Max(R) := { P | P is a maximal ideal of R } .

The set Spec(R) contains in some sense all irreducible “subvarieties” of the “geometric

model” of R. Let S be an arbitrary subset of R. We de¬ne the associated subset of

Spec(R) as the set consisting of the prime ideals which contain S:

V (S) := { P ∈ Spec(R) | P ⊇ S } . (2-3)

The subsets of Spec(R) obtained in this way are called the closed subsets. It is obvious

that S ⊆ T implies V (S) ⊇ V (T ) . Clearly, V (S) depends only of the ideal generated

by S: V ((S)) = V (S).

This de¬nes a topology on Spec(R) the Zariski topology.

(1) The whole space and the empty set are closed: V (0) = Spec(R) and V (1) = ….

(2) Arbitrary intersections of closed sets are again closed:

V (Si ) = V ( S) . (2-4)

i∈J i∈J

(3) Finite unions of closed set are again closed:

V (S1 ) ∪ V (S2 ) = V ((S1 ) © (S2 )) . (2-5)

Let me just show (2-5) here. Because (S1 ), (S2 ) ⊇ (S1 ) © (S2 ) we get V (S1 ) ∪ V (S2 ) ⊆

V ((S1 ) © (S2 )). Take P ∈ V ((S1 ) © (S2 )).This says P ⊇ (S1 ) © (S2 ). If P ⊇ (S1 ) we get

P ∈ V (S1 ) and we are done. Hence, assume P (S1 ). Then there is a y ∈ (S1 ) such

that y ∈ P . But now y · (S2 ) is a subset of both (S1 ) and (S2 ) because they are ideals.

Hence, y · (S2 ) ⊆ P . By the prime ideal condition (S2 ) ⊆ P which we had to show.

Remark 1. The closed points in Spec(R) are the prime ideals which are maximal ideals.

Remark 2. If we take any prime ideal P then the (topological) closure of P in Spec(R)

is given as

V (P ) = { Q ∈ Spec(R) | Q ⊇ P } .

CONCEPTS OF MODERN ALGEBRAIC GEOMETRY 15

Hence, the closure of P consists of P and all “subvarieties” of P together. In particular

the closure of a curve consists of the curve as geometric object and all points lying on

the curve.

At the end of this lecture let me return to the a¬ne line over a ¬eld K, resp. its

algebraic model the polynomial ring in one variable K[X]. We saw already that we

have the non-closed point corresponding to the prime ideal {0} and the closed points

corresponding to the prime ideals (f ) (which are automatically maximal) where f is

an irreducible polynomial of degree ≥ 1. If K is an algebraically closed ¬eld the only

irreducible polynomials are the linear polynomials X ’ ±. Hence, the closed points of

Spec(K[X]) indeed correspond to the geometric points ± ∈ K. The non-closed point

corresponds to the whole a¬ne line.

Now we want to drop the condition that K is algebraically closed. As example let

us consider R[X]. We have two di¬erent types of irreducible polynomials. Of type (i)

are the linear polynomials X ’ ± (with a real zero ±) and of type (ii) are the quadratic

polynomials X 2 + 2aX + b with pairs of conjugate complex zeros. The maximal ideals

generated by the polynomials of type (i) correspond again to the geometric points of

R. There is no such relation for type (ii). In this case we have V (X 2 + 2aX + b) = ….

Hence, there is no subvariety at all associated to this ideal. But if we calculate the

coordinate ring R(A) of this (not existing) subvariety A we obtain

R(A) = R[X]/(X 2 + 2aX + b) ∼ R • RX

¯

=

¯ ¯

with the relation X 2 = ’2aX ’b. In particular, R(A) is a two-dimensional vector space.

It is easy to show that R(A) is isomorphic to C. Instead of describing the “point” A

as non-existing we should better describe it as a point of the real a¬ne line which is

C’valued. (Recall that for the points of type (i) R(A) ∼ R.) This corresponds to the

=

fact that the polynomial splits over the complex numbers C into two factors

a2 ’ b))(X + (a ’ a2 ’ b)).

(X + (a +

In this sense, the ideals of type (ii) correspond to conjugate pairs of complex numbers.

Note that there is no way to distinguish between the two numbers from our point of

view.

In the general situation for K one has to consider L’valued points, where L is allowed

to be any ¬nite-dimensional ¬eld extension of K.

16 MARTIN SCHLICHENMAIER

3. Homomorphisms

Part 1. Let V and W be algebraic sets (not necessarily irreducible), resp.

R(V ) = K[X1 , X2 , . . . , Xn ]/I, R(W ) = K[Y1 , Y2 , . . . , Ym ]/J

their coordinate rings. If ¦ : V ’ W is an arbitrary map and f : W ’ K is a function

then the pull-back ¦— (f ) := f —¦ ¦ is a function V ’ K. If we interpret the elements

of R(W ) as functions we want to call ¦ an algebraic map if ¦— (f ) ∈ R(V ) for every

f ∈ R(W ). Roughly speaking this is equivalent to the fact that ¦ “comes” from an

algebra homomorphism R(W ) ’ R(V ). In this sense the coordinate rings are the dual

objects to the algebraic varieties.

To make this precise, especially also to take care of the multiplicities, we should start

from the other direction. Let Ψ : R(W ) ’ R(V ) be an algebra homomorphism. This

homomorphism de¬nes a homomorphism Ψ (where ν is the natural quotient map)

Ψ = Ψ —¦ ν : K[Y1 , Y2 , . . . , Ym ] ’ R(V ) with Ψ(J) = 0 mod I .

Such a homomorphism is given if we know the elements Ψ(Yj ). Conversely, if we ¬x

elements r1 , r2 , . . . , rm ∈ R(V ) then Ψ(Yj ) := rj , for j = 1, . . . , m de¬nes an algebra

homomorphism Ψ : K[Y1 , Y2 , . . . , Ym ] ’ R(V ). If f (r1 , r2 , . . . , rm ) = 0 mod I for all

f ∈ J then Ψ factorizes through R(W ). Such a map indeed de¬nes a map Ψ— on the

set of geometric points,

Ψ— : V ’ W, Ψ— (±1 , ±2 , . . . .±n ) := (β1 , β2 , . . . , βm )

where the βj are de¬ned as

βj = Yj (Ψ— (±1 , ±2 , . . . , ±n )) := Ψ(Yi )(±1 , ±2 , . . . , ±n ).

We have to check whether Ψ— (±) = β ∈ Km lies on the algebraic set W for ± ∈ V . For

this we have to show that for all f ∈ J we get f (Ψ— (±)) = 0 for ± ∈ V . But

f (Ψ— (±)) = f Y1 (Ψ— (±)), . . . , Ym (Ψ— (±)) = f Ψ(Y1 )(±), . . . , Ψ(Ym )(±) = Ψ(f )(±) .

Now Ψ(f ) = 0 , hence the claim.

Example 1. A function V ’ K is given on the dual objects as a K’algebra homomor-

phism

¦ : K[T ] ’ R(V ) = K[X1 , X2 , . . . , Xn ]/I.

CONCEPTS OF MODERN ALGEBRAIC GEOMETRY 17

Such a ¦ is uniquely given by choosing an arbitrary element a ∈ R(V ) and de¬ning

¦(T ) := a. Here again you see the (now complete) interpretation of the elements of

R(V ) as functions on V .

Example 2. The geometric process of choosing a (closed) point ± on V can alternatively

be described as giving a map from the algebraic variety consisting just of one point to

the variety. Changing to the dual objects such a map is given as a map ¦± from R(V )

to the ¬eld K which is the coordinate ring of a point. In this sense points correspond

to homomorphisms of the coordinate ring to the base ¬eld K. Such a homomorphism

has of course a kernel ker ¦± which is a maximal ideal. Again, it is the ideal de¬ning

the closed point ±.

We will study this relation later. But ¬rst we take a di¬erent look on the situation.

Part 2. Let R be a K-algebra where K is a ¬eld. The typical examples are the quotients

of the polynomial ring K[X1 , X2 , . . . , Xn ]. Let M be a module over R, i.e. a linear

structure over R. In particular, M is a vector space over K. Some standard examples

of modules are obtained in the following manner. Let I be an ideal of R, ν : R ’ R/I

the quotient map then R/I is a module over R by de¬ning r · ν(m) := ν(r · m).

De¬nition. Let M be a module over R. The annulator ideal is de¬ned to be

Ann(M ) := { r ∈ R | r · m = 0, ∀m ∈ M } .

That Ann(M ) is an ideal is easy to check. It is also obvious that M is a module over