the ringed space structure on each Vi induced

by that of V is the original one.

Products of varieties. Let V and W be objects in a category C. A triple

(V × W, p: V ×W V , q : V ×W W)

is said to be the product of V and W if, for all objects Z in C, themap ϕ _ (pϕ, qϕ)

is a bijection

Hom(Z, V ×W) Hom(Z, V ) × Hom(Z,W),

i.e., if every pair of morphisms Z V , Z W factors uniquely through V ×W :

Z

_

_

_

❅❘

V V ×W

.....

!

W.

Clearly, the product, if it exists, is uniquely determined up to a unique isomorphism11.

For example, the product of two sets (in the category of sets) is the usual cartesion

product of the sets, and the product of two topological spaces (in the category of

topological spaces) is the cartesian product of the spaces (as sets) with the usual

product topology.

We shall show that products exist in the category of algebraic varieties. Suppose,

for the moment, that V × W exists. It follows from (2.17b) that for any prevariety

Z, Mor(A0, Z) is the underlying set of Z, i.e., for any z Z, the map A0 Z with

image z is regular, and these are all the regular maps. Thus, from the definition of

products we have

(underlying set of V ×W) = Mor(A0, V ×W)

= Mor(A0, V ) × Mor(A0,W)

= (underlying set of V ) × (underlying set of W).

Thus our problem can be restated as follows:g iven two prevarieties V and W, define

on the set V ×W the structure of a prevariety such that the projection maps p, q : V ×

W V,W are regular, and such that a map ϕ: T V × W of sets (with T an

algebraic prevariety) is regular if and only if its components p ϕ, q ϕ are regular.

Clearly, there can be at most one such structure on the set V × W (because the

identity map will identify any two structures having these properties).

11If (P, p_ : P V, q_ : P W) also has this property, then there exists a unique morphism

γ : P V ×W such that p γ = p_ and q γ = q_ (universal property of V ×W), and there exists

a unique morphism γ_ : V × W P such that p_ γ_ = p and q_ γ_ = q (universal property of

P). The composite γ γ_ is the unique morphism V × W V × W such that p γ γ_ = p and

q γ γ_ = q. But we already know one such morphism, namely, the identity morphism, and so

γ γ_ = id. Similarly γ_ γ = id, and so γ and γ_ are inverse isomorphisms.

Algebraic Geometry: 3. Algebraic Varieties 49

Before we can define products of algebraic varieties, we need to review tensor

products.

Review of tensor products. Let A and B be k-algebras. A k-algebra C together

with homomorphisms i : A C and j : B C is called the tensor product of A and

B if it has the following universal mapping property:fo r every pair of homomorphisms

(of k-algebras) α: A R and β : B R, there is a unique homomorphism γ : C R

such that γ i = α and γ j = β:

A

i C j

B

α ❅❘ _

_

_

β

R.

!

.....

γ

Clearly, if the tensor product exists, it is uniquely determined up to a unique isomorphism

(same argument as in the above footnote). We write it A k B.

Construction. Let C be the k-vector space with basis A × B. Thus the elements of

C are finite sums

_

ci(ai, bi) with ci k, ai A, bi B. Let D be the subspace of

C generated by the following elements,

(a + a_, b) (a, b) (a_, b), a, a_ A, b B,

(a, b + b

_

) (a, b) (a, b

_

), a A, b, b

_ B,

(ca, b) c(a, b), a A, b B, c k,

(a, cb) c(a, b), a A, b B, c k,

and define C = C/D. Write ab for the class of (a, b) in C we have imposed the

fewest conditions forcing (a, b) _ a b to be k-bilinear. Every element of C can be

written as a finite sum,

_

ai bi, ai A, bi B, and the map

A × B C, (a, b) _ a b

is k-bilinear. By definition, C is a k-vector space, and there is a product structure on

C such that (a b)(a_ b_) = aa_ bb_ for this one has to check that the map

C

× C

C, ((a, b), (a

_

, b

_

)) _ aa

_ bb

_

factors through C × C. It becomes a k-algebra by means of the homomorphism

c _ c(1 1) = c 1 = 1 c. The maps

a _ a 1: A C and b _ 1 b : B C

are homomorphisms, and it is routine to check that they make C into the tensor

product of A and B in the above sense.

Example 3.13. The algebra B, together with the given map k B and the

identity map B B, has the universal property characterizing k k B. In terms of

the constructive definition of tensor products, the map cb _ cb : kk B B is an

isomorphism.

50 Algebraic Geometry: 3. Algebraic Varieties

Example 3.14. (a) The ring k[X1, . . . ,Xm,Xm+1, . . . ,Xm+n], together with the

maps

k[X1, . . . ,Xm] obviousinclusion k[X1, . . . ,Xm+n] obvious inclusion k[Xm+1, . . . ,Xm+n]

is the tensor product of k[X1, . . . ,Xm] and k[Xm+1, . . . ,Xm+n]. To verify this we

only have to check that, for every k-algebra R, the map

Homk-alg(k[X1, . . . ,Xm+n], R) Homk-alg(k[X1, . . . ], R) × Homk-alg(k[Xm+1, . . . ], R)

induced by the inclusions is a bijection. But this map can be identified with the

bijection

Rm+n Rm × Rn.

In terms of the constructive definition of tensor products, the map

f g _ fg: k[X1, . . . ,Xm] k k[Xm+1, . . . ,Xm+n] k[X1, . . . ,Xm+n]

is an isomorphism.

(b) Let a and b be ideals in k[X1, . . . ,Xm] and k[Xm+1, . . . ,Xm+n] respectively,

and let (a, b) be the ideal in k[X1, . . . ,Xm+n] generated by the elements of a and b.

Then there is an isomorphism

f g _ fg:

k[X1, . . . ,Xm]

a

k

k[Xm+1, . . . ,Xm+n]

b

k[X1, . . . ,Xm+n]

(a, b)

.

Again this comes down to checking that the natural map from

Homk-alg(k[X1, . . . ,Xm+n]/(a, b), R) to

Homk-alg(k[X1, . . . ,Xm]/a, R) × Homk-alg(k[Xm+1, . . . ,Xm+n]/b, R)

is a bijection. But the three sets are respectively

V (a, b) = zero-set of (a, b) in Rm+n,

V (a) = zero-set of a in Rm,

V (b) = zero-set of b in Rn,

and so this is obvious.

Remark 3.15. (a) If (bα) is a family of generators (resp. basis) for B as a k-vector

space, then (1bα) is a family of generators (resp. basis) for AkB as an A-module.

(b) Let k 8 Ω be fields. Then

Ω k k[X1, . . . ,Xn]

=

Ω[1 X1, . . . , 1 Xn]

=

Ω[X1, . . . ,Xn].

If A = k[X1, . . . ,Xn]/(g1, . . . , gm), then

Ω k A

=

Ω[X1, . . . ,Xn]/(g1, . . . , gm).

For more details on tensor products, see Atiyah and MacDonald 1969, Chapter 2

(but note that the description there (p31) of the homomorphism A D making the

tensor product into an A-algebra is incorrect the map is a _ f(a) 1 = 1 g(a).

Algebraic Geometry: 3. Algebraic Varieties 51

Products of affine varieties. The tensor product of two k-algebras A and B has the

universal property to be a product, but with the arrows reversed. Because of the

category anti-equivalence (2.14), this will show that Specm(A k B) is the product

of SpecmA and SpecmB in the category of affine algebraic varieties once we have

shown that A k B is an affine k-algebra.

Proposition 3.16. Let A and B be finitely generated k-algebras; if A and B are

reduced, then so also is A k B; if A and B are integral domains, then so also is

A k B.

Proof. Assume A and B to be reduced, and let α AkB. Then α =

_n

i=1 aibi,

some ai A, bi B. If one of the bis is a linear combination of the remaining bs,

say, bn =

_n1

i=1 cibi, ci k, then, using the bilinearity of , we find that

α =

_n1

i=1

ai bi +

_n1

i=1

cian bi =

_n1

i=1

(ai + cian) bi.

Thus we can suppose that in the original expression of α, the bis are linearly independent

over k.

Now suppose that α is nilpotent, and let m be a maximal ideal in A. From a _

a: A A/m = k we obtain homomorphisms

a b _ a b _ ab : A k B k k B

B

The image

_

aibi of α under this homomorphism is a nilpotent element of B, and

hence is zero (because B is reduced). As the bis are linearly independent over k, this

means that the ai are all zero. Thus, for all i, ai lies in every maximal ideal m of A,

and so is zero (by 2.12). Hence α = 0. This shows that A k B is reduced.

Assume A and B to be integral domains, and let α, α_ AB be such that αα_ = 0.

As before, we can write α =

_

ai bi and α_ =

_

a_

i

b_

i with the sets {b1, b2, . . . }

and {b_

1, b_

2, . . . } each linearly independent over k. For each maximal ideal m of A, we

know (

_

aibi)(

_

a_

ib_

i) = 0 in B, and so either (

_

aibi) = 0 or (

_

a_

ib_

i) = 0. Thus

either all the ai m or all the a_

i

m. This shows that

specm(A) = V (a1, . . . , am) V (a

_

1, . . . , a

_

n).

Since specm(A) is irreducible (see 1.15), we must have specm(A) = V (a1, . . . , am) or

V (a_

1, . . . , a_

n). In the first case α = 0, and in the second α_ = 0.

Example 3.17. We give some examples to illustrate that k must be taken to be

algebraically closed in the proposition.

(a) Suppose k is nonperfect of characteristic p. To say that k is not perfect means

that there is an element α in an algebraic closure of k such that α / k but αp k.

Let k_ = k[α], αp = a k, α / k. Then (α 1 1 α) _= 0 in k_ k k_ (in fact, the

elements αi αj, 0 i, j p 1, form a basis for k_ k k_ as a k-vector space), but

(α 1 1 α)p = (a 1 1 a) = (1 a 1 a) = 0.

Thus k_ k k_ is not reduced, even though k_ is a field.

(b) Let K be a finite separable extension of k and let Ω be a big field containing k

(for example an algebraic closure of k). Write K = k[α] = k[X]/(f(X)), and assume

52 Algebraic Geometry: 3. Algebraic Varieties

f(X) splits in Ω[X], say, f(X) =

_

X αi. Because K/k is separable, the αi are

distinct, and so

K k Ω

=

Ω[X]/(f(X))

=

 

Ω[X]/(X αi),

and so it is not an integral domain. (The second isomorphism follows from the Chinese

remainder theorem.)

Having (3.16), we can make the following definition:l et V and W be affine varieties,

and let Γ(V,OV) = A and Γ(W,OW) = B; then V × W = Specm(A k B) with

the projection maps p: V × W V and q : V × W W defined by the maps

a _ a 1: A A k B and b _ 1 b : B A k B.

Proposition 3.18. Let V and W be affine varieties; the projection maps p: V ×

W V , q : V × W W are regular, and a map ϕ: U V × W is regular if and

only if p ϕ and q ϕ are regular. Therefore (V ×W, p, q) is the product of V and W

in the category of algebraic prevarieties. If V and W are irreducible, then so also is

V ×W.

Proof. The projection maps are regular because they correspond to the k-algebra

homomorphisms k[V ] k[V ]k k[W] and k[W] k[V ]k k[W]. Let ϕ: U V ×W

be a map (of sets) such that pϕ and qϕ are regular. If U is affine, then ϕ corresponds

to the map k[V ] k[W] k[U] induced by

f _ f (p ϕ) : k[V ] k[U] and f _ f (q ϕ) : k[W] k[U],

and so is regular. This shows that, for a general U, the restriction of ϕ to every open

affine of U is regular, and this implies that ϕ is regular (see 3.5).

The final statement follows from the second statement in 3.16.

Example 3.19. (a) It follows from (3.14a) that

Am p

Am+n q An,

where

p(a1, . . . , am+n) = (a1, . . . , am),

q(a1, . . . , am+n) = (am+1, . . . , am+n),

is the product of Am and An.

(b) It follows from (3.14b) that

V (a)

p

V (a, b)

q

V (b)

is the product of V (a) and V (b).

Warning! The topology on V ×W is not the product topology; for example, the

topology on A2 = A1 × A1 is not the product topology.

Products in general. Now let V and W be two algebraic prevarieties V and W.

We define their product as follows:A s a set, we take V × W. Now write V and W

as unions of open affines, V = Vi, W = Wj. Then V × W = Vi × Wj, and we

give V ×W the topology for which U V ×W is open if and only if U (Vi ×Wj)

is open for all i and j. Finally, we define a ringed space structure by saying that a

Algebraic Geometry: 3. Algebraic Varieties 53

function f : U k on an open subset U is regular if its restriction to U (Ui × Vj)

is regular for all i and j.

Proposition 3.20. With the above structure, V ×W is a prevariety, the projection

maps

p: V ×W V , q : V ×W W

are regular, and a map ϕ: U V × W is regular if and only if p ϕ and q ϕ

are regular. Therefore (V × W, p, q) is the product of V and W in the category of

prevarieties.

Proof. Straightforward.

Proposition 3.21. If V and W are separated, then so also is V ×W.

Proof. Straightforward.

Example 3.22. An algebraic group is a variety G together with regular maps

mult : G × G G, inverse: G G,

and an element e G that make G into a group in the usual sense. For example, SLn

and GLn are algebraic groups, and any finite group can be regarded as an algebraic

group. Connected affine algebraic groups are called linear algebraic groups because

they can all be realized as closed subgroups of GLn for some n, and connected algebraic

groups that can be realized as closed algebraic subvarieties of a projective space are

called abelian because they are related to the integrals studied by Abel.

Coarse Classification:e very algebraic group contains a sequence of normal subgroups

G G0 G1 {e} with G/G0 a finite group, G0/G1 an abelian variety, and

G1 a linear algebraic group.

The separation axiom. Now that we have the notion of the product of varieties,

we can restate the separation axiom in terms of the diagonal.

By way of motivation, consider a topological space V and the diagonal Δ V ×V ,

Δ df = {(x, x) | x V }.

If Δ is closed (for the product topology), then every pair of points (x, y) /

Δ has a

neighbourhood U × U_ such that U × U_ Δ = . In other words, if x and y are

distinct points in V then there are neighbourhoods U and U_ of x and y respectively

such that U U_ = . Thus V is Hausdorff. Conversely, if V is Hausdorff, the reverse

argument shows that Δ is closed.

For a variety V , we let Δ = ΔV (the diagonal) be the subset {(v, v) | v V } of

V × V .

Proposition 3.23. An algebraic prevariety V is separated if and only if ΔV is

closed.

Proof. Assume Δ to be closed, and let ϕ and ψ be regular maps Z V. The

map

(ϕ, ψ) : Z V × V , z _ (ϕ(z), ψ(z))

54 Algebraic Geometry: 3. Algebraic Varieties

is regular, because its composites with the projections to V are ϕ and ψ. In particular,

it is continuous, and so (ϕ, ψ)1(Δ) is closed. But this is precisely the subset on which

ϕ and ψ agree.

Conversely, suppose V is separated. By definition, this means that for any prevariety

Z and regular maps ϕ, ψ: Z V , the set on which ϕ and ψ agree is closed in

Z. Apply this with ϕ and ψ the two projection maps V ×V V , and note that the

set on which they agree is Δ.

Corollary 3.24. For any prevariety V , the diagonal is a locally closed subset of

V × V .

Proof. Let P V , and let U be an open affine neighbourhood of P. Then U ×U

is a neighbourhood of (P, P) in V × V , and ΔV (U × U) = ΔU, which is closed in

U × U because U is separated.

Thus ΔV is always a subvariety of V × V , and it is closed if and only if V is

separated.

The graph Γϕ of a regular map ϕ: V W is defined to be

{(v,ϕ(v)) V ×W | v V }.

At this point, the reader should draw a picture, suggested by calculus.

Corollary 3.25. For any morphism ϕ: V W of prevarieties, the graph Γϕ of

ϕ is locally closed in V ×W, and it is closed if W is separated. The map v _ (v,ϕ(v))

is an isomorphism of V onto Γϕ.

Proof. The first statement follows from the preceding corollary because the graph

is the inverse image of the diagonal of W ×W under the regular map

(v,w) _ (ϕ(v), w) : V ×W W × W.

The second follows from the fact that the regular map Γϕ 8 V × W

p

V is an

inverse to v _ (v,ϕ(v)) : V Γϕ.

Theorem 3.26. The following three conditions on a prevariety are equivalent:

(a) V is separated;

(b) for every pair of open affines U and U_ in V , U U_ is an open affine, and

Γ(U U_,OV ) is generated by the functions P _ f(P)g(P), f Γ(U,OV ),

g Γ(U_,OV ), i.e., the map k[U] k k[U_] k[U U_] is surjective;

(c) the condition in (b) holds for the sets in some open affine covering of V .

Proof. Let Ui and Uj be open affines in V . We shall prove:

(i) Δ closed Ui Uj affine.

(ii) If Ui Uj is affine, then

(Ui × Uj) Δ is closed ⇐⇒ the map k[Ui] k k[Uj ] k[Ui Uj ] is surjective.

If {Ui × Uj}(i,j)I×J is an open covering of V × V , Δ is closed in V × V ⇐⇒

Δ (Ui × Uj) is closed in Ui × Uj for each pair (i, j). Thus these statements show

that (a)(b) and (c)(a). Since the implication (b)(c) is trivial, this shows that

(i) and (ii) imply the theorem.

Algebraic Geometry: 3. Algebraic Varieties 55

Proof of (i):T he graph of the inclusion ι : Ui Uj 8 V is Γι = (Ui × Uj) Δ

(Ui Uj)×V . If Δ is closed, (Ui ×Uj) Δ is a closed subvariety of an affine variety,

and hence is affine (see p45). Since Ui Uj Γι, it also is affine.

Proof of (ii):No w assume that Ui Uj is affine. Then (Ui × Uj) ΔV is closed in

Ui×Uj ⇐⇒ v _ (v, v) : UiUj Ui×Uj is a closed immersion ⇐⇒ the morphism

k[Ui ×Uj ] k[Ui Uj ] is surjective (see 2.21). Since k[Ui ×Uj] = k[Ui]k k[Uj ], this

completes the proof of (ii).

Example 3.27. (a) Let V = P1, and let U0 and U1 be the standard open subsets

(Ui = A1). Then U0 U1 = A1 {0}, and the maps on rings corresponding to the

inclusions Ui 8 U0 U1 are k[X] k[X,X1], X _ X, and k[X] k[X,X1],

X _ X1. Thus the sets U0 and U1 satisfy the condition in (b).

(b) Let V be A1 with the origin doubled (see 3.10), and let U and U_ be the upper

and lower copies of A1 in V. Then U U_ is affine, but k[U] k[U_] k[U U_] is

not surjective. In fact the map is

k[X] k[Y ] = k[X, Y ] k[X,X1], X_ X, Y _ X.

(c) Let V be A2 with the origin doubled, and let U and U_ be the upper and lower

copies of A2 in V . Then U U_ is not affine (see 2.20).

Dimension. Let V be an irreducible algebraic variety. Then every open subset of V

is dense, and is irreducible. If U U_ are open affines in V , then we have

k[U] k[U

_

] k(U).

Therefore k(U) is also the field of fractions of k[U_]. This remark shows that we can

attach to V a field k(V ), called the field of rational functions on V , such that for

every open affine U in V , k(V ) is the field of fractions of k[U]. The dimension of V

is defined to be the transcendence degree of k(V ) over k. Note the dim(V) = dim(U)

for any open subset U of V . In particular, dim(V) = dim(U) for U an open affine in

V . It follows that some of the results in §1 carry over for example, if Z is a proper

closed subvariety of V , then dim(Z) < dim(V ).

Proposition 3.28. Let V and W be irreducible varieties. Then

dim(V ×W) = dim(V ) + dim(W).

Proof. We can assume V and W to be affine, and write k[V ] = k[x1, . . . , xm] and

k[W] = k[y1, . . . , yn] where {x1, . . . , xd} and {y1, . . . , ye} are maximal algebraically

independent sets of elements of k[V ] and k[W]. Thus d = dim(V ) and e = dim(W).

Then12

k[V ×W] = k[V ] k k[W] k[x1, . . . , xd] k k[y1, . . . , ye] k[x1, . . . , xd, y1, . . . , ye].

Therefore {x1 1, . . . , xd 1, 1y1, . . . , 1ye} will be algebraically independent in

k[V ]k k[W]. Obviously k[V ×W] is generated as a k-algebra by the elements xi1,

1 yj, 1 i m, 1 j n, and all of them are algebraic over

k[x1, . . . , xd] k k[y1, . . . , ye].

12In general, it is not true that if M_ and N_ are R-submodules of M and N, then M_ R N_ is

an R-submodule of M R N. However, this is true if R is a field, because then M_ and N_ will be

direct summands of M and N, and tensor products preserve direct summands.

56 Algebraic Geometry: 3. Algebraic Varieties

Thus the transcendence degree of k(V ×W) is d + e.

We extend the definition to an arbitrary variety V as follows. A variety is a

finite union of Noetherian topological spaces, and so is Noetherian. Consequently

(see 1.17), V is a finite union V = Vi of its irreducible components, and we define

dim(V ) = maxdim(Vi).

An algebraic variety as a functor of affine k-algebras. Let A be an affine kalgebra,

and let V be an algebraic variety. We define a point of V with coordinates

in A to be a regular map Specm(A) V . For example, if V = V (a) kn, then

V (A) = {(a1, . . . , an) An | f(a1, . . . , an) = 0 all f a},

which is what you expect. In particular V (k) = V (as a set), i.e., V (as a set) can be

identified with the set of points of V with coordinates in k. Note that (V ×W)(A) =

V (A) ×W(A).

Theorem 3.29. A regular map ϕ: V W of algebraic varieties defines a family

of maps of sets, ϕ(A) : V (A) W(A), one for each affine k-algebra A, such that for

every homomorphism α: A B of k-algebras,

A V(A)

ϕ(A)

>W(A)

(*)

B

α

V (B)

V (α)

ϕ(B)

> V (B)

W(α)

commutes. Every family of maps with this property arises from a unique morphism

of algebraic varieties.

The proof is trivial, once one has made the correct definitions, which we do in the

next subsection.

Categories and functors. A category C consists of

(a) a class of objects ob(C);

(b) for each pair (A,B) of objects, a set Mor(A,B), whose elements are called

morphisms from A to B, and are written α: A B;

(c) for each triple of objects (A,B,C) a map (called composition)

(α, β) _ β α:Mo r(A,B) × Mor(B,C) Mor(A,C).

Composition is required to be associative, i.e., (γ β) α = γ (β α), and for each

object A there is required to be an element idA Mor(A,A) such that idA α = α,

β idA = β, for all (appropriate) α and β. The sets Mor(A,B) are required to be

disjoint (so that a morphism α determines its source and target).

Example 3.30. (a) There is a category of sets, Sets, whose objects are the sets

and whose morphisms are the usual maps of sets.

(b) There is a category Affk of affine k-algebras, whose objects are the affine kalgebras

and whose morphisms are the homomorphisms of k-algebras.

(c) There is a category Vark of algebraic varieties over k, whose objects are the

algbraic varieties over k and whose morphisms are the regular maps.

Algebraic Geometry: 3. Algebraic Varieties 57

The objects in a category need not be sets with structure, and the morphisms need

not be maps.

Exercise 3.31. List twenty more examples of categories.

Let C and D be categories. A covariant functor F from C to D consists of

(a) a map A _ F(A), sending each object of C to an object of D, and,

(b) for each pair of objects A,B of C, a map

α _ F(α):M or(A,B) Mor(F(A), F(B))

such that F(idA) = idF(A) and F(β α) = F(β) F(α).

A contravariant functor is defined similarly, except that the map on morphisms is

α _ F(α):M or(A,B) Mor(F(B), F(A))

A functor F : C D is fully faithful if, for all objects A and B of C, the map

Mor(A,B) Mor(F(A), F(B))

is a bijection. Then F defines an equivalence of C with the full subcategory of D

whose objects are isomorphic to F(A) for some object A of C (see p42). For example,

the functor A _ SpecmA is fully faithful contravariant functor Affk Vark, and

defines an equivalence of the first category with the subcategory of the second whose

objects are the affine algebraic varieties.

Example 3.32. (a) For any object V of a category C, we have a contravariant

functor

hV : C ____________Sets,

which sends an object A to the set Mor(A, V ) and sends a morphism α: A B to

ϕ _ ϕ α: hV (B) hV (A),

i.e., hV () =Mor(, V ) and hV (α) = α.

(b) We have a contravariant functor

V _ Γ(V,OV ) : Vark Affk.

Let F and G be two functors C D. A morphism α: F G is a collection of

morphisms α(A) : F(A) G(A), one for each object A of C, such that, for every

morphism u: A B in C, the following diagram commutes:

A F(A)

α(A)

>G(A)

(**)

B

u

F(B)

F(u)

α(B)

>G(B)

G(u)

.

Example 3.33. Let α: V W be a morphism in C. The collection of maps

hα(A) : hV (A) hW(A), ϕ_ α ϕ

is a morphism of functors.

With this notion of morphism, the functors C D form a category Fun(C,D)

(we ignore the problem that Mor(F,G) may not be a set only a class).

58 Algebraic Geometry: 3. Algebraic Varieties

Proposition 3.34 (Yoneda Lemma). The functor

V _ hV : C Fun(C, Sets)

is fully faithful.

Proof. Let A,B be objects of C. We construct an inverse to

α _ hα:Mo r(A,B) Mor(hA, hB).

For a morphism of functors γ : hA hB, define β(γ) = γ(idA)it is morphism

A B. Then

β(hα) df = hα(idA) df = α idA = α,

and

hβ(γ)(α) df = β(γ) α df = γ(idA) α = γ(α)

because of the commutativity of (**):

A hA(A)

γ

> hB(A)

(***)

B

α

hB(B)

∗◦α

γ

> hB(B)

∗◦α

Thus α hα and γ _ β(γ) are inverse maps.

Algebraic varieties as functors (continued). The Yoneda lemma shows that the

functor V _ hV embeds the category of affine algebraic varieties as a full subcategory

of the category of covariant functors Affk Sets, and it is not difficult to deduce

that it embeds the category of all algebraic varieties in to the category of such functors

(use 3.12 for example). This proves (3.29).

It is not unusual for a variety to be most naturally defined in terms of its points

functor. For example, for any affine k-algebra, let SLn(A) be the group of n × n

matrices with coefficients in A having determinant 1. A homomorphism A B

induces a homomorphism SLn(A) SLn(B), and so SLn(A) is a functor. In fact, it

is the points functor of the affine variety:

Specm k[X11, . . . ,Xnn]/(det(Xij) 1).

Matrix multiplication defines a morphism of functors

SLn ×SLn SLn

which, because of (3.29), arises from a morphism of algebraic varieties. In fact, SLn

is an algebraic group.

Instead of defining varieties to be ringed spaces, it is possible to define them to be

functors Affk Sets satisfying certain conditions.

Dominating maps. A regular map α: V W is said to be dominating if the

image of α is dense in W. Suppose V and W are irreducible. If V _ and W_ are open

affine subsets of V and W such that ϕ(V _) W_, then (2.21) implies that the map

f _ f ϕ: k[W_] k[V _] is injective. Therefore it extends to a map on the fields of

fractions, k(W) k(V ), and this map is independent of the choice of V _ and W_.

Algebraic Geometry: 4. Local Study 59