an integral domain: (« + t)(« ’ t) = 0. Let us calculate nil(Rt ). For this we take an

element 0 = z = a + b« and calculate

n n’1 1 n n’2 2 2

0 = (a + b«)n = an + b « + ··· .

a b «+ a

1 2

CONCEPTS OF MODERN ALGEBRAIC GEOMETRY 11

Replacing «2 by the positive real number t2 we obtain

n n’2 2 2 n n’4 4 4

0 = (a + b«)n = an + b t + ···

a bt + a +

2 4

n n’1 1 n n’3 3 2

b t + ··· .

+« a b+ a

1 3

From this we conclude that all terms in the ¬rst and in the second sum have to vanish

(all terms have the same sign). This implies a = 0. Regarding the last element in both

sums we see that for t = 0 we get b = 0. Hence nil(Rt ) = {0}, for t = 0 and the ring Rt

is reduced. For t = 0 the value of b is arbitrary. Hence nil(R0 ) = («), which says that

R0 is not a reduced ring. This is the typical situation: a non-reduced coordinate ring

R(V ) corresponds to a variety V which should be considered with higher multiplicity.

For the polynomial ring we have the following very important result.

Hilbertscher Nullstellensatz. Let I be an ideal in Rn = K[X1 , X2 , . . . , Xn ]. If K is

algebraically closed then I(V (I)) = Rad(I).

The proof of this theorem is not easy. The main tool is the following version of the

Nullstellensatz which more resembles his name

Hilbertscher Nullstellensatz. Let I be an ideal in Rn = K[X1 , X2 , . . . , Xn ]. I = Rn .

If K is algebraically closed then V (I) = …. In other words given a set of polynomials

such that the constant polynomial 1 cannot be represented as a Rn ’linear sum in these

polynomials then there is a simultaneous zero of these polynomials.

For the proof let me refer to [Ku].

The Nullstellensatz gives us a correspondence between algebraic sets in Kn and the

radical ideals of Rn = K[X1 , X2 , . . . , Xn ]. If we consider the prime ideals we get

Proposition. Let P be a radical ideal. Then P is a prime ideal if and only if V (P ) is

a variety.

Before we come to the proof of the proposition let me state the following simple obser-

vation. For arbitrary subsets S and T of Kn the ideals I(S) and I(T ) can be de¬ned

completely in the same way as in (1-4), i.e.,

I(S) := {f ∈ Rn | f (x) = 0, ∀x ∈ S} . (2-1)

It is easy to show that

I(S ∪ T ) = I(S) © I(T ), and V (I(S)) = S . (2-2)

12 MARTIN SCHLICHENMAIER

Here S denotes the topological closure of S, which is the smallest (Zariski-)closed subset

of Kn containing S.

Proof of the above proposition. Let P be a prime ideal and set Y = V (P ) then

I(V (P )) = Rad(P ) = P by the Nullstellensatz. Assuming Y = Y1 ∪ Y2 a closed

decomposition of Y then I(Y ) = I(Y1 ∪ Y2 ) = I(Y1 ) © I(Y2 ) = P . Because P is prime

either P = I(Y1 ) or P = I(Y2 ). Assume the ¬rst then Y1 = V (I(Y1 )) = V (P ) = Y

(using that Y1 is closed).

Conversely: let Y = V (P ) be irreducible with P a radical ideal. By the Nullstellensatz

P = Rad(P ) = I(Y ). Let f · g ∈ P then f · g vanishes on Y . We can decompose

Y = (Y © V (f )) ∪ (Y © V (g)) into closed subset of Y . By the irreducibility it has to

coincide with one of them. Assume with the ¬rst. But this implies that V (f ) ⊇ Y and

hence f is identically zero on Y . We get f ∈ I(Y ) = P . This shows that P is a prime

ideal.

Note the fact that we restricted the situation to radical ideals corresponds to the fact

that varieties as sets have always multiplicity 1, hence they are always “reduced”. To

incorporate all ideals and hence “nonreduced structures” we have to use the language

of schemes (see below).

Let us look at the maximal ideals of Rn = K[X1 , X2 , . . . , Xn ]. (Still K is assumed

to be algebraically closed). The same argument as in the two-dimensional case shows

that the ideals

M± = (X1 ’ ±1 , X2 ’ ±2 , . . . , Xn ’ ±n )

are maximal and that Rn /M± ∼ K. This is even true if the ¬eld K is not algebraically

=

closed. Now let M ′ be a maximal ideal. By the Nullstellensatz (here algebraically

closedness is important) there is a common zero ± for all elements f ∈ M ′ . Take f ∈ M ′

then Rn = (f, M ′ ). Now f (±) = 0 would imply that ± is a zero of all polynomials in

Rn which is impossible. Hence, every polynomial f which vanishes at ± lies in M ′ . All

elements in M± have ± as a zero. This implies M± ⊆ M ′ Rn . By the maximality of

M± we conclude M± = M ′ .

Everything can be generalized to an arbitrary variety A over an algebraically closed

¬eld. The points of A correspond to the maximal ideals of Rn lying above the de¬ning

prime ideal P of A. They correspond exactly to the maximal ideals in R(A). All of

them can be given as M± /P . This can be extended to the varieties of Kn lying on A.

They correspond to the prime ideals of Rn lying between the prime ideal P and the

whole ring. They in turn can be identi¬ed with the prime ideals of R(A).

CONCEPTS OF MODERN ALGEBRAIC GEOMETRY 13

Coming back to arbitrary rings it is now quite useful to talk about dimensions.

De¬nition. Let R be a ring. The (Krull-) dimension dim R of a ring R is de¬ned as

the maximal length r of all strict chains of prime ideals Pi in R

P0 P1 P2 . . . Pr R.

Example 1. For a ¬eld K the only (prime) ideals are {0} ‚ K. Hence dim K = 0.

Example 2. The dimension of Rn = K[X1 , X2 , . . . , Xn ] = R(Kn ) is n. This result one

should expect from a reasonable de¬nition of dimension. Indeed we have the chain of

prime ideals

··· (X1 ’ ±1 , X2 ’ ±2 , . . . , Xn ’ ±n )

(0) (X1 ) (X1 , X2 ) Rn .

Hence dim Rn ≥ n. With some more commutative algebra it is possible to show the

equality, see [Ku,S.54].

Example 3. As a special case one obtains dim K[X] = 1. Here the reason is a quite

general result. Recall that K[X] is a principal ideal ring without zero divisors. Hence,

every ideal I can be generated by one element f . Assume I to be a prime ideal, I = {0}

and let M = (g) be a maximal ideal lying above I. We show that I is already maximal.

Because (f ) ⊆ (g) we get f = r · g. But I is prime. This implies either r or g lies in I. If

g ∈ I we are done. If r ∈ I then r = s · f and f = f · s · g. In a ring without zero divisor

one is allowed to cancel common factors. We obtain 1 = s · g. Hence, 1 ∈ M which

contradicts the fact that M is not allowed to be the whole ring. From this it follows

that dim K[X] = 1. Note that we did not make any reference to the special nature of

the polynomial ring here.

What are the conditions on f assuring that the ideal (f ) is prime. The necessary

and su¬cient condition is that f is irreducible but not a unit. This says if there is

decomposition f = g · h then either g or h has to be a unit (i.e. to be invertible)

which in our situation says that g or h must be a constant. This can be seen in the

following way. From the decomposition it follows (using (f ) is prime) that either g or

h has to be in (f ) hence is a multiple of f . By considering the degree we see that the

complementary factor has degree zero and hence is a constant.

Conversely, let f be irreducible but not a unit. Assume g · h ∈ (f ), then g · h = f · r. In

the polynomial ring we have unique factorization (up to units) into irreducible elements.

Hence, the factor f is contained either in g or h. This shows the claim.

Example 4. The ring of integers Z is also a principal ideal ring without zero divisor.

Again we obtain dim Z = 1. In fact, the integers behave very much (at least from

14 MARTIN SCHLICHENMAIER

the point of view of algebraic geometry) like the a¬ne line over a ¬eld. What are the

“points” of Z? As already said the points should correspond to the maximal ideals.

Every prime ideal in Z is maximal. An ideal (n) is prime exactly if n is a prime number.

Hence, the “points” of Z are the prime numbers.

Now we want to introduce the Zariski topology on the set of all prime ideals of a

ring. First we introduce the sets