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1. Algebraic Sets

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We now take k to be an algebraically closed field.

Definition of an algebraic set. An algebraic subset V (S) of kn is the set of common

zeros of some set S of polynomials in k[X1, . . . ,Xn]:

V (S) = {(a1, . . . , an) ∈ kn | f(a1, . . . , an) = 0 all f(X1, . . . ,Xn) ∈ S}.

Note that

S ⊂ S

_ ⇒ V (S) ⊃ V (S

);

— the more equations we have, the fewer solutions.

Recall that the ideal a generated by a set S consists of all finite sums

figi, fi ∈ k[X1, . . . ,Xn], gi ∈ S.

Such a sum

figi is zero at any point at which the gi are zero, and so V (S) ⊂ V (a),

but the reverse conclusion is also true because S ⊂ a. Thus V (S) = V (a)—the zero

set of S is the same as that of the ideal generated by S. Hence the algebraic sets can

also be described as the sets of the form V (a), a an ideal in k[X1, . . . ,Xn].

Example 1.1. (a) If S is a system of homogeneous linear equations, then V (S) is

a subspace of kn. If S is a system of nonhomogeneous linear equations, V (S) is either

empty or is the translate of a subspace of kn.

(b) If S consists of the single equation

Y 2 = X3 + aX + b, 4a3 + 27b2 _= 0,

then V (S) is an elliptic curve. For more on elliptic curves, and their relation to

Fermat’s last theorem, see my notes on Elliptic Curves. The reader should sketch the

curve for particular values of a and b. We generally visualize algebraic sets as though

the field k were R.

(d) The algebraic subsets of k are the finite subsets (including ∅) and k itself.

(e) Some generating sets for an ideal will be more useful than others for determining

what the algebraic set is. For example, a Gr¨obner basis for the ideal

a = (X2 + Y 2 + Z2 − 1, X2 + Y 2 − Y, X − Z)

is (according to Maple)

X − Z, Y 2 − 2Y + 1, Z2 −1 + Y.

The middle polynomial has (double) root 1, and it follows easily that V (a) consists

of the single point (0, 1, 0).

The Hilbert basis theorem. In our definition of an algebraic set, we didn’t require

the set S of polynomials to be finite, but the Hilbert basis theorem shows that every

algebraic set will also be the zero set of a finite set of polynomials. More precisely,

the theorem shows that every ideal in k[X1, . . . ,Xn] can be generated by a finite set

of elements, and we have already observed that any set of generators of an ideal has

the same zero set as the ideal.

Algebraic Geometry: 1. Algebraic Sets 15

We sketched an algorithmic proof of the Hilbert basis theorem in the last section.

Here we give the slick proof.

Theorem 1.2 (Hilbert Basis Theorem). The ring k[X1, . . . ,Xn] is Noetherian,

i.e., every ideal is finitely generated.

Proof. For n = 1, this is proved in advanced undergraduate algebra courses:

k[X] is a principal ideal domain, which means that every ideal is generated by a

single element. We shall prove the theorem by induction on n. Note that the obvious

map

k[X1, . . . ,Xn−1][Xn] → k[X1, . . . ,Xn]

is an isomorphism—this simply says that every polynomial f in n variables

X1, . . . ,Xn can be expressed uniquely as a polynomial in Xn with coefficients in

k[X1, . . . ,Xn−1] :

f(X1, . . . ,Xn) = a0(X1, . . . ,Xn−1)Xrn

+ · · · + ar(X1, . . . ,Xn−1).

Thus the next lemma will complete the proof.

Lemma 1.3. If A is Noetherian, then so also is A[X].

Proof. For a polynomial

f(X) = a0Xr + a1Xr−1 + · · · + ar, ai ∈ A, a0 _= 0,

r is called the degree of f, and a0 is its leading coefficient. We call 0 the leading

coefficient of the polynomial 0.

Let a be an ideal in A[X]. The leading coefficients of the polynomials in a form an

ideal a_ in A, and since A is Noetherian, a_ will be finitely generated. Let g1, . . . , gm

be elements of a whose leading coefficients generate a_, and let r be the maximum

degree of the gi.

Now let f ∈ a, and suppose f has degree s > r, say, f = aXs + · · ·. Then a ∈ a_,

and so we can write

a =

biai, bi ∈ A, ai = leading coefficient of gi.

Now

f −

bigiXs−ri, ri = deg(gi),

has degree < deg(f). By continuing in this way, we find that

f ≡ ft mod (g1, . . . , gm)

with ft a polynomial of degree t < r.

For each d < r, let ad be the subset of A consisting of 0 and the leading coefficients

of all polynomials in a of degree d; it is again an ideal in A. Let gd,1, . . . , gd,md

be polynomials of degree d whose leading coefficients generate ad. Then the same

argument as above shows that any polynomial fd in a of degree d can be written

fd ≡ fd−1 mod (gd,1, . . . , gd,md )

with fd−1 of degree ≤ d − 1. On applying this remark repeatedly we find that

ft ∈ (gr−1,1, . . . , gr−1,mr−1, . . . , g0,1, . . . , g0,m0).

16 Algebraic Geometry: 1. Algebraic Sets

Hence

f ∈ (g1, . . . , gm, gr−1,1, . . . , gr−1,mr−1, . . . , g0,1, . . . , g0,m0),

and so the polynomials g1, . . . , g0,m0 generate a.

Aside 1.4. One may ask how many elements are needed to generate an ideal a in

k[X1, . . . ,Xn], or, what is not quite the same thing, how many equations are needed

to define an algebraic set V. When n = 1, we know that every ideal is generated

by a single element. Also, if V is a linear subspace of kn, then linear algebra shows

that it is the zero set of n − dim(V ) polynomials. All one can say in general, is that

at least n − dim(V ) polynomials are needed to define V (see §6), but often more are

required. Determining exactly how many is an area of active research. Chapter V of

Kunz 1985 contains a good discussion of this problem.

The Zariski topology.

Proposition 1.5. There are the following relations:

(a) a ⊂ b ⇒ V (a) ⊃ V (b);

(b) V (0) = kn; V (k[X1, . . . ,Xn]) = ∅;

(d) V (

ai) = ∩V (ai).

Proof. The first two statements are obvious. For (c), note that

ab ⊂ a ∩ b ⊂ a, b ⇒ V (ab) ⊃ V (a ∩ b) ⊃ V (a) ∪ V (b).

For the reverse inclusions, observe that if a /∈ V (a) ∪ V (b), then there exist f ∈ a,

g ∈ b such that f(a) _= 0, g(a) _= 0; but then (fg)(a) _= 0, and so a /∈ V (ab). For (d)

recall that, by definition,

ai consists of all finite sums of the form

fi, fi ∈ ai.

Thus (d) is obvious.

Statements (b), (c), and (d) show that the algebraic subsets of kn satisfy the axioms

to be the closed subsets for a topology on kn:b oth the whole space and the empty set

are closed; a finite union of closed sets is closed; an arbitrary intersection of closed sets

is closed. This topology is called the Zariski topology. It has many strange properties

(for example, already on k one sees that it not Hausdorff), but it is nevertheless of

great importance.

The closed subsets of k are just the finite sets and k. Call a curve in k2 the set

of zeros of a nonzero irreducible polynomial f(X, Y ) ∈ k[X, Y ]. Then we shall see

in (1.25) below that, apart from k2 itself, the closed sets in k2 are finite unions of

(isolated) points and curves. Note that the Zariski topologies on C and C2 are much

coarser (have many fewer open sets) than the complex topologies.

The Hilbert Nullstellensatz. We wish to examine the relation between the algebraic

subsets of kn and the ideals of k[X1, . . . ,Xn], but first we consider the question

of when a set of polynomials has a common zero, i.e., when the equations

g(X1, . . . ,Xn) = 0, g∈ a,

are “consistent”. Obviously, equations

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

Algebraic Geometry: 1. Algebraic Sets 17

are inconsistent if there exist fi ∈ k[X1, . . . ,Xn] such that

figi = 1,

i.e., if 1 ∈ (g1, . . . , gm) or, equivalently, (g1, . . . , gm) = k[X1, . . . ,Xn]. The next

theorem provides a converse to this.

Theorem 1.6 (Hilbert Nullstellensatz). Every proper ideal a in k[X1, . . . ,Xn] has

a z ero in kn.

Proof. A point a ∈ kn defines a homomorphism “evaluate at a”

k[X1, . . . ,Xn] → k, f(X1, . . . ,Xn) _→ f(a1, . . . , an),

and clearly

a ∈ V (a) ⇐⇒ a ⊂ kernel of this map.

Conversely, if ϕ: k[X1, . . . ,Xn] → k is a homomorphism of k-algebras such that

Ker(ϕ) ⊃ a, then

(a1, . . . , an) df = (ϕ(X1), . . . , ϕ(Xn))

lies in V (a). Thus, to prove the theorem, we have to show that there exists a k-algebra

homomorphism k[X1, . . . ,Xn]/a → k.

Since every proper ideal is contained in a maximal ideal, it suffices to prove this for

a maximal ideal m. Then K df = k[X1, . . . ,Xn]/m is a field, and it is finitely generated

as an algebra over k (with generators X1 + m, . . . ,Xn + m). To complete the proof,

we must show K = k. The next lemma accomplishes this.

Although we shall apply the lemma only in the case that k is algebraically closed,

in order to make the induction in its proof work, we need to allow arbitrary k’s in

the statement.

Lemma 1.7 (Zariski’s Lemma). Let k ⊂ K be fields (k not necessarily algebraically

closed). If K is finitely generated as an algebra over k, then K is algebraic over k.

(Hence K = k if k is algebraically closed.)

Proof. We shall prove this by induction on r, the minimum number of elements

required to generate K as a k-algebra. Suppose first that r = 1, so that K = k[x] for

some x ∈ K. Write k[X] for the polynomial ring over k in the single variable X, and

consider the homomorphism of k-algebras k[X] → K, X _→ x. If x is not algebraic

over k, then this is an isomorphism k[X] → K, which contradicts the condition that

K be a field. Therefore x is algebraic over k, and this implies that every element of

K = k[x] is algebraic over k (because it is finite over k).

For the general case, we need to use results about integrality (see the Appendix to

this Section). Consider an integral domain A with field of fractions K, and a field L

containing K. An element of L is said to be integral over A if it satisfies an equation

of the form

Xn + a1Xn−1 + · · · + an = 0, ai ∈ A.

We shall need three facts:

(a) The elements of L integral over A form a subring of L.

18 Algebraic Geometry: 1. Algebraic Sets

(b) If β ∈ L is algebraic over K, then aβ is integral over A for some a ∈ A.

over A lies in A.

Now suppose that K can be generated (as a k-algebra) by r elements, say, K =

k[x1, . . . , xr]. If the conclusion of the lemma is false for K/k, then at least one xi,

say x1, is not algebraic over k. Thus, as before, k[x1] is a polynomial ring in one

variable over k (≈ k[X]), and its field of fractions k(x1) is a subfield of K. Clearly K

is generated as a k(x1)-algebra by x2, . . . , xr, and so the induction hypothesis implies

that x2, . . . , xr are algebraic over k(x1). From (b) we find there exist di ∈ k[x1] such

that dixi is integral over k[x1], i = 2, . . . , r. Write d =

di.

Let f ∈ K; by assumption, f is a polynomial in the xi with coefficients in k.

For a sufficiently large N, dNf will be a polynomial in the dixi. Then (a) implies

that dNf is integral over k[x1]. When we apply this to an element f of k(x1), (c)

shows that dNf ∈ k[x1]. Therefore, k(x1) = ∪Nd−Nk[x1], but this is absurd, because

k[x1] (≈ k[X]) has infinitely many distinct irreducible polynomials7 that can occur

as denominators of elements of k(x1).

The correspondence between algebraic sets and ideals. For a subset W of kn,

we write I(W) for the set of polynomials that are zero on W:

I(W) = {f ∈ k[X1, . . . ,Xn] | f(a) = 0 all a ∈ W}.

It is an ideal in k[X1, . . . ,Xn]. There are the following relations:

(a) V ⊂ W ⇒ I(V ) ⊃ I(W);

(b) I(∅) = k[X1, . . . ,Xn]; I(kn) = 0;

Only the statement I(kn) = 0, i.e., that every nonzero polynomial is nonzero at

some point of kn, is nonobvious. It is not difficult to prove this directly by induction

on the number of variables—in fact it’s true for any infinite field k—but it also follows

easily from the Nullstellensatz (see (1.11a) below).

Example 1.8. Let P be the point (a1, . . . , an). Clearly I(P) ⊃ (X1−a1, . . . ,Xn−

an), but (X1−a1, . . . ,Xn−an) is a maximal ideal, because “evaluation at (a1, . . . , an)”

defines an isomorphism

k[X1, . . . ,Xn]/(X1 − a1, . . . ,Xn − an) → k.

As I(P) _= k[X1, . . . ,Xn], we must have I(P) = (X1 − a1, . . . ,Xn − an).

The radical rad(a) of an ideal a is defined to be

{f | fr ∈ a, some r ∈ N, r>0}.

It is again an ideal, and rad(rad(a)) = rad(a).

An ideal is said to be radical if it equals its radical, i.e., fr ∈ a ⇒ f ∈ a. Equivalently,

a is radical if and only if A/a is a reduced ring, i.e., a ring without nonzero

nilpotent elements (elements some power of which is zero). Since an integral domain

is reduced, a prime ideal (a fortiori a maximal ideal) is radical.

7If k is infinite, then consider the polynomials X − a, and if k is finite, consider the minimum

polynomials of generators of the extension fields of k. Alternatively, and better, adapt Euclid’s proof

that there are infinitely many prime numbers.

Algebraic Geometry: 1. Algebraic Sets 19

If a and b are radical, then a ∩ b is radical, but a + b need not be — consider, for

example, a = (X2 −Y ) and b = (X2 +Y ); they are both prime ideals in k[X, Y ], but

X2 ∈ a + b, X /∈ a + b.

As fr(a) = f(a)r, fr is zero wherever f is zero, and so I(W) is radical. In

particular, IV (a) ⊃ rad(a). The next theorem states that these two ideals are equal.

Theorem 1.9 (Strong Hilbert Nullstellensatz). (a) The ideal IV (a) is the radical

of a; in particular, IV (a) = a if a is a radical ideal.

(b) The set V I(W) is the smallest algebraic subset of kn containing W; in particular,

V I(W) = W if W is an algebraic set.

Proof. (a) We have already noted that IV (a) ⊃ rad(a). For the reverse inclusion,

consider h ∈ IV (a); we have to show that some power of h belongs to a. We may

assume h _= 0 as 0 ∈ a. We are given that h is identically zero on V (a), and we have

to show that hN ∈ a for some N > 0. Let g1, . . . , gm be a generating set for a, and

consider the system of m + 1 equations in n + 1 variables, X1, . . . ,Xn, Y, _

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

1 − Y h(X1, . . . ,Xn) = 0.

If (a1, . . . , an, b) satisfies the first m equations, then (a1, . . . , an) ∈ V (a); consequently,

h(a1, . . . , an) = 0, and (a1, . . . , an, b) doesn’t satisfy the last equation. Therefore,

the equations are inconsistent, and so, according to the original Nullstellensatz,

there exist fi ∈ k[X1, . . . ,Xn, Y ] such that

1 =

i=1

figi + fm+1 · (1 − Y h).

On regarding this as an identity in the field k(X1, . . . ,Xn, Y ) and substituting 1/h

for Y , we obtain the identity

1 =

i=1

fi(X1, . . . ,Xn,

) · gi(X1, . . . ,Xn)

in k(X1, . . . ,Xn). Clearly

fi(X1, . . . ,Xn,

) =

polynomial in X1, . . . ,Xn

hNi

for some Ni. Let N be the largest of the Ni. On multiplying the identity by hN we

obtain an equation

hN =

(polynomial in X1, . . . ,Xn) · gi(X1, . . . ,Xn),

which shows that hN ∈ a.

(b) Let V be an algebraic set containing W, and write V = V (a). Then a ⊂ I(W),

and so V (a) ⊃ V I(W).

Corollary 1.10. The map a _→ V (a) defines a one-to-one correspondence between

the set of radical ideals in k[X1, . . . ,Xn] and the set of algebraic subsets of kn;

its inverse is I.

Proof. We know that IV (a) = a if a is a radical ideal, and that V I(W) = W if

W is an algebraic set.

20 Algebraic Geometry: 1. Algebraic Sets

Remark 1.11. (a) Note that V (0) = kn, and so

I(kn) = IV (0) = rad(0) = 0,

as claimed above.

(b) The one-to-one correspondence in the corollary is order inverting. Therefore

the maximal proper radical ideals correspond to the minimal nonempty algebraic sets.

But the maximal proper radical ideals are simply the maximal ideals in k[X1, . . . ,Xn],

and the minimal nonempty algebraic sets are the one-point sets. As I((a1, . . . , an)) =

(X1 − a1, . . . ,Xn − an), this shows that the maximal ideals of k[X1, . . . ,Xn] are

precisely the ideals of the form (X1 − a1, . . . ,Xn − an).

empty ⇒ rad(a) = k[X1, . . . ,Xn] ⇒ 1 ∈ rad(a) ⇒ 1 ∈ a.

(d) Let W and W_ be algebraic sets. Then W ∩W_ is the largest algebraic subset

contained in both W and W_, and so I(W ∩ W_) must be the smallest radical ideal

containing both I(W) and I(W_). Hence I(W ∩W_) = rad(I(W) + I(W_)).

For example, let W = V (X2 − Y) and W_ = V (X2 + Y ); then I(W ∩ W_) =

rad(X2, Y) = (X, Y ) (assuming characteristic _= 2). Note that W ∩ W_ = {(0, 0)},

but when realized as the intersection of Y = X2 and Y = −X2, it has “multiplicity

2”. [The reader should draw a picture.]

Finding the radical of an ideal. Typically, an algebraic set V will be defined

by a finite set of polynomials {g1, . . . , gs}, and then we shall need to find I(V) =

rad((g1, . . . , gs)).

Proposition 1.12. The polynomial h ∈ rad(a) if and only if 1 ∈ (a, 1−Y h) (the

ideal in k[X1, . . . ,Xn, Y ] generated by the elements of a and 1 − Y h).

Proof. We saw that 1 ∈ (a, 1 − Y h) implies h ∈ rad(a) in the course of proving

(1.9). Conversely, if hN ∈ a, then

1 = Y NhN + (1 − Y NhN)

= Y NhN + (1 − Y h) · (1 + Y h + · · · + Y N−1hN−1) ∈ a + (1 − Y h).

Thus we have an algorithm for deciding whether h ∈ rad(a), but not yet an algorithm

for finding a set of generators for rad(a). There do exist such algorithms (see

Cox et al. 1992, p177 for references), and one has been implemented in the computer

algebra system Macaulay. To start Macaulay on most computers, type: Macaulay;

type <radical to find out the syntax for finding radicals.

The Zariski topology on an algebraic set. We now examine the Zariski topology

on kn and on an algebraic subset of kn more closely. The Zariski topology on Cn

is much coarser than the complex topology. Part (b) of (1.9) says that, for each

subset W of kn, V I(W) is the closure of W, and (1.10) says that there is a oneto-

one correspondence between the closed subsets of kn and the radical ideals of

k[X1, . . . ,Xn].

Let V be an algebraic subset of kn, and let I(V) = a. Then the algebraic subsets

of V correspond to the radical ideals of k[X1, . . . ,Xn] containing a.

Algebraic Geometry: 1. Algebraic Sets 21

Proposition 1.13. Let V be an algebraic subset of kn.

(a) The points of V are closed for the Zariski topology (thus V is a T1-space).

(b) Every descending chain of closed subsets of V becomes constant, i.e., given

V1 ⊃ V2 ⊃ V3 ⊃ · · · (closed subsets of V ),

eventually VN = VN+1 = . . . . Alternatively, every ascending chain of open sets

becomes constant.

Proof. (a) We have already observed that {(a1, . . . , an)} is the algebraic set defined

by the ideal (X1 − a1, . . . ,Xn − an).

(b) A sequence V1 ⊃ V2 ⊃ · · · gives rise to a sequence of radical ideals I(V1) ⊂

I(V2) ⊂ . . . , which eventually becomes constant because k[X1, . . . ,Xn] is Noetherian.

i∈I Ui with each Ui open. Choose an i0 ∈ I; if Ui0

_= V , then there

exists an i1 ∈ I such that Ui0 _ Ui0

∪Ui1. If Ui0

∪Ui1

_= V , then there exists an i2 ∈ I

etc.. Because of (b), this process must eventually stop.

A topological space having the property (b) is said to be Noetherian. The condition

is equivalent to the following:ev ery nonempty set of closed subsets of V has a minimal

element. A space having property (c) is said to be quasi-compact (by Bourbaki at

least; others call it compact, but Bourbaki requires a compact space to be Hausdorff).

The coordinate ring of an algebraic set. Let V be an algebraic subset of kn, and

let I(V) = a. An element f(X1, . . . ,Xn) of k[X1, . . . ,Xn] defines a mapping kn → k,

a _→ f(a) whose restriction to V depends only on the coset f +a of f in the quotient

ring

k[V ] = k[X1, . . . ,Xn]/a = k[x1, . . . , xn].

Moreover, two polynomials f1(X1, . . . ,Xn) and f2(X1, . . . ,Xn) restrict to the same

function on V only if they define the same element of k[V ]. Thus k[V ] can be identified

with a ring of functions V → k.

We call k[V ] the ring of regular functions on V , or the coordinate ring of V . It

is a finitely generated reduced k-algebra (because a is radical), but need not be an

integral domain.

For an ideal b in k[V ], we set

V (b) = {a ∈ V | f(a) = 0, all f ∈ b}.

Let W = V (b). The maps

k[X1, . . . ,Xn] → k[V] =

k[X1, . . . ,Xn]

→ k[W] =

k[V ]

should be regarded as restricting a function from kn to V , and then restricting that

function to W.

Write π for the map k[X1, . . . ,Xn] → k[V ]. Then b _→ π−1(b) is a bijection

from the set of ideals of k[V ] to the set of ideals of k[X1, . . . ,Xn] containing a,

under which radical, prime, and maximal ideals correspond to radical, prime, and

22 Algebraic Geometry: 1. Algebraic Sets

maximal ideals (each of these conditions can be checked on the quotient ring, and

k[X1, . . . ,Xn]/π−1(b) ≈ k[V ]/b). Clearly

V (π

−1(b)) = V (b),

and so b _→ V (b) gives a bijection between the set of radical ideals in k[V ] and the

set of algebraic sets contained in V .

For h ∈ k[V ], we write

D(h) = {a ∈ V | h(a) _= 0}.

It is an open subset of V , because it is the complement of V ((h)).

Proposition 1.14. (a) The points of V are in one-to-one correspondence with

the maximal ideals of k[V ].

(b) The closed subsets of V are in one-to-one correspondence with the radical ideals

of k[V ].

open, and each open set is a union (in fact, a finite union) of D(h)’s.

Proof. (a) and (b) are obvious from the above discussion. For (c), we have already

observed that D(h) is open. Any other open set U ⊂ V is the complement of a set of

the form V (b), b an ideal in k[V ]. If f1, . . . , fm generate b, then U = ∪D(fi).

The D(h) are called the basic (or principal) open subsets of V . We sometimes

write Vh for D(h). Note that D(h) ⊂ D(h_) ⇐⇒ V (h) ⊃ V (h_) ⇐⇒ rad((h)) ⊂

rad((h_)) ⇐⇒ hr ∈ (h_) some r ⇐⇒ hr = h_g, some g.

Some of this should look familiar:if V is a topological space, then the zero set of a

family of continuous functions f : V → R is closed, and the set where such a function

is nonzero is open.

Irreducible algebraic sets. A nonempty subset W of a topological space V is said

to be irreducible if it satisfies any one of the following equivalent conditions:

(a) W is not the union of two proper closed subsets;

(b) any two nonempty open subsets of W have a nonempty intersection;

The equivalences (a) ⇐⇒ (b) and (b) ⇐⇒ (c) are obvious. Also, one sees that

if W is irreducible, and W = W1 ∪ . . . ∪ Wr with each Wi closed, then W = Wi for

some i.

This notion is not useful for Hausdorff topological spaces, because such a space

is irreducible only if it consists of a single point — otherwise any two points have

disjoint open neighbourhoods, and so (b) fails.

Proposition 1.15. An algebraic set W is irreducible and only if I(W) is prime.

Proof. ⇒:Supp ose fg ∈ I(W). At each point of W, either f or g is zero, and

so W ⊂ V (f) ∪ V (g). Hence

W = (W ∩ V (f)) ∪ (W ∩ V (g)).

As W is irreducible, one of these sets, say W ∩ V (f), must equal W. But then

f ∈ I(W). Thus I(W) is prime.

Algebraic Geometry: 1. Algebraic Sets 23

⇐=:Supp ose W = V (a) ∪ V (b) with a and b radical ideals—we have to show

that W equals V (a) or V (b). Recall that V (a) ∪ V (b) = V (a ∩ b), and that a ∩ b is

radical; hence I(W) = a ∩ b. If W _= V (a), then there is an f ∈ a, f /∈ I(W). But

fg ∈ a ∩ b = I(W) for all g ∈ b, and, because f /∈ I(W) and I(W) is prime, this

implies that b ⊂ I(W); therefore W ⊂ V (b).

Thus, there are one-to-one correspondences

radical ideals ↔ algebraic subsets

prime ideals ↔ irreducible algebraic subsets

maximal ideals ↔ one-point sets.

These correspondences are valid whether we mean ideals in k[X1, . . . ,Xn] and algebraic

subsets of kn, or ideals in k[V ] and algebraic subsets of V. Note that

the last correspondence implies that the maximal ideals in k[V ] are of the form

(x1 − a1, . . . , xn − an), (a1, . . . , an) ∈ V .

Example 1.16. Let f ∈ k[X1, . . . ,Xn]. As we noted in §0, k[X1, . . . ,Xn] is a

unique factorization domain, and so (f) is a prime ideal ⇐⇒ f is irreducible. Thus

V (f) is irreducible ⇐⇒ f is irreducible.

On the other hand, suppose f factors, f =

fmi

i , with the fi distinct irreducible

polynomials. Then (f) = ∩(fmi

i ), rad((f)) = (

fi) = ∩(fi), and V (f) = ∪V (fi)

with V (fi) irreducible.

Proposition 1.17. Let V be a Noetherian topological space. Then V is a finite

union of irreducible closed subsets, V = V1 ∪. . .∪Vm. Moreover, if the decomposition

is irredundant in the sense that there are no inclusions among the Vi, then the Vi are

uniquely determined up to order.

Proof. Suppose the first assertion is false. Then, because V is Noetherian, there

will be a closed subset W of V that is minimal among those that cannot be written as

a finite union of irreducible closed subsets. But such a W cannot itself be irreducible,

and so W = W1∪W2, with each Wi a proper closed subset of W. From the minimality

of W, it follows that each Wi is a finite union of irreducible closed subsets, and so

therefore is W. We have arrived at a contradiction.

Suppose that V = V1∪. . .∪Vm = W1∪. . .∪Wn are two irredundant decompositions.

Then Vi = ∪j(Vi ∩ Wj), and so, because Vi is irreducible, Vi ⊂ Vi ∩ Wj for some j.

Consequently, there is a function f : {1, . . . ,m} → {1, . . . , n} such that Vi ⊂ Wf(i) for

each i. Similarly, there is a function g : {1, . . . , n} → {1, . . . ,m} such that Wj ⊂ Vg(j).

Since Vi ⊂ Wf(i) ⊂ Vgf(i), we must have gf(i) = i and Vi = Wf(i); similarly fg = id.

Thus f and g are bijections, and the decompositions differ only in the numbering of

the sets.

The Vi given uniquely by the proposition are called the irreducible components of

V . They are the maximal closed irreducible subsets of V . In Example 1.16, the V (fi)

are the irreducible components of V (f).

Corollary 1.18. A radical ideal a of k[X1, . . . ,Xn] is a finite intersection of

prime ideals, a = p1 ∩ . . . ∩ pn; if there are no inclusions among the pi, then the pi

are uniquely determined up to order.

24 Algebraic Geometry: 1. Algebraic Sets

Proof. Write V (a) = ∪Vi, and take pi = I(Vi).

Remark 1.19. (a) In a Noetherian ring, every ideal a has a decomposition into

primary ideals: a = ∩qi (see Atiyah and MacDonald 1969, IV, VII). For radical ideals,

this becomes a much simpler decomposition into prime ideals, as in the corollary.

(b) In k[X], (f(X)) is radical if and only if f is square-free, in which case f is a

product of distinct irreducible polynomials, f = p1 . . . pr, and (f) = (p1) ∩ . . . ∩ (pr)

(a polynomial is divisible by f if and only if it is divisible by each pi).

irreducible components are the one-point sets.

Dimension. We briefly introduce the notion of the dimension of an algebraic variety.

In Section 7 we shall discuss this in more detail.

Let V be an irreducible algebraic subset. Then I(V ) is a prime ideal, and so k[V ]

is an integral domain. Let k(V ) be its field of fractions—k(V ) is called the field of

rational functions on V. The dimension of V is defined to be the transcendence

degree of k(V ) over k.

For those who know some commutative algebra, according to the last theorem in

Atiyah and MacDonald 1969, this is equal to the Krull dimension of k[V ]; we shall

prove this later.

Example 1.20. (a) Let V = kn; then k(V) = k(X1, . . . ,Xn), and so dim(V) =

n. Later we shall see that the Noether normalization theorem implies that V has

dimension n if and only if there is a surjective finite-to-one map V → kn.

(b) If V is a linear subspace of kn (or a translate of such a subspace), then it is

an easy exercise to show that the dimension of V in the above sense is the same as

its dimension in the sense of linear algebra (in fact, k[V ] is canonically isomorphic to

k[Xi1, . . . ,Xid ] where the Xij are the “free” variables in the system of linear equations

defining V ).

In linear algebra, we justify saying V has dimension n by pointing out that its

elements are parametrized by n-tuples; unfortunately, it is not true in general that

the points of an algebraic set of dimension n are parametrized by n-tuples; the most

one can say is that there is a finite-to-one map to kn.

point. Certainly, for any point P ∈ kn, k[P] = k, and so k(P) = k. Conversely,

suppose V = V (p), p prime, has dimension 0. Then k(V ) is an algebraic extension of

k, and so equals k. From the inclusions

k ⊂ k[V ] ⊂ k(V) = k

we see that k[V ] = k. Hence p is maximal, and we saw in (1.11b) that this implies

that V (p) is a point.

The zero set of a single nonconstant nonzero polynomial f(X1, . . . ,Xn) is called a

hypersurface in kn.

Proposition 1.21. An irreducible hypersurface in kn has dimension n − 1.

Proof. Let k[x1, . . . , xn] = k[X1, . . . ,Xn]/(f), xi = Xi + p, and let k(x1, . . . , xn)

be the field of fractions of k[x1, . . . , xn]. Since x1, . . . , xn generate k(x1, . . . , xn) and

Algebraic Geometry: 1. Algebraic Sets 25

they are algebraically dependent, the transcendence degree must be < n (because

{x1, . . . , xn} contains a transcendence basis — see 6.12 of my notes on Fields and

Galois Theory). To see that it is not < n− 1, note that if Xn occurs in f, then it

occurs in all nonzero multiples of f, and so no nonzero polynomial in X1, . . . ,Xn−1

belongs to (f). This means that x1, . . . , xn−1 are algebraically independent.

For a reducible algebraic set V , we define the dimension of V to be the maximum of

the dimensions of its irreducible components. When these all have the same dimension

d, we say that V has pure dimension d.

Proposition 1.22. If V is irreducible and Z is a proper closed subvariety of V ,

then dim(Z) < dim(V ).

Proof. We may assume that Z is irreducible. Then Z corresponds to a nonzero

prime ideal p in k[V ], and k[Z] = k[V ]/p.

Suppose V ⊂ kn, so that k[V] = k[X1, . . . ,Xn]/I(V) = k[x1, . . . , xn]. If Xi is

regarded as a function on kn, then its image xi in k[V ] is the restriction of this

function to V .

Let f ∈ k[V ]. The image ￣ f of f in k[V ]/p = k[Z] can be regarded as the restriction

of f to Z. With this notation, k[Z] = k[￣x1, . . . , ￣xn]. Suppose that dimZ = d

and that ￣x1, . . . , ￣xd are algebraically independent. I will show that, for any nonzero

f ∈ p, the d + 1 elements x1, . . . , xd, f are algebraically independent, which implies

that dimV ≥ d + 1.

Suppose otherwise. Then there is a nontrivial algebraic relation among the xi and

f, which we can write

a0(x1, . . . , xd)fm + a1(x1, . . . , xd)fn−1 + · · · + am(x1, . . . , xd) = 0,

with ai(x1, . . . , xd) ∈ k[x1, . . . , xd]. Because the relation is nontrivial, at least one of

the ai is nonzero (in the polynomial ring k[x1, . . . , xd]). After cancelling by a power

of f if necessary, we can assume am(x1, . . . , xd) _= 0 (in this step, we use that k[V ]

is an integral domain). On restricting the functions in the above equality to Z, i.e.,

applying the homomorphism k[V ] → k[Z], we find that

am(￣x1, . . . , ￣xd) = 0,

which contradicts the algebraic independence of ￣x1, . . . , ￣xd.

Example 1.23. Let F(X, Y ) and G(X, Y ) be nonconstant polynomials with no

common factor. Then V (F(X, Y )) has dimension 1 by (1.21), and so V (F(X, Y )) ∩

V (G(X, Y )) must have dimension zero; it is therefore a finite set.

Remark 1.24. Later we shall show that if, in the situation of (1.22), Z is a maximal

proper irreducible subset of V , then dimZ = dimV − 1. This implies that the

dimension of an algebraic set V is the maximum length of a chain

V0 _ V1 _ · · · _ Vd

with each Vi closed and irreducible and V0 an irreducible component of V. Note that

this description of dimension is purely topological—it makes sense for any Noetherian

topological space.

26 Algebraic Geometry: 1. Algebraic Sets

On translating the description in terms of ideals, we see immediately that the

dimension of V is equal to the Krull dimension of k[V ]—the maximal length of a

chain of prime ideals,

pd _ pd−1 _ · · · _ p0.

Example 1.25. We classify the irreducible closed subsets V of k2. If V has dimension

2, then (by 1.22) it can’t be a proper subset of k2, so it is k2. If V has

dimension 1, then V _= k2, and so I(V ) contains a nonzero polynomial, and hence

a nonzero irreducible polynomial f (being a prime ideal). Then V ⊃ V (f), and so

equals V (f). Finally, if V has dimension zero, it is a point. Correspondingly, we can

make a list of all the prime ideals in k[X, Y ]:the y have the form (0), (f) (with f

irreducible), or (X − a, Y − b).

Appendix A: Integrality. Throughout this subsection, A is an integral domain.

An element α of a field L containing A is said to be integral over A if it is a root of

a monic polynomial with coefficients in A, i.e., if it satisfies an equation

αn + a1αn−1 + . . . + an = 0, ai ∈ A.

Before proving that the elements of L integral over A form a ring, we need to review

symmetric polynomials.

Symmetric polynomials. A polynomial P(X1, ...,Xr) ∈ A[X1, . . . ,Xr] is said to

be symmetric if it is unchanged when its variables are permuted, i.e., if

P(Xσ(1), . . . ,Xσ(r)) = P(X1, . . . ,Xr), all σ ∈ Symr.

For example

S1 =

Xi, S2 =

i<j

XiXj, . . . , Sr = X1 · · · Xr,

are all symmetric. These particular polynomials are called the elementary symmetric

polynomials.

Theorem 1.26 (Symmetric function theorem). Let A be a ring. Every symmetric

polynomial P(X1, ...,Xr) in A[X1, ...,Xr] is equal to a polynomial in the symmetric

elementary polynomials with coefficients in A, i.e., P ∈ A[S1, ..., Sr].

Proof. We define an ordering on the monomials in the Xi by requiring that

Xi1

1 Xi2

· · ·Xir

r >Xj1

1 Xj2

· · ·Xjr

if either

i1 + i2 + · · · + ir > j1 + j2 + · · · + jr

or equality holds and, for some s,

i1 = j1, . . . , is = js, but is+1 > js+1.

Let Xk1

· · ·Xkr

r be the highest monomial occurring in P with a coefficient c _= 0.

Because P is symmetric, it contains all monomials obtained from Xk1

· · ·Xkr

r by

permuting the X’s. Hence k1 ≥ k2 ≥ · · · ≥ kr.

Algebraic Geometry: 1. Algebraic Sets 27

Clearly, the highest monomial in Si is X1 · · ·Xi, and it follows that the highest

monomial in Sd1

· · · Sdr

r is

Xd1+d2+···+dr

1 Xd2+···+dr

· · ·Xdr

r .

Therefore

P(X1, . . . ,Xr) − cSk1−k2

1 Sk2−k3

· · · Skr

r < P(X1, . . . ,Xr).

We can repeat this argument with the polynomial on the left, and after a finite number

of steps, we will arrive at a representation of P as a polynomial in S1, . . . , Sr. (For

more details, see Jacobson, Basic Algebra I, 2.20, p139.)

Let f(X) = Xn + a1Xn−1 + · · · + an ∈ A[X], and let α1, . . . , αn be the roots of

f(X) in some ring containing A, i.e., f(X) =

(X − αi). Then

a1 = −S1(α1, . . . , αn), a2 = S2(α1, . . . , αn), . . . , an = ±Sn(α1, . . . , αn).

Thus the elementary symmetric polynomials in the roots of f(X) lie in A, and so the

theorem implies that every symmetric polynomial in the roots of f(X) lie in A.

Integral elements.

Theorem 1.27. The set of elements of L integral over A forms a ring.

Proof. Let α and β be integral over A; we have to show that α ± β and αβ are

integral over A. Let Ω be an algebraically closed field containing L.

We are given that α is a root of a polynomial f(X) = Xm + a1Xm−1 + · · · + am,

ai ∈ A. Write

f(X) =

(X − αi), αi ∈ Ω.

Similarly, β is a root of polynomial g(X) = Xn + b1Xn−1 + ... +bn, bi ∈ A, and we

write

f(X) =

(X − βi), βi ∈ Ω.

Let γ1, γ2, ..., γmn be the family of numbers of the form αi + βj (or αi − βj, or αiβj).

I claim that

h(X) df =

1≤i≤m, 1≤j≤n

(X − γij)

has coefficients in A. This will prove that α+β is integral over A because h(α+β) = 0.

The coefficients of h are symmetric in the αi and βj. Let P(α1, ..., αm, β1, ..., βn) be

one of these coefficients, and regard it as a polynomial Q(β1, ..., βn) in the β’s with

coefficients in A[α1, ..., αm]; then its coefficients are symmetric in the αi, and so lie in

A. Thus P(α1, ..., αm, β1, ..., βn) is a symmetric polynomial in the β’s with coefficients

in A—it therefore lies in A, as claimed.

Definition 1.28. The ring of elements of L integral over A is called the integral

closure of A in L.

Proposition 1.29. Let K be the field of fractions of A, and let L be a field containing

K. If α ∈ L is algebraic over K, then there exists a d ∈ A such that dα is

integral over A.

28 Algebraic Geometry: 1. Algebraic Sets

Proof. By assumption, α satisfies an equation

αm + a1αm−1 + · · · + am = 0, ai ∈ K.

Let d be a common denominator for the ai, so that dai ∈ A, all i, and multiply

through the equation by dm:

dmαm + a1dmαm−1 + · · · + amdm = 0.

We can rewrite this as

(dα)m + a1d(dα)m−1 + · · · + amdm = 0.

As a1d, . . . , amdm ∈ A, this shows that dα is integral over A.

Corollary 1.30. Let A be an integral domain with field of fractions K, and let

L be an algebraic extension of K. If B is the integral closure of A in L, then L is the

field of fractions of B.

Proof. The proposition shows that every α ∈ L can be written α = β/d with

β ∈ B, d ∈ A.

Definition 1.31. A ring A is integrally closed if it is its own integral closure in

its field of fractions K, i.e., if

α ∈ K, α integral over A ⇒ α ∈ A.

Proposition 1.32. A unique factorization domain (e.g. a principal ideal domain)

is integrally closed.

Proof. Suppose a/b, a, b ∈ A, is an element of the field of fractions of A and is

integral over A. If b is a unit, then a/b ∈ A. Otherwise we may suppose that there

is an irreducible element p of A dividing b but not a. As a/b is integral over A, it

satisifies an equation

(a/b)n + a1(a/b)n−1 + · · · + an = 0, ai ∈ A.

On multiplying through by bn, we obtain the equation

an + a1an−1b + ... + anbn = 0.

The element p then divides every term on the left except an, and hence must divide

an. Since it doesn’t divide a, this is a contradiction.

Proposition 1.33. Let K be the field of fractions of A, and let L be an extension

of K of finite degree. Assume A is integrally closed. An element α of L is integral

over A if and only if its minimum polynomial over K has coefficients in A.

Proof. Assume α is integral over A, so that

αm + a1αm−1 + ... + am = 0, some ai ∈ A.

Let α_ be a conjugate of α, i.e., a root of the minimum polynomial of α over K. Then

there is an K-isomorphism8

σ : K[α] → K[α

], σ(α) = α

8If f(X) is the minimum polynomial of α, hence also of α_, over K, then the map h(X) _→

h(α) : K[X] → K[α] induces an isomorphism τ : K[X]/(f(X)) → K[α]. Similarly, h(X) _→ h(α_) :

K[X] → K[α_] induces an isomorphism τ _ : K[X]/(f(X)) → K[α_], and we set σ = τ _ ◦ τ−1.

On applying σ to the above equation we obtain the equation

_m + a1α

_m−1 + ... + am = 0,

which shows that α_ is integral over A. Hence all the conjugates of α are integral over

A, and it follows from (1.27) that the coefficients of f(X) are integral over A. They

lie in K, and A is integrally closed, and so they lie in A. This proves the “only if”

part of the statement, and the “if” part is obvious.

Appendix B: Transcendence degree. I have deleted this subsection from the

notes since it was merely a copy of Section 6 of my notes Fields and Galois Theory.

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