examples of principal ideal rings: Z the integers, and K[X] the polynomial ring in one

variable over the ¬eld K. Let me recall the proof for Z. Take I an ideal of Z. If I = {0}

we are done. Hence assume I = {0} then there is a n ∈ N with n ∈ I minimal. We

now claim I = (n). To see this take m ∈ I. By the division algorithm of Euklid there

are q, r ∈ Z with 0 ¤ r < n such that m = qn + r . Hence, with m and n in I we

get r = m ’ qn ∈ I. But n was chosen minimal, hence r = 0 and m ∈ (n). Note that

the proof for K[X] is completely analogous if we replace the division algorithm for the

integers by the division algorithm for polynomials.

Now we have

Hilbertscher Basissatz. Let R be a noetherian Ring. Then R[X] is also noetherian.

As a nice exercise you may try to proof it by yourself (maybe guided by [Ku]).

Remark 3. If R is a noncommutative ring one has to deal with left, right and two-sided

ideals. It is also necessary to de¬ne left, right, and two-sided noetherian.

It is time to give some examples of algebraic sets:

(1) The whole a¬ne space is the zero set of the zero ideal: Kn = V (0).

(2) The empty set is the zero set of the whole ring Rn : … = V ((1)).

(3) Let ± = (±1 , ±2 , . . . , ±n ) ∈ Kn be a point given by its coordinates. De¬ne the ideal

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

then {±} = V (I± ).

(4) Now take 2 points ±, β and their associated ideals I± , Iβ as de¬ned in (3). Then

I± © Iβ is again an ideal and we get {±, β} = V (I± © Iβ ) .

This is a general fact. Let A = V (I) and B = V (J) be two algebraic sets then the

union A ∪ B is again an algebraic set because A ∪ B = V (I © J). Let me give a proof

of this. Obviously, we get for two ideals K and L with K ⊆ L for their vanishing sets

CONCEPTS OF MODERN ALGEBRAIC GEOMETRY 5

V (K) ⊇ V (L). Hence because I © J ⊆ I and I © J ⊆ J we obtain V (I © J) ⊇

V (I) ∪ V (J). To proof the other inclusion assume that there is an x ∈ V (I) ∪ V (J)

then there are f ∈ I and g ∈ J with f (x) = 0 and g(x) = 0. Now f · g ∈ I © J but

(f · g)(x) = f (x) · g(x) = 0. Hence x ∈ V (I © J). Let me repeat the result for further

/

reference:

V (I) ∪ V (J) = V (I © J) . (1-2)

(5) A hypersurface H is the vanishing set of the ideal generated by a single polynomial

f : H = V ((f )). An example in C2 is given by I = (Y 2 ’ 4X 3 + g2 X + g3 ) where

g2 , g3 ∈ C. The set V (I) de¬nes a cubic curve in the plane. For general g2 , g3 this curve

is isomorphic to a (complex) one-dimensional torus with the point 0 removed.

(6) Linear a¬ne subspaces are algebraic sets. A linear a¬ne subspace of Kn is the

set of solutions of a system of linear equations A · x = b with

« «

b1

a1,—

ai,— ∈ Kn ,

A = ··· , b = ··· , bi ∈ K , i = 1, . . . , r .

br

ar,—

The solutions (by de¬nition) are given as the elements of the vanishing set of the ideal

I = (a1,— · X ’ b1 , a2,— · X ’ b2 , · · · , ar,— · X ’ br ) .

(7) A special case are the straight lines in the plane. For this let li = ai,1 X +ai,2 Y ’bi ,

i = 1, 2 be two linear forms. Then Li = V ((li )), i = 1, 2 are lines. For the union of the

two lines we obtain by (1-2)

L1 ∪ L2 = V ((l1 ) © (l2 )) = V ((l1 · l2 )) .

Note that I do not claim (l1 ) © (l2 ) = (l1 · l2 ). The reader is encouraged to search for

conditions when this will hold. For the intersection of the two lines we get L1 © L2 =

V ((l1 , l2 )) which can be written as V ((l1 ) + (l2 )). Of course, this set consists just of one

point if the linear forms l1 and l2 are linearly independent. Again, there is the general

fact

V (I) © V (J) = V (I + J), (1-3)

where

I + J := { f + g | f ∈ I, g ∈ J } .

You see there is a ample supply of examples for algebraic sets. Now we introduce

for Kn a topology, the Zariski-Topology. For this we call a subset U open if it is a

complement of an algebraic set, i.e. U = Kn \ V (I) where I is an ideal of Rn . In

6 MARTIN SCHLICHENMAIER

other words: the closed sets are the algebraic sets. It is easy to verify the axioms for a

topology:

(1) Kn and … are open.

(2) Finite intersections are open:

U1 © U2 = (Kn \ V (I1 )) © (Kn \ V (I2 )) = Kn \ (V (I1 ) ∪ V (I2 )) = Kn \ V (I1 © I2 ) .

(3) Arbitrary unions are open:

(Kn \ V (Ii )) = Kn \ V (Ii ) = Kn \ V ( Ii ) .

i∈S i∈S i∈S

Here S is allowed to be an in¬nite index set. The ideal i∈S Ii consists of elements

in Rn which are ¬nite sums of elements belonging to di¬erent Ii . The claim (1-3) easily

extends to this setting.

Let us study the a¬ne line K. Here R1 = K[X]. All ideals in K[X] are principal

ideals, i.e. generated by just one polynomial. The vanishing set of an ideal consists just

of the ¬nitely many zeros of this polynomial (if it is not identically zero). Conversely,

for every set of ¬nitely many points there is a polynomial vanishing exactly at these

points. Hence, beside the empty-set and the whole line the algebraic sets are the sets

of ¬nitely many points. At this level there is already a new concept showing up. The

polynomial assigned to a certain point set is not unique. For example it is possible to

increase the vanishing order of the polynomial at a certain zero without changing the

vanishing set. It would be better to talk about point sets with multiplicities to get a

closer correspondence to the polynomials. Additionally, if K is not algebraically closed

then there are non-trivial polynomials without any zero at all. These ideas we will take

up in later lectures. The other important observation is that the open sets in K are

either empty or dense. The latter says that the closure U of U , i.e. the smallest closed

set which contains U is the whole space K. Assuming the whole space to be irreducible

this is true in a more general context.

De¬nition.

(a) Let V be a closed set. V is called irreducible if for every decomposition V = V1 ∪V2

with V1 , V2 closed we have V1 = V or V2 = V .

(b) An algebraic set which is irreducible is called a variety.

Now let U be an open subset of an irreducible V . The two set V \ U and U are

closed and V = (V \ U ) ∪ U . Hence, V has to be one of these sets. Hence, either U = …

or V = U . As promised, this shows that every open subset of an irreducible space is

either empty or dense. Note that this has nothing to do with our special situation. It

CONCEPTS OF MODERN ALGEBRAIC GEOMETRY 7

follows from general topological arguments. In the next section we will see that the

spaces Kn are irreducible.

Up to now we were able to describe our geometric objects with the help of the ring

of polynomials. This ring plays another important role in the whole theory. We need

it to study polynomial (algebraic) functions on Kn . If f ∈ Rn is a polynomial then

x ’ f (x) de¬nes a map from Kn to K. This can be extended to functions on algebraic

sets A = V (I). We associate to A the quotient ring

R(A) := K[X1 , X2 , . . . , Xn ]/I .

This ring is called the coordinate ring of A. The elements of R(A) can be considered

¯ ¯

as functions on A. Take x ∈ A, and f ∈ R(A) then f (x) := f (x) is a well-de¬ned

¯

element of K. Assume f = g then there is an h ∈ I with f = g + h hence f (x) =

¯

g(x) + h(x) = g(x) + 0. You might have noticed that it is not really correct to call this

ring the coordinate ring of A. It is not clear, in fact it is not even true that the ideal I is

¬xed by the set A. But R(A) depends on I. A ¬rst way to avoid these complications

is to assign to every A a unique de¬ning ideal,