7. Dimension Theory

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Recall that to an irreducible variety V , we attach a field k(V ) — it is the field of

fractions of k[U] for any open affine subvariety U of V , and also the field of fractions

of OP for any point P in V . We defined the dimension of V to be the transcendence

degree of k(V ) over k. Note that, directly from this definition, dimV = dimU for

any open subvariety U of V . Also, that if W → V is a finite surjective map, then

dimW = dimV (because k(W) is a finite field extension of k(V )).

When V is not irreducible, we defined the dimension of V to be the maximum

dimension of an irreducible component of V , and we said that V is pure of dimension

d if the dimensions of the irreducible components are all equal to d.

In §1 and §3 we proved the following results:

7.1. (a) The dimension of a linear subvariety of An (that is, a subvariety defined

by linear equations) has the value predicted by linear algebra (see 1.20b, 4.11).

In particular, dimAn = n. As a consequence, dimPn = n.

(b) Let Z be a proper closed subset of An; then Z has pure codimension one in An if

and only if I(Z) is generated by a single nonconstant polynomial. Such a variety

is called an affine hypersurface (see 1.21 and 4.25)21.

(c) If V is irreducible and Z is a proper closed subset of V , then dimZ < dimV

(see 1.22).

Affine varieties. The fundamental additional result that we need is that, when we

impose additional polynomial conditions on an algebraic set, the dimension doesn’t

go down by more than linear algebra would suggest.

Theorem 7.2. Let V be an irreducible affine variety, and let f ∈ k[V ]. If f is not

zero or a unit in k[V ], then V (f) is pure of dimension dim(V ) − 1.

Alternatively we can state this as follows:l et V be a closed subvariety of An and

let F ∈ k[X1, . . . ,Xn]; then

V ∩ V (f) =



V if F is identically zero on V

∅ if F has no zeros on V

hypersurface otherwise.

where by hypersurface we mean a closed subvariety of codimension 1.

We can also state it in terms of the algebras:l et A be an affine k-algebra; let

f ∈ A be neither zero nor a unit, and let p be a prime ideal that is minimal among

those containing (f); then

tr degkA/p = tr degkA − 1.

Proof. We begin the proof of Theorem 7.2. Note that we know it already in the

case that V = An (see 7.1b).

We first show that it suffices to prove the theorem in the case that V (f) is irreducible.

Suppose Z0, . . . , Zn are the irreducible components of V (f). We can choose a point

P ∈ Z0 that does not lie on any other Zi (otherwise the decomposition V (f) = ∪Zi

21The cautious reader will check that we didn’t use 4.18 or 4.19i n the proof of 4.25.

110 Algebraic Geometry: 7. Dimension Theory

would be redundant). As Z1, . . . , Zn are closed, there is an open neighbourhood U

of P, which we can take to be affine, that does not meet any Zi except Z0. Now

V (f|U) = Z0 ∩ U, which is irreducible.

As V (f) is irreducible, rad(f) is a prime ideal p ⊂ k[V ]. According to the Noether

normalization theorem (6.14), there is a finite surjective map π : V → Ad, which

realizes k(V ) is a finite extension of the field k(Ad). The idea of the proof is to show

that π(V ) is the zero set of a single element f0 ∈ k[Ad], and to use that we already

know the theorem for Ad.

Lemma 7.3. Let A be an integral domain, and let L be a finite extension of the

field of fractions K of A. If α ∈ L is integral over A, then so also is NmL/Kα.

Hence, if A is integrally closed (e.g., if A is a unique factorization domain), then

NmL/Kα ∈ A. In this last case, α divides NmL/K α in the ring A[α].

Proof. Let g(X) be the minimum polynomial of α over K,

g(X) = Xr + ar−1Xr−1 + · · · + a0.

Then Nmα = ±a

n

r

0 = a

n

r

0 , where n = [L : K]. In some extension field E of L, g(X)

will split

g(X) =

 

(X − αi), α1 = α,

 

αi = ±a0.

Because α is integral over A, g(X) has coefficients in A (see 1.33), and so each αi

is integral over A. Since the elements of E integral over A form a subring of E, it

follows that Nmα is integral over A.

Now suppose A is integrally closed, so that a = Nmα ∈ A. From the equation

0 = α(αr−1 + ar−1αr−2 + · · · + a1) + a0

we see that α divides a0 in A[α], and therefore it also divides a = a

n

r

0 .

Proof. (of 7.2 continued) Let f0 = Nmk(V )/k(Ad) f. According to the lemma, f0

lies in k[Ad], and I claim that p ∩ k[Ad] = rad(f0). The lemma shows that f divides

f0 in k[V ], and so f0 ∈ (f) ⊂ p. Hence rad(f0) ⊂ p∩ k[Ad]. For the reverse inclusion,

suppose that g ∈ p ∩ k[Ad]. Then g ∈ rad(f), and so gm = fh for some h ∈ k[V ],

m ∈ N. Taking norms, we find that gme = Nm(fh) = f0 · Nm(h) ∈ (f0), where

e = [k(V ) : k(An)], which proves the claim.

The inclusion k[V ] 8→ k[Ad] therefore induces an inclusion

k[Ad]/ rad(f0) = k[Ad]/p ∩ k[Ad] 8→ k[V ]/p,

which makes k[V ]/p into a finite algebra over k[Ad]/ rad(f0). Hence

dimV (p) = dimV (f0).

Clearly f _= 0 ⇒ f0 _= 0, and f0 ∈ p ⇒ f0 is not a nonzero constant. Therefore

dimV (f0) = d − 1 by (7.1b).

Corollary 7.4. Let V be an irreducible variety, and let Z be a maximal proper

closed irreducible subset of V. Then dim(Z) = dim(V ) − 1.

Algebraic Geometry: 7. Dimension Theory 111

Proof. For any open affine subset U of V meeting Z, dimU = dimV and dimU ∩

Z = dimZ. We may therefore assume that V itself is affine. Let f be a nonzero

regular function on V vanishing on Z, and let V (f) be the set of zeros of f (in

V ). Then Z ⊂ V (f) ⊂ V , and Z must be an irreducible component of V (f) for

otherwise it wouldn’t be maximal in V . Thus we can apply the theorem to obtain

that dimZ = dimV − 1.

Corollary 7.5 (Topological Characterization of Dimension). Suppose V is irreducible

and that

V  V1  · · ·  Vd _= ∅

is a maximal chain of closed irreducible subsets of V. Then dim(V) = d. (Maximal

means that the chain can’t be refined.)

Proof. From (7.4) we know that

dimV = dimV1 + 1 = dimV2 + 2 = · · · = dimVd + d = d.

Remark 7.6. (a) Recall that the Krull dimension of a ring A is the sup of the

lengths of chains of prime ideals in A. It may be infinite, even when A is Noetherian

(for an example of this, see Nagata, Local Rings, 1962, Appendix A.1). However a

local Noetherian ring has finite Krull dimension, and so

Krull dimA= sup

m maximal

Krull dimAm.

In Nagata’s nasty example, there is a sequence of maximal ideals m1, m2, m3, ... in

A such that the Krull dimension of Ami tends to infinity.

The corollary shows that, when V is affine, dim V = Krull dimk[V ], but it shows

much more. Note that each Vi in a maximal chain (as above) has dimension d − i,

and that any closed irreducible subset of V of dimension d − i occurs as a Vi in a

maximal chain. These facts translate into statements about ideals in affine k-algebras

that do not hold for all Noetherian rings. For example, if A is an affine k-algebra

that is an integral domain, then Krull dimAm is the same for all maximal ideals of A

— all maximal ideals in A have the same height (we have proved 4.19). Moreover, if

p is an ideal in k[V ] with height i, then there is a maximal (i.e., nonrefinable) chain

of prime ideals

(0) _ p1 _ p2 _ · · · _ pd _ k[V ]

with pi = p.

(b) Now that we know that the two notions of dimension coincide, we can restate

(7.2) as follows:l et A be an affine k-algebra; let f ∈ A be neither zero nor a

unit, and let p be a prime ideal that is minimal among those containing (f); then

Krull dimA/p =Krull dimA − 1.

This statement does hold for all Noetherian local rings (see Atiyah and MacDonald

1969, 11.18), and is called Krull’s principal ideal theorem.

Corollary 7.7. Let V be an irreducible variety, and let Z be an irreducible component

of V (f1, . . .fr), where the fi are regular functions on V. Then codim(Z) ≤ r.

112 Algebraic Geometry: 7. Dimension Theory

Proof. As in the proof of (7.4), we can assume V to be affine. We use induction

on r. Because Z is a closed irreducible subset of V (f1, . . .fr−1), it is contained in

some irreducible component Z_ of V (f1, . . . fr−1). By induction, codim(Z_) ≤ r − 1.

Also Z is an irreducible component of Z_ ∩ V (fr) because

Z ⊂ Z_ ∩ V (fr) ⊂ V (f1, . . . , fr)

and Z is a maximal closed irreducible subset of V (f1, . . . , fr). If fr vanishes identically

on Z_, then Z = Z_ and codim(Z) = codim(Z_) ≤ r−1; otherwise, the theorem shows

that Z has codimension 1 in Z_, and codim(Z) = codim(Z_) + 1 ≤ r.

Proposition 7.8. Let V and W be closed subvarieties of An; for any (nonempty)

irreducible component Z of V ∩ W,

dim(Z) ≥ dim(V ) + dim(W) − n;

that is,

codim(Z) ≤ codim(V ) + codim(W).

Proof. In the course of the proof of (3.26), we showed that V ∩W is isomorphic

to Δ ∩ (V ×W), and this is defined by the n equations Xi = Yi in V ×W. Thus the

statement follows from (7.7).

Remark 7.9. (a) The example

_

X2 + Y 2 = Z2

Z = 0

shows that Proposition 7.8 becomes false if one only looks at real points. Also, that

the pictures we draw can mislead.

(b) The statement of (7.8) is false if An is replaced by an arbitrary affine variety.

Consider for example the affine cone V

X1X4 − X2X3 = 0.

It contains the planes,

Z : X2 = 0 = X4; Z = {(∗, 0, ∗, 0)}

Z_ : X1 = 0 = X3; Z_ = {(0, ∗, 0, ∗)}

and Z ∩ Z_ = {(0, 0, 0, 0)}. Because V is a hypersurface in A4, it has dimension 3,

and each of Z and Z_ has dimension 2. Thus

codimZ ∩ Z_ = 3 _ 1 + 1 = codimZ + codimZ_.

The proof of (7.8) fails because the diagonal in V ×V cannot be defined by 3 equations

(it takes the same 4 that define the diagonal in A4)—thus the diagonal is not a settheoretic

complete intersection.

Remark 7.10. In (7.7), the components of V (f1, . . . , fr) need not all have the

same dimension, and it is possible for all of them to have codimension < r without

any of the fi being redundant.

For example, let V be the same affine cone as in the above remark. Note that

V (X1) ∩ V is a union of the planes:

V (X1) ∩ V = {(0, 0, ∗, ∗)} ∪ {(0, ∗, 0, ∗)}.

Algebraic Geometry: 7. Dimension Theory 113

Both of these have codimension 1 in V (as required by (7.2)). Similarly, V (X2) ∩ V

is the union of two planes,

V (X2) ∩ V = {(0, 0, ∗, ∗)} ∪ {(∗, 0, ∗, 0)},

but V (X1,X2) ∩ V consists of a single plane {(0, 0, ∗, ∗)}:it is still of codimension 1

in V , but if we drop one of two equations from its defining set, we get a larger set.

Proposition 7.11. Let Z be a closed irreducible subvariety of codimension r in an

affine variety V . Then there exist regular functions f1, . . . , fr on V such that Z is an

irreducible component of V (f1, . . . , fr) and all irreducible components of V (f1, . . . , fr)

have codimension r.

Proof. We know that there exists a chain of closed irreducible subsets

V ⊃ Z1 ⊃ · · · ⊃ Zr = Z

with codim Zi = i. We shall show that there exist f1, . . . , fr ∈ k[V ] such that, for all

s ≤ r, Zs is an irreducible component of V (f1, . . . , fs) and all irreducible components

of V (f1, . . . , fs) have codimension s.

We prove this by induction on s. For s = 1, take any f1 ∈ I(Z1), f1 _= 0, and apply

Theorem 7.2. Suppose f1, . . . , fs−1 have been chosen, and let Y1 = Zs−1, . . . , Ym, be

the irreducible components of V (f1, . . . , fs−1). We seek an element fs that is identically

zero on Zs but is not identically zero on any Yi—for such an fs, all irreducible

components of Yi ∩V (fs) will have codimension s, and Zs will be an irreducible component

of Y1 ∩ V (fs). But Yi _ Zs for any i (Zs has smaller dimension than Yi), and

so I(Zs) _ I(Yi). Now the prime avoidance lemma (see below) tells us that there

is an element fs ∈ I(Zs) such that fs /∈

I(Yi) for any i, and this is the function we

want.

Lemma 7.12 (Prime Avoidance Lemma). If an ideal a of a ring A is not contained

in any of the prime ideals p1, . . . , pr, then it is not contained in their union.

Proof. We may assume that none of the prime ideals is contained in a second,

because then we could omit it. Fix an i0 and, for each i _= i0, choose an fi ∈ pi, fi /∈

pi0, and choose fi0

∈ a, fi0 /∈

pi0. Then hi0

df =

_

fi lies in each pi with i _= i0 and a,

but not in pi0 (here we use that pi0 is prime). The element

_

hi is therefore in a but

not in any pi.

Remark 7.13. The proposition shows that for a prime ideal p in an affine kalgebra,

if p has height r, then there exist elements f1, . . . , fr ∈ A such that p is

minimal among the prime ideals containing (f1, . . . , fr). This statement is true for

all Noetherian local rings.

Remark 7.14. The last proposition shows that a curve C in A3 is an irreducible

component of V (f1, f2) for some f1, f2 ∈ k[X, Y,Z]. In fact C = V (f1, f2, f3) for

suitable polynomials f1, f2, and f3 — this is an exercise in Shafarevich 1994 (I.6,

Exercise 8); see also Hartshorne 1977, I, Exercise 2.17. Apparently, it is not known

whether two polynomials always suffice to define a curve in A3 — see Kunz 1985,

p136. The union of two skew lines in P3 can’t be defined by two polynomials (ibid.

p140), but it is unknown whether all connected curves in P3 can be defined by two

polynomials. Macaulay (the man, not the program) showed that for every r ≥ 1,

114 Algebraic Geometry: 7. Dimension Theory

there is a curve C in A3 such that I(C) requires at least r generators (see the same

exercise in Hartshorne for a curve whose ideal can’t be generated by 2 elements).

In general, a closed variety V of codimension r in An (resp. Pn) is said to be a settheoretic

complete intersection if there exist r polynomials fi ∈ k[X1, . . . ,Xn] (resp.

homogeneous polynomials fi ∈ k[X0, . . . ,Xn]) such that

V = V (f1, . . . , fr).

Such a variety is said to be an ideal-theoretic complete intersection if the fi can be

chosen so that I(V) = (f1, . . . , fr). Chapter V of Kunz’s book is concerned with

the question of when a variety is a complete intersection. Obviously there are many

ideal-theoretic complete intersections, but most of the varieties one happens to be

interested in turn out not to be. For example, no abelian variety of dimension > 1 is an

ideal-theoretic complete intersection (being an ideal-theoretic complete intersection

imposes constraints on the cohomology of the variety, which are not fulfilled in the

case of abelian varieties).

Let P be a point on an irreducible variety V ⊂ An. Then (7.11) shows that

there is a neighbourhood U of P in An and functions f1, . . . , fr on U such that

U ∩ V = V (f1, . . . , fr) (zero set in U). Thus U ∩ V is a set-theoretic complete

intersection in U. One says that V is a local complete intersection at P ∈ V if there

is an open affine neighbourhood U of P in An such that I(V ∩ U) can be generated

by r regular functions on U. Note that

ideal-theoretic complete intersection⇒ local complete intersection at all p.

It is not difficult to show that a variety is a local complete intersection at every

nonsingular point.

Proposition 7.15. Let Z be a closed subvariety of codimension r in variety V ,

and let P be a point of Z that is nonsingular when regarded both as a point on Z

and as a point on V . Then there is an open affine neighbourhood U of P and regular

functions f1, . . . , fr on U such that Z ∩ U = V (f1, . . . , fr).

Proof. By assumption

dimk TP (Z) = dimZ = dimV − r = dimk TP (V ) − r.

There exist functions f1, . . . , fr contained in the ideal of OP corresponding to Z such

that TP (Z) is the subspace of TP (V ) defined by the equations

(df1)P = 0, . . . , (dfr)P = 0.

All the fi will be defined on some open affine neighbourhood U of P (in V ), and clearly

Z is the only component of Z_ df = V (f1, . . . , fr) (zero set in U) passing through P.

After replacing U by a smaller neighbourhood, we can assume that Z_ is irreducible.

As f1, . . . , fr ∈ I(Z_), we must have TP (Z_) ⊂ TP (Z), and therefore dimZ_ ≤ dimZ.

But I(Z_) ⊂ I(Z∩U), and so Z_ ⊃ Z∩U. These two facts imply that Z_ = Z∩U.

Proposition 7.16. Let V be an affine variety such that k[V ] is a unique factorization

domain. Then every pure closed subvariety Z of V of codimension one is

principal, i.e., I(Z) = (f) for some f ∈ k[V ].

Proof. In (4.25) we proved this in the case that V = An, but the argument only

used that k[An] is a unique factorization domain.

Algebraic Geometry: 7. Dimension Theory 115

Example 7.17. The condition that k[V ] is a unique factorization domain is definitely

needed. Again let V be the cone

X1X4 − X2X3 = 0

in A4 and let Z and Z_ be the planes

Z = {(∗, 0, ∗, 0)} Z

_

= {(0, ∗, 0, ∗)}.

Then Z ∩ Z_ = {(0, 0, 0, 0)}, which has codimension 2 in Z_. If Z = V (f) for some

regular function f on V , then V (f|Z_) = {(0, . . . , 0)}, which is impossible (because

it has codimension 2, which violates 7.2). Thus Z is not principal, and so

k[X1,X2,X3,X4]/(X1X4 − X2X3)

is not a unique factorization domain.

Projective varieties. The results for affine varieties extend to projective varieties

with one important simplification:i f V and W are projective varieties of dimensions

r and s in Pn and r + s ≥ n, then V ∩W _= ∅.

Theorem 7.18. Let V = V (a) ⊂ Pn be a projective variety of dimension ≥ 1,

and let f ∈ k[X0, . . . ,Xn] be homogeneous, nonconstant, and /∈

a; then V ∩ V (f) is

nonempty and of pure codimension 1.

Proof. Since the dimension of a variety is equal to the dimension of any dense

open affine subset, the only part that doesn’t follow immediately from (7.2) is the

fact that V ∩ V (f) is nonempty. Let V aff (a) be the zero set of a in An+1 (that is,

the affine cone over V ). Then V aff (a)∩V aff (f) is nonempty (it contains (0, . . . , 0)),

and so it has codimension 1 in V aff (a). Clearly V aff (a) has dimension ≥ 2, and so

V aff (a) ∩ V aff (f) has dimension ≥ 1. This implies that the polynomials in a have a

zero in common with f other than the origin, and so V (a) ∩ V (f) _= ∅.

Corollary 7.19. Let f1, · · · , fr be homogeneous nonconstant elements of

k[X0, . . . ,Xn]; and let Z be an irreducible component of V ∩ V (f1, . . . fr). Then

codim(Z) ≤ r, and if dim(V ) ≥ r, then V ∩ V (f1, . . .fr) is nonempty.

Proof. Induction on r, as before.

Corollary 7.20. Let α: Pn → Pm be regular; if m < n, then α is constant.

Proof. Let π : An+1 − {origin} → Pn be the map (a0, . . . , an) _→ (a0 : . . . : an).

Then α ◦ π is regular, and there exist polynomials F0, . . . , Fm ∈ k[X0, . . . ,Xn] such

that α ◦ π is the map

(a0, . . . , an) _→ (F0(a) : . . . : Fm(a)).

As α ◦ π factors through Pn, the Fi must be homogeneous of the same degree. Note

that

α(a0 : . . . : an) = (F0(a) : . . . : Fm(a)).

If m < n and the Fi are nonconstant, then (7.18) shows they have a common zero

and so α is not defined on all of Pn. Hence the Fi’s must be constant.

116 Algebraic Geometry: 7. Dimension Theory

Proposition 7.21. Let Z be a closed irreducible subvariety of V ; if codim(Z) = r,

then there exist homogeneous polynomials f1, . . . , fr in k[X0, . . . ,Xn] such that Z is

an irreducible component of V ∩ V (f1, . . . , fr).

Proof. Use the same argument as in the proof (7.11).

Proposition 7.22. Every pure closed subvariety Z of Pn of codimension one is

principal, i.e., I(Z) = (f) for some f homogeneous element of k[X0, . . . ,Xn].

Proof. Follows from the affine case.

Corollary 7.23. Let V and W be closed subvarieties of Pn; if dim(V) +

dim(W) ≥ n, then V ∩ W _= ∅, and every irreducible component of it has

codim(Z) ≤codim(V )+codim(W).

Proof. Write V = V (a) and W = V (b), and consider the affine cones V _ = V (a)

and W_ = W(b) over them. Then

dim(V

_

) + dim(W

_

) = dim(V ) + 1+dim(W) + 1 ≥ n + 2.

As V _ ∩ W_ _= ∅, V _ ∩ W_ has dimension ≥ 1, and so it contains a point other than

the origin. Therefore V ∩ W _= ∅. The rest of the statement follows from the affine

case.

Proposition 7.24. Let V be a closed subvariety of Pn of dimension r < n; then

there is a linear projective variety E of dimension n − r − 1 (that is, E is defined by

r + 1 independent linear forms) such that E ∩ V = ∅.

Proof. Induction on r. If r = 0, then V is a finite set, and the next lemma shows

that there is a hyperplane in kn+1 not meeting V .

Lemma 7.25. Let W be a vector space of dimension d over an infinite field k, and

let E1, . . . , Er be a finite set of nonzero subspaces of W. Then there is a hyperplane

H in W containing none of the Ei.

Proof. Pass to the dual space V of W. The problem becomes that of showing

V is not a finite union of proper subspaces E∨

i . Replace each E∨

i by a hyperplane

Hi containing it. Then Hi is defined by a nonzero linear form Li. We have to show

that

_

Lj is not identically zero on V . But this follows from the statement that a

polynomial in n variables, with coefficients not all zero, can not be identically zero

on kn. (See the first homework exercise.)

Suppose r > 0, and let V1, . . . , Vs be the irreducible components of V . By assumption,

they all have dimension ≤ r. The intersection Ei of all the linear projective

varieties containing Vi is the smallest such variety. The lemma shows that there is a

hyperplane H containing none of the nonzero Ei; consequently, H contains none of

the irreducible components Vi of V , and so each Vi ∩H is a pure variety of dimension

≤ r − 1 (or is empty). By induction, there is an linear subvariety E_ not meeting

V ∩ H. Take E = E_ ∩ H.

Let V and E be as in the theorem. If E is defined by the linear forms L0, . . . , Lr

then the projection a _→ (L0(a) : · · · : Lr(a)) defines a map V → Pr. We shall see

later that this map is finite, and so it can be regarded as a projective version of the

Noether normalization theorem.

Algebraic Geometry: 8. Regular Maps and Their Fibres 117

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