9. Algebraic Geometry over an Arbitrary Field

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We now explain how to extend the theory in the preceding sections to a nonalgebraically

closed base field. Fix a field k, and let kal be an algebraic closure of

k.

Sheaves. We shall need a more abstract notion of a ringed space and of a sheaf.

A presheaf F on a topological space V is a map assigning to each open subset U

of V a set F(U) and to each inclusion U ⊃ U_ a “restriction” map

a _→ a|U_ : F(U)→F(U_);

the restriction map F(U) → F(U) is required to be the identity map, and if U__ ⊃

U_ ⊃ U, then the composite of the restriction maps F(U) → F(U_) and F(U_) →

F(U__) is required to be the restriction map F(U) → F(U__). In other words, a

presheaf is a contravariant functor to the category of sets from the category whose

objects are the open subsets of V and whose morphisms are the inclusions . A

homomorphism of presheaves α: F →F_ is a family of maps

α(U) : F(U)→F_(U)

commuting with the restriction maps.

A presheaf F is a sheaf if for every open covering {Ui} of an open subset U of V

and family of elements ai ∈ F(Ui) agreeing on overlaps (that is, such that ai|Ui∩Uj =

aj |Ui ∩ Uj for all i, j), there is a unique element a ∈ F(U) such that ai = a|Ui for all

i. A homomorphism of sheaves on V is a homomorphism of presheaves.

If the sets F(U) are abelian groups and the restriction maps are homomorphisms,

then the sheaf is a sheaf of abelian groups. Similarly one defines a sheaf of rings, a

sheaf of k-algebras, and a sheaf of modules over a sheaf of rings.

For v ∈ V , the stalk of a sheaf F (or presheaf) at v is

Fv = lim

←

F(U) (limit over open neighbourhoods of v).

A ringed space is a pair (V,O) consisting of topological space V together with a

sheaf of rings. If the stalk Ov of O at v is a local ring for all v ∈ V , then (V,O) is

called a locally ringed space. A morphism (V,O) → (V _,O_) of ringed spaces is a pair

(ϕ, ψ) with ϕ a continuous map V → V _ and ψ a family of maps

ψ(U

_

) : O_

(U

_

)→O(ϕ

−1(U

_

)), U

_

open in V

_

,

commuting with the restriction maps. Such a pair defines homomorphism of rings

ψv : O_

ϕ(v)

→Ov for all v ∈ V. A morphism of locally ringed spaces is a morphism of

ringed space such that ψv is a local homomorphism for all v.

Extending scalars. Recall that a ring A is reduced if it has no nonzero nilpotents.

If A is reduced, then A⊗k kal need not be reduced. Consider for example the algebra

A = k[X, Y ]/(Xp + Y p + a) where p = char(k) and a /∈ kp. Then A is reduced (even

an integral domain) because Xp + Y p + a is irreducible in k[X, Y ], but

A ⊗k kal = kal[X, Y ]/(Xp + Y p + a) = kal[X, Y ]/((X + Y + α)p), αp = a,

which is not reduced because x + y + α _= 0 but (x + y + α)p = 0.

132 Algebraic Geometry: 9. Algebraic Geometry over an Arbitrary Field

The next proposition shows that problems of this kind arise only because of inseparability;

in particular, they don’t occur if k is perfect.

Recall that the characteristic exponent of a field is p if k has characteristic p _= 0,

and it is 1 is k has characteristic zero. For p equal to the characteristic exponent of

k, let

k

1

p = {α ∈ kal | αp ∈ k}.

It is a subfield of k al , and k

1

p = k if and only if k is perfect.

Proposition 9.1. Let A be a reduced finitely generated k-algebra. The following

statements are equivalent:

(a) A ⊗k k

1

p is reduced;

(b) A ⊗k kal is reduced;

(c) A ⊗k K is reduced for all fields K ⊃ k.

Proof. Clearly c=⇒b=⇒a. The implication a =⇒c follows from Zariski and

Samuel 1958, III.15, Theorem 39 (localize A at a minimal prime to get a field).

Even when A is an integral domain and A⊗k kal is reduced, the latter need not be

an integral domain. Suppose, for example, that A is a finite separable field extension

of k. Then A ≈ k[X]/(f(X)) with f(X) an irreducible separable polynomial. Hence

A ⊗k kal ≈ kal[X]/(f(X)) = kal/(Π(X − ai)) ≈ Π kal/(X − ai)

(by the Chinese remainder theorem). This shows that if A contains a finite separable

field extension of k, then A⊗k kal can’t be an integral domain. The next proposition

gives a converse.

Proposition 9.2. Let A be a finitely generated k-algebra, and assume that A is

an integral domain, and that A⊗k kal is reduced. Then A⊗k kal is an integral domain

if and only if k is algebraically closed in A (i.e., if a ∈ A is algebraic over k, then

a ∈ k).

Proof. Ibid. III.15.

After these preliminaries, it is possible rewrite all of the preceding sections with k

not necessarily algebraically closed. I indicate briefly how this is done.

Affine algebraic varieties. An affine k-algebra A is a finitely generated k-algebra

A such that A ⊗k kal is reduced. Since A ⊂ A ⊗k kal, A itself is then reduced.

Proposition 9.1 has the following consequence.

Corollary 9.3. Let A be a reduced finitely generated k-algebra.

(a) If k is perfect, then A is an affine k-algebra.

(b) If A is an affine k-algebra, then A⊗k K is reduced for all fields K containing k.

Let A be a finitely generated k-algebra. The choice of a set {x1, ..., xn} of generators

for A, determines isomorphisms

A

∼=

k[x1, ..., xn]

∼=

k[X1, ...,Xn]/(f1, ..., fm),

Algebraic Geometry: 9. Algebraic Geometry over an Arbitrary Field 133

and

A ⊗k kal ∼=

kal[X1, ...,Xn]/(f1, ..., fm).

Thus A is an affine algebra if the elements f1, ..., fm of k[X1, ...,Xn] generate a radical

ideal when regarded as elements of kal[X1, ...,Xn]. From the above remarks, we see

that this condition implies that they generate a radical ideal in k[X1, ...,Xn], and the

converse implication holds when k is perfect.

Let A be an affine k-algebra. Define specm(A) to be the set of maximal ideals in A,

and endow it with the topology having as basis the sets D(f), D(f) = {m | f /∈ m}.

There is a unique sheaf of k-algebras O on specm(A) such that O(D(f)) = Af for all

f (recall that Af is the ring obtained from A by inverting f). Here O is a sheaf in the

above abstract sense — the elements of O(U) are not functions on U with values in k,

although we may wish to think of them as if they were. If f ∈ A and mv ∈ specm(A),

then we can define f(v) to be the image of f in the κ(v) df = A/mv, and it does make

sense to speak of the zero set of f in V . The ringed space

Specm(A) = (specm(A),O)

is called an affine variety over k. The stalk at m ∈ V is the local ring Am, and so

Specm(A) is a locally ringed space.

If k = kal, and

A = k[x1, ..., xn] = k[X1, ...,Xn]/(f1, ..., fm),

then the Nullstellensatz allows us to identify specm(A) with the set V (f1, ..., fm) of

common zeros of the fi, via

(x1 − a1, ..., xn − an) _−→ (a1, ..., an).

Moreover, in this case, the elements of O(U) can be identified with k-valued functions

on U.

A morphism of affine algebraic varieties over k is defined to be a morphism

(V,OV ) → (W,OW) of ringed spaces of k-algebras — it is automatically a morphism

of locally ringed spaces.

A homomorphism of k-algebras A → B defines a morphism of affine k-varieties,

SpecmB → Specm A

in a natural way, and this gives a bijection:

Homk−alg(A,B)

∼=

Homk(W, V ), V = Specm A, W = Specm B.

Therefore A _→ Specm(A) is an equivalence of from the category of affine k-algebras

to that of affine algebraic varieties over k; its quasi-inverse is V _→ k[V ] df = Γ(V,OV ).

If A = k[X1, ...,Xm]/a and B = k[Y1, ..., Yn]/b, a homomorphism A → B is determined

by a family of polynomials, Pi(Y1, ..., Yn), i = 1, ...,m; the homomorphism

sends xi to Pi(y1, ..., yn); in order to define a homomorphism, the Pi must be such

that

F ∈ a =⇒ F(P1, ..., Pn) ∈ b;

two families P1, ..., Pm and Q1, ...,Qm determine the same map if and only if Pi ≡ Qi

mod b for all i.

134 Algebraic Geometry: 9. Algebraic Geometry over an Arbitrary Field

Let A be an affine k-algebra, and let V = SpecmA. For any field K ⊃ k, A ⊗k K

is an affine algebra over K, and hence we get a variety VK

df = Specm(A ⊗k K) over

K. We say that VK has been obtained from V by extension of scalars. Note that

if A = k[X1, ...,Xn]/(f1, ..., fm) then A ⊗k K = K[X1, ...,Xn]/(f1, ..., fm). The map

V _→ VK is a functor from affine varieties over k to affine varieties over K.

Let V0 = Specm(A0) be an affine variety over k, and let W = V (b) be a closed

subvariety of V df = V0,kal. Then W arises by extension of scalars from a closed subvariety

W0 of V0 if and only if the ideal b of A0 ⊗k kal is generated by elements A0.

Except when k is perfect, this is stronger than saying W is the zero set of a family of

elements of A.

Algebraic varieties. A ringed space (V,O) is a prevariety over k if there is a finite

covering (Ui) of V by open subsets such that (Ui, O|Ui) is an affine variety over k for

all i. A morphism of prevarieties over k is a morphism of ringed spaces of k-algebras.

A prevariety V over k is separated if for all pairs of morphisms of k-varieties α, β :

Z → V , the subset of Z on which α and β agree is closed. A variety is a separated

prevariety.

Products: Let A and B be finitely generated k-algebras. It is possible for A and B

to be reduced but for A ⊗k B fail to be reduced — consider for example,

A = k[X, Y ]/(Xp + Y p + a), B= k[Z]/(Zp − a), a /∈ kp.

However, if A and B are affine k-algebras, then A ⊗k B is again an affine k-algebra.

To see this, note that (by definition), A⊗k kal and B ⊗k kal are affine k-algebras, and

therefore so also is their tensor product over kal (3.16); but

(A ⊗k kal) ⊗

kal (kal ⊗k B) = ((A ⊗k kal) ⊗

kal kal) ⊗k B = (A ⊗k B) ⊗k kal.

Thus we can define the product of two affine algebraic varieties, V = Specm A and

W = Specm B, over k by

V ×k W = Specm(A ⊗k B).

It has the universal property expected of products, and the definition extends in a

natural way to (pre)varieties.

Just as in (3.18), the diagonal Δ is locally closed in V × V , and it is closed if and

only if V is separated.

Extension of scalars: Let V be a variety over k, and let K be a field containing

k. There is a natural way of defining a variety VK, said to be obtained from V by

extension of scalars:if V is a union of open affines, V = ∪Ui, then VK = ∪Ui,K and

the Ui,K are patched together the same way as the Ui. The dimension of a variety

doesn’t change under extension of scalars.

When V is a variety over kal obtained from a variety V0 over k by extension of

scalars, we sometimes call V0 a model for V over k. More precisely, a model of V over

k is a variety V0 over k together with an isomorphism ϕ: V0,kal → V.

Of course, V need not have a model over k — for example, an elliptic curve

E : Y 2Z = X3 + aXZ2 + bZ3

Algebraic Geometry: 9. Algebraic Geometry over an Arbitrary Field 135

over kal will have a model over k ⊂ kal if and only if its j-invariant j(E) df = 1728(4a)3

−16(4a3+27b2)

lies in k. Moreover, when V has a model over k, it will usually have a large number

of them, no two of which are isomorphic over k. Consider, for example, the quadric

surface in P3over Qal,

V : X2 + Y 2 + Z2 +W2 = 0.

The models over V over Q are defined by equations

aX2 + bY 2 + cZ2 + dW2 = 0, a, b, c, d ∈ Q.

Classifying the models of V over Q is equivalent to classifying quadratic forms over

Q in 4 variables. This has been done, but it requires serious number theory. In

particular, there are infinitely many (see Chapter VIII of my notes on Class Field

Theory).

Exercise 9.4. Show directly that, up to isomorphism, the curve X2 + Y 2 = 1

over C has exactly two models over R.

The points on a variety. Let V be a variety over k. A point of V with coordinates

in k, or a point of V rational over k, is a morphism Specmk → V. For

example, if V is affine, say V = Specm(A), then a point of V with coordinates in

k is a k-homomorphism A → k. If A = k[X1, ...,Xn]/(f1, ..., fm), then to give a

k-homomorphism A → k is the same as to give an n-tuple (a1, ..., an) such that

fi(a1, ..., an) = 0, i = 1, ..., m.

In other words, of V is the affine variety over k defined by the equations

fi(X1, . . . ,Xn) = 0, i= 1, . . . ,m

then a point of V with coordinates in k is a solution to this system of equations in k.

We write V (k) for the points of V with coordinates in k.

We extend this notion to obtain the set of points V (R) of a variety V with coordinates

in any k-algebra R. For example, when V = Specm(A), we set

V (R) = Homk−alg(A,R).

Again, if

A = k[X1, ...,Xn]/(f1, ..., fm),

then

V (R) = {(a1, ..., an) ∈ Rn | fi(a1, ..., an) = 0, i = 1, 2, ...,m}.

What is the relation between the elements of V and the elements of V (k)? Suppose

V is affine, say V = Specm(A). Let v ∈ V . Then v corresponds to a maximal ideal mv

in A (actually, it is a maximal ideal), and we write κ(v) for the residue field Ov/mv.

Then κ(v) is a finite extension of k, and we call the degree of κ(v) over k the degree

of v. Let K be a field algebraic over k. To give a point of V with coordinates in K is

to give a homomorphism of k-algebras A → K. The kernel of such a homomorphism

is a maximal ideal mv in A, and the homomorphisms A → k with kernel mv are in

one-to-one correspondence with the k-homomorphisms κ(v) → K. In particular, we

see that there is a natural one-to-one correspondence between the points of V with

136 Algebraic Geometry: 9. Algebraic Geometry over an Arbitrary Field

coordinates in k and the points v of V with κ(v) = k, i.e., with the points v of V of

degree 1. This statement holds also for nonaffine algebraic varieties.

Assume now that k is perfect. The kal-rational points of V with image v ∈ V are

in one-to-one correspondence with the k-homomorphisms κ(v) → kal — therefore,

there are exactly deg(v) of them, and they form a single orbit under the action of

Gal(kal/k). Thus there is a natural bijection from V to the set of orbits for Gal(kal/k)

acting on V (kal).

Local Study. Let V = V (a) ⊂ An, and let a = (f1, ..., fr). The singular locus Vsing

of V is defined by the vanishing of the (n − d) × (n − d) minors of the matrix

Jac(f1, f2, . . . , fr) =



∂f1

∂x1

∂f1

∂x2

· · · ∂f1

∂xr

∂f2

∂x1 ...

∂fr

∂x1

∂fr

∂xr



.

We say that v is nonsingular if some (n − d) × (n − d) minor doesn’t vanish at v.

We say V is nonsingular all its singular locus is empty. If V is nonsingular, then it is

regular, but not conversely. Obviously V is nonsingular ⇐⇒ Vkal is nonsingular.

Note that Vsing is compatible with extension of scalars. Therefore (Theorem 4.21)

it is a proper subvariety of V .

Projective varieties; complete varieties. It is possible to associate projective

varieties to certain graded rings over k. An algebraic variety over k is complete if for

all varieties W, the projection map V ×W → W is closed, and this property persists

under extension of scalars to kal. A projective variety is complete.

Finite maps. The Noether normalization theorem needs a different proof when the

field is finite.

Algebraic Geometry: 10. Divisors and Intersection Theory 137

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