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8. Regular Maps and Their Fibres.

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Throughout this section, k is an algebraically closed field. Consider again the

regular map ϕ: A2 → A2, (x, y) _→ (x, xy). We have seen that its image

C = (A2 {y-axis}) ∪ {(0, 0)}

is neither open nor closed, and, in fact, is not even locally closed. The fibre

ϕ−1(x, y) = (A2 {y-axis}) ∪ {(0, 0)}.

From this unpromising example, it would appear that it is not possible to say anything

about the image of a regular map, nor about the dimension or number of elements in

its fibres. However, it turns out that (almost) everything that can go wrong already

goes wrong for this map. We shall show:

(a) the image of a regular map is a finite union of locally closed sets;

(b) the dimensions of the fibres can jump only on closed subsets;

provided the map is finite, the target variety is normal, and k has characteristic

zero.

Constructible sets. Let W be a topological space. A subset C of W is said to

constructible if it is a finite union of sets of the form U ∩ Z with U open and Z

closed. Obviously, if C is constructible and V ⊂ W, then C ∩ V is constructible. A

constructible set in An is definable by a finite number of polynomials; more precisely,

it is defined by a finite number of statements of the form

f(X1, · · · ,Xn) = 0, g(X1, · · · ,Xn) _= 0

combined using only “and” and “or” (or, better, statements of the form f = 0 combined

using “and”, “or”, and “not”). The next proposition shows that a constructible

set C that is dense in an irreducible variety V must contain a nonempty open subset

of V . Contrast Q, which is dense in R (real topology), but does not contain an open

subset of R, or any infinite subset of A1 that omits an infinite set.

Proposition 8.1. Let C be a constructible set whose closure ￣ C is irreducible.

Then C contains a nonempty open subset of ￣ C.

Proof. We are given that C = ∪(Ui ∩ Zi) with each Ui open and each Zi closed.

We may assume that each set Ui ∩ Zi in this decomposition is nonempty. Clearly

￣ C ⊂ ∪Zi, and as ￣ C is irreducible, it must be contained in one of the Zi. For this i

C ⊃ Ui ∩ Zi ⊃ Ui ∩ ￣ C ⊃ Ui ∩ C ⊃ Ui ∩ (Ui ∩ Zi) = Ui ∩ Zi.

Thus Ui ∩ Zi = Ui ∩ ￣ C is a nonempty open subset of ￣ C contained in C.

Theorem 8.2. A regular map ϕ: W → V sends constructible sets to constructible

sets. In particular, if U is a nonempty open subset of W, then ϕ(U) contains a

nonempty open subset of its closure in V .

The key result we shall need from commutative algebra is the following. (In the next

two results, A and B are arbitrary commutative rings—they need not be k-algebras.)

118 Algebraic Geometry: 8. Regular Maps and Their Fibres

Proposition 8.3. Let A ⊂ B be integral domains with B finitely generated as an

algebra over A, and let b be a nonzero element of B. Then there exists an element

a _= 0 in A with the following property: every homomorphism α: A → Ω from A into

an algebraically closed field Ω such that α(a) _= 0 can be extended to a homomorphism

β : B → Ω such that β(b) _= 0.

Consider, for example, the rings k[X] ⊂ k[X,X−1]. A homomorphism α: k[X] → k

extends to a homomorphism k[X,X−1] → k if and only if α(X) _= 0. Therefore, for

b = 1, we can take a = X. In the application we make of Proposition 8.3, we only

really need the case b = 1, but the more general statement is needed so that we can

prove it by induction.

Lemma 8.4. Let B ⊃ A be integral domains, and assume B = A[t] ≈ A[T ]/a.

Let c ⊂ A be the set of leading coefficients of the polynomials in a. Then every

homomorphism α: A → Ω from A into an algebraically closed field Ω such that α(c) _=

0 can be extended to a homomorphism of B into Ω.

Proof. Note that c is an ideal in A. If a = 0, then c = 0, and there is nothing to

prove (in fact, every α extends). Thus we may assume a _= 0. Let f = amTm+· · ·+a0

be a nonzero polynomial of minimumdegree in a such that α(am) _= 0. Because B _= 0,

we have that m ≥ 1.

Extend α to a homomorphism ˜α: A[T ] → Ω[T ] by sending T to T. The Ω-

submodule of Ω[T ] generated by ˜α(a) is an ideal (because T ·

ci ˜α(gi) =

ci ˜α(giT )).

Therefore, unless ˜α(a) contains a nonzero constant, it generates a proper ideal in Ω[T ],

which will have a zero c in Ω. The homomorphism

A[T ]→αe Ω[T ] → Ω, T _→ T _→ c

then factors through A[T ]/a = B and extends α.

In the contrary case, a contains a polynomial

g(T) = bnT n + · · · + b0, α(bi) = 0 (i > 0), α(b0) _= 0.

On dividing f(T) into g(T ) we find that

mg(T) = q(T )f(T)+ r(T ), d∈ N, q,r∈

A[T ], degr <m.

On applying ˜α to this equation, we obtain

α(am)dα(b0) = ˜α(q)˜α(f) + ˜α(r).

Because ˜α(f) has degree m > 0, we must have ˜α(q) = 0, and so ˜α(r) is a nonzero

constant. After replacing g(T) with r(T ), we may assume n < m. If m = 1, such a

g(T ) can’t exist, and so we may suppose m > 1 and (by induction) that the lemma

holds for smaller values of m.

For h(T) = crT r+cr−1T r−1+· · ·+c0, let h_(T) = cr+· · ·+c0T r. Then the A-module

generated by the polynomials T sh_(T ), s ≥ 0, h ∈ a, is an ideal a_ in A[T ]. Moreover,

a_ contains a nonzero constant if and only if a contains a nonzero polynomial cT r,

which implies t = 0 and A = B (since B is an integral domain).

If a_ does not contain nonzero constants, then set B_ = A[T ]/a_ = A[t_]. Then a_

contains the polynomial g_ = bn + · · · + b0T n, and α(b0)_= 0. Because deg g_ < m, the

Algebraic Geometry: 8. Regular Maps and Their Fibres 119

induction hypothesis implies that α extends to a homomorphism B_ → Ω. Therefore,

there is a c ∈ Ω such that, for all h(T) = crT r + cr−1T r−1 + · · · + c0 ∈ a,

On taking h = g, we see that c = 0, and on taking h = f, we obtain the contradiction

α(am) = 0.

Proof. (of 8.3) Suppose that we know the proposition in the case that B is generated

by a single element, and write B = A[x1, . . . , xn]. Then there exists an element

bn−1 such that any homomorphism α: A[x1, . . . , xn−1] → Ω such that α(bn−1) _= 0 extends

to a homomorphism β : B → Ω such that β(b) _= 0. Continuing in this fashion,

we obtain an element a ∈ A with the required property.

Thus we may assume B = A[x]. Let a be the kernel of the homomorphism X _→ x,

A[X] → A[x].

Case (i). The ideal a = (0). Write

b = f(x) = a0xn + a1xn−1 + · · · + an, ai ∈ A,

and take a = a0. If α: A → Ω is such that α(a0) _= 0, then there exists a c ∈ Ω such

that f(c) _= 0, and we can take β to be the homomorphism

dixi _→ _

α(di)ci.

Case (ii). The ideal a _= (0). Let f(T) = amTm + · · · , am _= 0, be an element of a

of minimum degree. Let h(T ) ∈ A[T] represent b. Since b _= 0, h /∈ a. Because f is

irreducible over the field of fractions of A, it and h are coprime over that field. Hence

there exist u, v ∈ A[T] and c ∈ A −{0} such that

uh + vf = c.

It follows now that cam satisfies our requirements, for if α(cam) _= 0, then α can be

extended to β : B → Ω by the previous lemma, and β(u(x) · b) = β(c) _= 0, and so

β(b) _= 0.

Aside 8.5. In case (ii) of the above proof, both b and b−1 are algebraic over A,

and so there exist equations

a0bm + · · · + am = 0, ai ∈ A, a0 _= 0;

−n + · · · + a

n = 0, a

i ∈ A, a

_= 0.

One can show that a = a0a_

0 has the property required by the Proposition—see Atiyah

and MacDonald, 5.23.

Proof. (of 8.2) We first prove the “in particular” statement of the Theorem.

By considering suitable open affine coverings of W and V , one sees that it suffices to

prove this in the case that both W and V are affine. If W1, . . . ,Wr are the irreducible

components of W, then the closure of ϕ(W) in V , ϕ(W)− = ϕ(W1)− ∪ . . . ∪ ϕ(Wr)−,

and so it suffices to prove the statement in the case that W is irreducible. We may

also replace V with ϕ(W)−, and so assume that both W and V are irreducible. Then

ϕ corresponds to an injective homomorphism A → B of affine k-algebras. For some

b _= 0, D(b) ⊂ U. Choose a as in the lemma. Then for any point P ∈ D(a), the

homomorphism f _→ f(P) : A → k extends to a homomorphism β : B → k such that

β(b) _= 0. The kernel of β is a maximal ideal corresponding to a point Q ∈ D(b) lying

over P.

120 Algebraic Geometry: 8. Regular Maps and Their Fibres

We now prove the theorem. Let Wi be the irreducible components of W. Then

C ∩Wi is constructible in Wi, and ϕ(W) is the union of the ϕ(C ∩Wi); it is therefore

constructible if the ϕ(C ∩ Wi) are. Hence we may assume that W is irreducible.

Moreover, C is a finite union of its irreducible components, and these are closed in C;

they are therefore constructible. We may therefore assume that C also is irreducible;

￣ C is then an irreducible closed subvariety of W.

We shall prove the theorem by induction on the dimension of W. If dim(W) = 0,

then the statement is obvious because W is a point. If ￣ C _= W, then dim( ￣ C) <

dim(W), and because C is constructible in ￣ C, we see that ϕ(C) is constructible (by

induction). We may therefore assume that ￣ C = W. But then ￣ C contains a nonempty

open subset of W, and so the case just proved shows that ϕ(C) contains an nonempty

open subset U of its closure. Replace V be the closure of ϕ(C), and write

ϕ(C) = U ∪ ϕ(C ∩ ϕ

−1(V − U)).

Then ϕ−1(V −U) is a proper closed subset of W (the complement of V − U is dense

in V and ϕ is dominating). As C ∩ϕ−1(V −U) is constructible in ϕ−1(V −U), the set

ϕ(C ∩ϕ−1(V −U)) is constructible in V by induction, which completes the proof.

The fibres of morphisms. We wish to examine the fibres of a regular map ϕ: W →

V . Clearly, we can replace V by the closure of ϕ(W) in V and so assume ϕ to be

dominating.

Theorem 8.6. Let ϕ: W → V be a dominating regular map of irreducible varieties.

Then

(a) dim(W) ≥ dim(V );

(b) if P ∈ ϕ(W), then

dim(ϕ

−1(P)) ≥ dim(W) − dim(V )

for every P ∈ V , with equality holding exactly on a nonempty open subset U of

V .

Vi = {P ∈ V | dim(ϕ−1(P)) ≥ i}

are closed ϕ(W).

Example 8.7. Consider the subvariety W ⊂ V ×Am defined by r linear equations

j=1

aijXj = 0, aij ∈ k[V ], i= 1, . . . , r,

and let ϕ be the projection W → V. For P ∈ V , ϕ−1(P) is the set of solutions of

j=1

aij(P)Xj = 0, aij(P) ∈ k, i = 1, . . . , r,

and so its dimension is m−rank(aij(P)). Since the rank of the matrix (aij(P)) drops

on closed subsets, the dimension of the fibre jumps on closed subsets.

Algebraic Geometry: 8. Regular Maps and Their Fibres 121

Proof. (a) Because the map is dominating, there is a homomorphism k(V ) 8→

k(W), and obviously tr degkk(V ) ≤ tr degkk(W) (an algebraically independent subset

of k(V ) remains algebraically independent in k(W)).

(b) In proving the first part of (b), we may replace V by any open neighbourhood

of P. In particular, we can assume V to be affine. Let m be the dimension of

V . From (7.11) we know that there exist regular functions f1, . . . , fm such that

P is an irreducible component of V (f1, . . . , fm). After replacing V by a smaller

neighbourhood of P, we can suppose that P = V (f1, . . . , fm). Then ϕ−1(P) is

the zero set of the regular functions f1 ◦ ϕ, . . . , fm ◦ ϕ, and so (if nonempty) has

codimension ≤ m in W (see 7.7). Hence

dimϕ

−1(P) ≥ dimW − m = dim(W) − dim(V ).

In proving the second part of (b), we can replace both W and V with open affine

subsets. Since ϕ is dominating, k[V ] → k[W] is injective, and we may regard it as an

inclusion (we identify a function x on V with x◦ϕ on W). Then k(V ) ⊂ k(W). Write

k[V ] = k[x1, . . . , xM] and k[W] = k[y1, . . . , yN], and suppose V and W have dimensions

m and n respectively. Then k(W) has transcendence degree n−m over k(V ), and

we may suppose that y1, . . . , yn−m are algebraically independent over k[x1, . . . , xm],

and that the remaining yi are algebraic over k[x1, . . . , xm, y1, . . . , yn−m]. There are

therefore relations

Fi(x1, . . . , xm, y1, . . . , yn−m, yi) = 0, i= n − m + 1, . . . ,N. (*)

with Fi(X1, . . . ,Xm, Y1, . . . , Yn−m, Yi) a nonzero polynomial. We write ￣yi for the

restriction of yi to ϕ−1(P). Then

k[ϕ−1(P)] = k[￣y1, . . . , ￣yN].

The equations (*) give an algebraic relation among the functions x1, . . . , yi on W.

When we restrict them to ϕ−1(P), they become equations:

Fi(x1(P), . . . , xm(P), ￣y1, . . . , ￣yn−m, ￣yi) = 0, i= n − m + 1, . . . ,N. (**).

If these are nontrivial algebraic relations, i.e., if none of the polynomials

Fi(x1(P), . . . , xm(P), Y1, . . . , Yn−m, Yi)

is identically zero, then the transcendence degree of k(￣y1, . . . , ￣yN) over k will be

≤ n − m.

Thus, regard Fi(x1, . . . , xm, Y1, . . . , Yn−m, Yi) as a polynomial in the Y ’s with coefficients

polynomials in the x’s. Let Vi be the closed subvariety of V defined by the

simultaneous vanishing of the coefficients of this polynomial—it is a proper closed

subset of V. Let U = V −∪Vi—it is a nonempty open subset of V . If P ∈ U, then

none of the polynomials Fi(x1(P), . . . , xm(P), Y1, . . . , Yn−m, Yi) is identically zero,

and so for P ∈ U, the dimension of ϕ−1(P) is ≤ n − m, and hence = n −m by (a).

Finally, if for a particular point P, dimϕ−1(P) = n − m, then one can modify the

above argument to show that the same is true for all points in an open neighbourhood

of P.

122 Algebraic Geometry: 8. Regular Maps and Their Fibres

dimV = 0. We know from (b) that there is an open subset U of V such that

dimϕ−1(P) = n − m ⇐⇒ P ∈ U.

Let Z be the complement of U in V ; thus Z = Vn−m+1. Let Z1, . . . , Zr be the

irreducible components of Z. On applying the induction to the restriction of ϕ to the

map ϕ−1(Zj) → Zj for each j, we obtain the result.

Proposition 8.8. Let ϕ: W → V be a regular surjective closed mapping of varieties

(e.g., W complete or ϕ finite). If V is irreducible and all the fibres ϕ−1(P) are

irreducible of dimension n, then W is irreducible of dimension dim(V ) + n.

Proof. Let Z be a closed irreducible subset of W, and consider the map ϕ|Z : Z →

V ; it has fibres (ϕ|Z)−1(P) = ϕ−1(P) ∩ Z. There are three possibilities.

(a) ϕ(Z) _= V. Then ϕ(Z) is a proper closed subset of V .

(b) ϕ(Z) = V, dim(Z) < n + dim(V ). Then (b) of (8.6) shows that there is a

nonempty open subset U of V such that for P ∈ U,

dim(ϕ

−1(P) ∩ Z) = dim(Z) − dim(V ) < n;

thus for P ∈ U, ϕ−1(P) _ Z.

dim(ϕ−1(P) ∩ Z) ≥ dim(Z) − dim(V ) ≥ n

for all P; thus ϕ−1(P) ⊂ Z for all P ∈ V , and so Z = W; moreover dimZ = n.

Now let Z1, . . . , Zr be the irreducible components of W. I claim that (iii) holds for

at least one of the Zi. Otherwise, there will be an open subset U of V such that for P

in U, ϕ−1(P) _ Zi for any i, but ϕ−1(P) is irreducible and ϕ−1(P) = ∪(ϕ−1(P)∪Zi),

and so this is impossible.

The fibres of finite maps. Let ϕ: W → V be a finite dominating morphism of

irreducible varieties. Then dim(W) = dim(V ), and so k(W) is a finite field extension

of k(V ). Its degree is called the degree of the map ϕ.

Lemma 8.9. An integral domain A is integrally closed if and only if Am is integrally

closed for all maximal ideals m of A.

Proof. ⇒:If A is integrally closed, then so is S−1A for any multiplicative subset

S (not containing 0), because if

bn + c1bn−1 + · · · + cn = 0, ci ∈ S−1A,

then there is an s ∈ S such that sci ∈ A for all i, and then

(sb)n + (sc1)(sb)n−1 + · · · + sncn = 0,

demonstrates that sb ∈ A, whence b ∈ S−1A.

⇐:If c is integral over A, it is integral over each Am, hence in each Am, and

A = ∩Am (if c ∈ ∩Am, then the set of a ∈ A such that ac ∈ A is an ideal in A, not

contained in any maximal ideal, and therefore equal to A itself).

Thus the following conditions on an irreducible variety V are equivalent:

(a) for all P ∈ V , OP is integrally closed;

Algebraic Geometry: 8. Regular Maps and Their Fibres 123

(b) for all irreducible open affines U of V , k[U] is integrally closed;

closed for all i.

An irreducible variety V satisfying these conditions is said to be normal. We also

call a disjoint union of such varieties normal. Thus a variety V is normal if and only

if OP is an integrally closed integral domain for all P ∈ V .

Theorem 8.10. Let ϕ: W → V be a finite surjective regular map of irreducible

varieties, and assume that V is normal.

(a) For all P ∈ V , #ϕ−1(P) ≤ deg(ϕ).

(b) The set of points P of V such that #ϕ−1(P) = deg(ϕ) is an open subset of V ,

and it is nonempty if k(W) is separable over k(V ).

Before proving the theorem, we give examples to show that we need W to be

separated and V to be normal in (a), and that we need k(W) to be separable over

k(V ) for the second part of (b).

Example 8.11. (a) Consider the map

{A1 with origin doubled } → A1.

The degree is one and that map is one-to-one except at the origin where it is two-toone.

(b) Let C be the curve Y 2 = X3 + X2, and let ϕ: A1 → C be the map t _→

(t2 − 1, t(t2 − 1)). The map corresponds to the inclusion k[x, y] 8→ k[T] and is of

degree one. The map is one-to-one except that the points t = ±1 both map to 0.

The ring k[x, y] is not integrally closed; in fact k[T ] is its integral closure in its field

of fractions.

1, . . . , ap

n), where

p = chark. This map has degree pn but it is one-to-one. The field extension corresponding

to the map is

k(X1, . . . ,Xn) ⊃ k(Xp

1 , . . . ,Xp

which is purely inseparable.

Lemma 8.12. Let Q1, . . . ,Qr be distinct points on an affine variety V . Then there

is a regular function f on V taking distinct values at the Qi.

Proof. We can embed V as closed subvariety of An, and then it suffices to prove

the statement with V = An — almost any linear form will do.

Proof. (of Theorem 8.10). In proving (a) of the theorem, we may assume that

V and W are affine, and so the map corresponds to a finite map of k-algebras,

k[V ] → k[W]. Let ϕ−1(P) = {Q1, . . . ,Qr}. According to the lemma, there exists an

f ∈ k[W] taking distinct values at the Qi. Let

F(T) = Tm + a1Tm−1 + · · · + am

be the minimumpolynomial of f over k(V ). It has degree m ≤ [k(W) : k(V )] = deg ϕ,

and it has coefficients in k[V ] because V is normal (see 1.33). Now F(f) = 0 implies

124 Algebraic Geometry: 8. Regular Maps and Their Fibres

F(f(Qi)) = 0, i.e.,

f(Qi)m + a1(P) · f(Qi)m−1 + · · · + am(P) = 0.

Therefore the f(Qi) are all roots of a single polynomial of degree m, and so r ≤ m ≤

deg(ϕ).

In order to prove the first part of (b), we show that, if there is a point P ∈ V

such that ϕ−1(P) has deg(ϕ) elements, then the same is true for all points in an open

neighbourhood of P. Choose f as in the last paragraph corresponding to such a P.

Then the polynomial

Tm + a1(P) · Tm−1 + · · · + am(P) = 0 (*)

has r = degϕ distinct roots, and so m = r. Consider the discriminant disc F of F.

Because (*) has distinct roots, disc(F)(P) _= 0, and so disc(F) is nonzero on an open

neighbourhood U of P. The factorization

k[V ] → k[V ][T ]/(F)

T→_→f k[W]

gives a factorization

W → Specm(k[V ][T ]/(F))→ V.

Each point P_ ∈ U has exactly m inverse images under the second map, and the

first map is finite and dominating, and therefore surjective (recall that a finite map

is closed). This proves that ϕ−1(P_) has at least deg(ϕ) points for P_ ∈ U, and part

(a) of the theorem then implies that it has exactly deg(ϕ) points.

We now show that if the field extension is separable, then there exists a point such

that #ϕ−1(P) has degϕ elements. Because k(W) is separable over k(V ), there exists

a f ∈ k[W] such that k(V )[f] = k(W). Its minimum polynomial F has degree deg(ϕ)

and its discriminant is a nonzero element of k[V ]. The diagram

W → Specm(A[T ]/(F))→ V

shows that #ϕ−1(P) ≥ deg(ϕ) for P a point such that disc(f)(P) _= 0.

When k(W) is separable over k(V ), then ϕ is said to be separable.

Remark 8.13. Let ϕ: W → V be as in the theorem, and let Vi = {P ∈ V |

#ϕ−1(P) ≤ i}. Let d = deg ϕ. Part (b) of the theorem states that Vd−1 is closed, and

is a proper subset when ϕ is separable. I don’t know under what hypotheses all the

sets Vi will closed (and Vi will be a proper subset of Vi−1). The obvious induction

argument fails because Vi−1may not be normal.

Lines on surfaces. As an application of some of the above results, we consider the

problem of describing the set of lines on a surface of degree m in P3. To avoid possible

problems, we assume for the rest of this chapter that k has characteristic zero.

We first need a way of describing lines in P3. Recall that we can associate with

each projective variety V ⊂ Pn an affine cone over ˜V in kn+1. This allows us to think

of points in P3 as being one-dimensional subspaces in k4, and lines in P3 as being

two-dimensional subspaces in k4. To such a subspace W ⊂ k4, we can attach a onedimensional

subspace

2W in

2 k4 ≈ k6, that is, to each line L in P3, we can attach

point p(L) in P5. Not every point in P5 should be of the form p(L)—heuristically,

the lines in P3 should form a four-dimensional set. (Fix two planes in P3; giving a

Algebraic Geometry: 8. Regular Maps and Their Fibres 125

line in P3 corresponds to choosing a point on each of the planes.) We shall show that

there is natural one-to-one correspondence between the set of lines in P3 and the set

of points on a certain hyperspace Π ⊂ P5. Rather than using exterior algebras, I shall

usually give the old-fashioned proofs.

Let L be a line in P3 and let x = (x0 : x1 : x2 : x3) and y = (y0 : y1 : y2 : y3) be

distinct points on L. Then

p(L) = (p01 : p02 : p03 : p12 : p13 : p23) ∈ P5, pij

df =

____

xi xj

yi yj

____

depends only on L. The pij are called the Pl¨ucker coordinates of L, after Pl¨ucker

(1801-1868).

In terms of exterior algebras, write e0, e1, e2, e3 for the canonical basis for k4,

so that x, regarded as a point of k4 is

xiei, and y =

yiei; then

2 k4 is a 6-

dimensional vector space with basis ei∧ej, 0 ≤ i < j ≤ 3, and x∧y =

pijei∧ej with

pij given by the above formula.

We define pij for all i, j, 0 ≤ i, j ≤ 3 by the same formula — thus pij = −pji.

Lemma 8.14. The line L can be recovered from p(L) as follows:

L = {(

ajp0j :

ajp1j :

ajp2j :

ajp3j) | (a0 : a1 : a2 : a3) ∈ P3}.

Proof. Let ˜L be the cone over L in k4—it is a two-dimensional subspace of k4—

and let x = (x0, x1, x2, x3) and y = (y0, y1, y2, y3) be two linearly independent vectors

in ˜L. Then

˜L

= {f(y)x − f(x)y | f : k4 → k linear}.

Write f =

ajXj; then

f(y)x − f(x)y = (

ajp0j ,

ajp1j ,

ajp2j ,

ajp3j).

Lemma 8.15. The point p(L) lies on the quadric Π ⊂ P5 defined by the equation

X01X23 − X02X13 + X03X12 = 0.

Proof. This can be verified by direct calculation, or by using that

0 =

________

x0 x1 x2 x3

y0 y1 y2 y3

x0 x1 x2 x3

y0 y1 y2 y3

________

= 2(p01p23 − p02p13 + p03p12)

(expansion in terms of 2 ×2 minors).

Lemma 8.16. Every point of Π is of the form p(L) for a unique line L.

Proof. Assume p03 _= 0; then the line through the points (0 : p01 : p02 : p03) and

(p03 : p13 : p23 :0) has Pl¨ucker coordinates

(−p01p03 : −p02p03 : −p2

03 : p01p23 − ! "# p02p1$3

−p03p12

: −p03p13 : −p03p23)

= (p01 : p02 : p03 : p12 : p13 : p23).

126 Algebraic Geometry: 8. Regular Maps and Their Fibres

A similar construction works when one of the other coordinates is nonzero, and this

way we get inverse maps.

Thus we have a canonical one-to-one correspondence

{lines in P3} ↔ {points on Π};

that is, we have identified the set of lines in P3 with the points of an algebraic variety.

We may now use the methods of algebraic geometry to study the set. [This is a

special case of the Grassmanians mentioned on p108.]

We next consider the set of homogeneous polynomials of degree m in 4 variables,

F(X0,X1,X2,X3) =

i0+i1+i2+i3=m

ai0i1i2i3Xi0

0 . . . Xi3

3 .

We don’t distinguish two polynomials if one is a nonzero multiple of the other.

Lemma 8.17. The set of homogeneous polynomials of degree m in 4 variables is a

vector space of dimension ( 3+m

m )

Proof. See a previous footnote page 89.

Let ν = (3+m

m ) = (m+1)(m+2)(m+3)

− 1; then we have a surjective map

Pν → {surfaces of degree m in P3},

(. . . : ai0i1i2i3 : . . . ) _→ V (F), F=

ai0i1i2i3Xi0

0 Xi1

1 Xi2

2 Xi3

3 .

The map is not quite injective—for example, X2Y and XY 2 define the same surface—

but nevertheless, we can (somewhat loosely) think of the points of Pν as being (possible

degenerate) surfaces of degree m in P3.

Let Γm ⊂ Π × Pν ⊂ P5 × Pν be the set of pairs (L, F) consisting of a line L in P3

lying on the surface F(X0,X1,X2,X3) = 0.

Theorem 8.18. The set Γm is a closed irreducible subset of Π×Pν; it is therefore

a projective variety. The dimension of Γm is m(m+1)(m+5)

6 + 3.

Example 8.19. For m = 1, Γm is the set of pairs consisting of a plane in P3 and

a line on the plane. The theorem says that the dimension of Γ1 is 5. Since there are

∞3 planes in P3, and each has ∞2 lines on it, this seems to be correct.

Proof. We first show that Γm is closed. Let

p(L) = (p01 : p02 : . . . ) F =

ai0i1i2i3Xi0

· · ·Xi3

3 .

From (8.14) we see that L lies on the surface F(X0,X1,X2,X3) = 0 if and only if

bjp0j :

bjp1j :

bjp2j :

bjp3j ) = 0, all (b0, . . . , b3) ∈ k4.

Expand this out as a polynomial in the bj’s with coefficients polynomials in the ai0i1i2i3

and pij ’s. Then F(...) = 0 for all b ∈ k4 if and only if the coefficients of the polynomial

are all zero. But each coefficient is of the form

P(. . . , ai0i1i2i3, . . . ; p01, p02 : . . . )

with P homogeneous separately in the a’s and p’s, and so the set is closed in Π × Pν

(cf. the discussion in 5.32).

Algebraic Geometry: 8. Regular Maps and Their Fibres 127

It remains to compute the dimension of Γm. We shall apply Proposition 8.8 to the

projection map

(L, F) Γm ⊂ Π × Pν

↓ ↓ϕ

L Π

For L ∈ Π, ϕ−1(L) consists of the homomogeneous polynomials of degree m such

that L ⊂ V (F) (taken up to nonzero scalars). After a change of coordinates, we can

assume that L is the line _

X0 = 0

X1 = 0,

i.e., L = {(0, 0, ∗, ∗)}. Then L lies on F(X0,X1,X2,X3) = 0 if and only if X0 or X1

occurs in each nonzero monomial term in F, i.e.,

F ∈ ϕ−1(L) ⇐⇒ ai0i1i2i3 = 0 whenever i0 = 0 = i1.

Thus ϕ−1(L) is a linear subspace of Pν ; in particular, it is irreducible. We now

compute its dimension. Recall that F has ν + 1 coefficients altogether; the number

with i0 = 0 = i1 is m+ 1, and so ϕ−1(L) has dimension

(m + 1)(m + 2)(m + 3)

− 1 − (m+ 1) =

m(m + 1)(m + 5)

− 1.

We can now deduce from (8.8) that Γm is irreducible and that

dim(Γm) = dim(Π)+dim(ϕ−1(L)) =

m(m + 1)(m + 5)

+ 3,

as claimed.

Now consider the other projection

(L, F) Γm ⊂ Π × Pν

↓ ↓ψ

F Pν

By definition

−1(F) = {L | L lies on V (F)}.

Example 8.20. Let m = 1. Then ν = 3 and dimΓ1 = 5. The projection

ψ: Γ1 → P3 is surjective (every plane contains at least one line), and (8.6) tells us

that dimψ−1(F) ≥ 2. In fact of course, the lines on any plane form a 2-dimensional

family, and so ψ−1(F) = 2 for all F.

Theorem 8.21. When m > 3, the surfaces of degree m containing no line correspond

to an open subset of Pν .

Proof. We have

dimΓm − dimPν =

m(m + 1)(m + 5)

+ 3 − (m+ 1)(m + 2)(m + 3)

+ 1 = 4 − (m + 1).

Therefore, ifm > 3, then dimΓm < dimPν, and so ψ(Γm) is a proper closed subvariety

of Pν. This proves the claim.

128 Algebraic Geometry: 8. Regular Maps and Their Fibres

We now look at the case m = 2. Here dimΓm = 10, and ν = 9, which suggests

that ψ should be surjective and that its fibres should all have dimension ≥ 1. We

shall see that this is correct.

A quadric is said to be nondegenerate if it is defined by an irreducible polynomial

of degree 2. After a change of variables, any nondegenerate quadric will be defined

by an equation

XW = Y Z.

This is just the image of the Segre mapping (see 5.21)

(a0 : a1), (b0 : b1) _→ (a0b0 : a0b1 : a1b0 : a1b1) : P1 × P1 → P3.

There are two obvious families of lines on P1 ×P1, namely, the horizontal family and

the vertical family; each is parametrized by P1, and so is called a pencil of lines. They

map to two families of lines on the quadric: _

t0X = t1X

t0Y = t1W

and

t0X = t1Y

t0Z = t1W.

Since a degenerate quadric is a surface or a union of two surfaces, we see that every

quadric surface contains a line, that is, that ψ: Γ2 → P9 is surjective. Thus (8.6) tells

us that all the fibres have dimension ≥ 1, and the set where the dimension is > 1 is

a proper closed subset. In fact the dimension of the fibre is > 1 exactly on the set of

reducible F’s, which we know to be closed (see the solution to Homework 9, Problem

1).

It follows from the above discussion that if F is nondegenerate, then ψ−1(F) is

isomorphic to the disjoint union of two lines, ψ−1(F) ≈ P1∪P1. Classically, one defines

a regulus to be a nondegenerate quadric surface together with a choice of a pencil of

lines. One can show that the set of reguli is, in a natural way, an algebraic variety

R, and that, over the set of nondegenerate quadrics, ψ factors into the composite of

two regular maps:

Γ2 − ψ−1(S) = pairs, (F,L) with L on F;

↓

R = set of reguli;

↓

P9 − S = set of nondegenerate quadrics.

The fibres of the top map are connected, and of dimension 1 (they are all isomorphic

to P1), and the second map is finite and two-to-one. Factorizations of this type occur

quite generally (see the Stein factorization theorem (8.25) below).

We now look at the case m = 3. Here dimΓ3 = 19; ν = 19 : we have a map

ψ: Γ3 → P19.

Theorem 8.22. The set of cubic surfaces containing exactly 27 lines corresponds

to an open subset of P19; the remaining surfaces either contain an infinite number of

lines or a nonzero finite number ≤ 27.

Example 8.23. (a) Consider the Fermat surface

0 + X3

1 + X3

2 + X3

3 = 0.

Algebraic Geometry: 8. Regular Maps and Their Fibres 129

Let ζ be a primitive cube root of one. There are the following lines on the surface,

0 ≤ i, j ≤ 2:_

X0 + ζiX1 = 0

X2 + ζjX3 = 0

X0 + ζiX2 = 0

X1 + ζjX3 = 0

X0 + ζiX3 = 0

X1 + ζjX2 = 0

There are three sets, each with nine lines, for a total of 27 lines.

(b) Consider the surface

X1X2X3 = X3

0 .

In this case, there are exactly three lines. To see this, look first in the affine space

where X0 _= 0—here we can take the equation to be X1X2X3 = 1. A line in A3 can

be written in parametric form Xi = ait + bi, but a direct inspection shows that no

such line lies on the surface. Now look where X0 = 0, that is, in the plane at infinity.

The intersection of the surface with this plane is given by X1X2X3 = 0 (homogeneous

coordinates), which is the union of three lines, namely,

X1 = 0; X2 = 0; X3 = 0.

Therefore, the surface contains exactly three lines.

1 + X3

2 = 0.

Here there is a pencil of lines: _

t0X1 = t1X0

t0X2 = −t1X0.

(In the affine space where X0 _= 0, the equation is X3 + Y 3 = 0, which contains the

line X = t, Y = −t, all t.)

We now discuss the proof of Theorem 8.22). If ψ: Γ3 → P19 were not surjective,

then ψ(Γ3) would be a proper closed subvariety of P19, and the nonempty fibres

would all have dimension ≥ 1 (by 8.6), which contradicts two of the above examples.

Therefore the map is surjective22, and there is an open subset U of P19 where the

fibres have dimension 0; outside U, the fibres have dimension > 0.

Given that every cubic surface has at least one line, it is not hard to show that

there is an open subset U_ where the cubics have exactly 27 lines (see Reid, 1988,

pp106–110); in fact, U_ can be taken to be the set of nonsingular cubics. According

to (6.24), the restriction of ψ to ψ−1(U) is finite, and so we can apply (8.10) to see

that all cubics in U − U_ have fewer than 27 lines.

Remark 8.24. The twenty-seven lines on a cubic surface were discovered in 1849

by Salmon and Cayley, and have been much studied—see A. Henderson, The Twenty-

Seven Lines Upon the Cubic Surface, Cambridge University Press, 1911. For example,

it is known that the group of permutations of the set of 27 lines preserving intersections

(that is, such that L∩L_ _=∅ ⇐⇒ σ(L)∩σ(L_) _= ∅) is isomorphic to the Weyl group

of the root system of a simple Lie algebra of type E6, and hence has 25920 elements.

It is known that there is a set of 6 skew lines on a nonsingular cubic surface V. Let

L and L_ be two skew lines. Then “in general” a line joining a point on L to a point

22According to Miles Reid (1988, p126) every adult algebraic geometer knows this proof that

every cubic contains a line.

130 Algebraic Geometry: 8. Regular Maps and Their Fibres

on L_ will meet the surface in exactly one further point. In this way one obtains an

invertible regular map from an open subset of P1 × P1 to an open subset of V , and

hence V is birationally equivalent to P2.

Stein factorization. The following important theorem shows that the fibres of a

proper map are disconnected only because the fibres of finite maps are disconnected.

Theorem 8.25. Let ϕ: W → V be a proper morphism of varieties. It is possible

to factor ϕ into W

→ϕ1 W_ →ϕ2 V with ϕ1 proper with connected fibres and ϕ2 finite.

Proof. This is usually proved at the same time as Zariski’s main theorem (if W

and V are irreducible, and V is affine, then W_ is the affine variety with k[W_] the

integral closure of k[V ] in k(W)).

Algebraic Geometry: 9. Algebraic Geometry over an Arbitrary Field 131

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