R/I. Hence, every ideal of R is the annulator ideal of a suitable R’module.

De¬nition. A module M is called a simple module if M = {0} and M has only the

trivial submodules {0} and M .

Claim. M is a simple module if and only if there is a maximal ideal P such that

M ∼ R/P .

=

Proof. Note that the submodules of R/P correspond to the ideals lying between R and

P . Hence, if P is maximal then R/P is simple. Conversely, given a simple module M

take m ∈ M, m = 0. Then R · m is a submodule of M . Because 1 · m = m the module

R · m = {0}, hence it is the whole module M . The map •(r) = r · m de¬nes a surjective

map • : R ’ M . This map is an R’module map where R is considered as a module

over itself. The kernel P of such a map is an R’submodule. But R’submodules of

R are nothing else than ideals of R. In view of the next lecture where we drop the

commutativity let us note already that submodules of a ring R are more precisely the

18 MARTIN SCHLICHENMAIER

left ideals of R. The kernel P has to be maximal otherwise the image of a maximal ideal

lying between P and R would be a non-trivial submodule of M . Hence, M ∼ R/P .

=

From this point of view the maximal ideals of R(V ) correspond to R(V )’module

homomorphisms to simple R(V )’modules. If R(V ) is a algebra over the ¬eld K, then

a simple module M is of course a vector space over K. By the above, we saw that it is

even a ¬eld extension of K. (Recall that M ∼ R/P with P a maximal ideal). Because

=

R(V ) is ¬nitely generated as K’algebra it is a ¬nite dimensional vector space over K

(see [Ku,S.56]) hence, a ¬nite (algebraic) ¬eld extension.

Observation. The maximal ideals (the “points”) of R = K[X1 , X2 , . . . , Xn ]/I cor-

respond to the K’algebra homomorphism from R to arbitrary ¬nite (algebraic) ¬eld

extensions L of the base ¬eld K. We call these homomorphisms L’valued points.

In particular, if the ¬eld K is algebraically closed there are no nontrivial algebraic ¬eld

extensions. Hence, there are only K’valued points. If we consider reduced varieties (i.e.

varieties whose coordinate rings are reduced rings) we get a complete dictionary. Let V

be a variety, P = I(V ) the associated prime ideal generated as P = (f1 , f2 , . . . , fr ) with

fi ∈ K[X1 , X2 , . . . , Xn ] suitable polynomials, and R(V ) the coordinate ring Rn /P .

The points can be given in 3 ways:

(1) As classical points. ± = (±1 , ±2 , . . . , ±n ) ∈ Kn with

f1 (±) = f2 (±) = · · · = fr (±) = 0.

(2) As maximal ideals in R(V ). They in turn can be identi¬ed with the maximal

ideals in K[X1 , X2 , . . . , Xn ] which contain the prime ideal P . In an explicit

manner these can be given as (X1 ’ ±1 , X2 ’ ±2 , . . . , Xn ’ ±n ) with the

condition f1 (±) = f2 (±) = · · · = fr (±) = 0.

(3) As surjective algebra homomorphisms φ : R(V ) ’ K. They are ¬xed by de¬ning

¯ ¯

Xi ’ φ(Xi ) = ±i , i = 1, . . . , n in such a way that

φ(f1 ) = φ(f2 ) = · · · = φ(fr ) = 0.

The situation is di¬erent if we drop the assumption that K is algebraically closed. The

typical changes can already be seen if we take the real numbers R and the real a¬ne line.

The associated coordinate ring is R[X]. There are only two ¬nite extension ¬elds of R,

either R itself or the complex number ¬eld C. If we consider R’algebra homomorphism

from R[X] to C then they are given by prescribing X ’ ± ∈ C. If ± ∈ R we are again

in the same situation as above (this gives us the type (i) maximal ideals). If ± ∈ R then

the kernel I of the map is a maximal ideal of type (ii) I = (f ) where f is a quadratic

polynomial. f has ± and ± as zeros. This says that the homomorphism Ψ± : X ’ ±

¯ ¯

¯

which is clearly di¬erent from Ψ± : X ’ ± has the same kernel. In particular, for

one maximal ideal of type (ii) we have two di¬erent homomorphisms. Note that the

map ± ’ ± is an element of the Galois group G(C/R) = {id, „ } where „ is complex

¯

conjugation. The two homomorphisms Ψ± and Ψ± are related as Ψ± = „ —¦ Ψ± .

¯ ¯

This is indeed the general situation for R(V ), a ¬nitely generated K’algebra. In

CONCEPTS OF MODERN ALGEBRAIC GEOMETRY 19

general, there is no 1-1 correspondence between (1) and (2) anymore. But there is a

1-1 correspondence between maximal ideals of R(V ) and orbits of K’algebra homo-

morphism of R(V ) onto ¬nite ¬eld extensions L of K under the action of the Galois

group

G(L/K) := { σ : L ’ L an automorphism of ¬elds with σ|K = id} .

4. Some Comments on the noncommutative situation

For the following let R be a (not necessarily commutative) algebra over the ¬eld K.

First, we have to distinguish in this more general context left ideals (e.g. subrings I

which are invariant under multiplication with R from the left), right ideals and two-

sided ideals (which are left and right ideals). To construct quotient rings two-sided

ideals are needed. If we use the term ideal without any additional comment we assume

the ideal to be a two-sided one.

We want to introduce the concepts of prime ideals, maximal ideals, etc.. A ¬rst

de¬nition of a prime ideal could be as follows. We call a two-sided ideal I prime if the

quotient R/I contains no zero-divisor. This de¬nition has the drawback that there are

rings without any prime ideal at all. Take for example the ring of 2 — 2 matrices. Beside

the ideal {0} and the whole ring the matrix ring does not contain any other ideal. To

see this assume there is an ideal I which contains a non-zero matrix A. By applying

elementary operations from the left and the right we can transform any matrix to normal

form which is a diagonal matrix with just 1 (at least one) and 0 on the diagonal. By

multiplication with a permutation matrix we can achieve any pattern in the diagonal.

These operations keep us inside the ideal. Adding suitable elements we see that the unit

matrix is in the ideal. Hence the ideal is the whole ring. But obviously, the matrix ring

has zero divisors. Hence, {0} is not prime in this de¬nition. We see that this ring does

not contain any prime ideal at all with respect to the de¬nition. We choose another

name for such ideals: they are called complete prime ideals.

De¬nition. A (two-sided) ideal I is called a prime ideal if for any two ideals J1 and

J2 with J1 · J2 ⊆ I it follows that J1 ⊆ I or J2 ⊆ I .

This de¬nition is equivalent to the following one.

De¬nition. A (two-sided) ideal I is called a prime ideal if for any two elements a, b ∈ R

with a · R · b ⊆ I it follows that a ∈ I or b ∈ I.

20 MARTIN SCHLICHENMAIER

Proof. (2. D) =’ (1. D): Take J1 I and J2 I ideals. We have to show that

J1 · J2 I. For this choose x ∈ J1 \ I and y ∈ J2 \ I. Then x · R · y ⊆ J1 · J2 but there

must be some r ∈ R such that x · r · y ∈ I due to the condition that I is prime with

respect to (2. D). Hence, J1 · J2 I which is the claim.

(1. D) =’ (2. D): Take a, b ∈ R. The ideals generated by these elements are RaR

and RbR. The product of these ”principal” ideals is not a principal ideal anymore. It

is RaR · RbR = RaRbR := (arb | r ∈ R). Assume arb ∈ I for all r ∈ R. Hence

(RaR)(RbR) ⊆ I and because I is prime we obtain by the ¬rst de¬nition (1. D) that

either RaR or RbR are in I. Taking as element of R the 1 we get a ∈ R or b ∈ R.

Every ideal which is a complete prime is prime. Obviously, the condition (2. D) is a

weaker condition than the condition that already from a · b ∈ I it follows that a ∈ R or

b ∈ R (which is equivalent to: R/I contains no zero-divisors). If R is commutative then

they coincide. In this case a · r · b = r · a · b, and with a · b ∈ I also r · a · b ∈ I which

is no additional condition. Here you see clearly where the noncommutativity enters the

picture. In the ring of matrices the ideal {0} is prime because if after ¬xing two matrices

A and B we obtain A · T · B = 0 for any matrix T then either A or B has to be the zero

matrix. This shows that the zero ideal in the matrix ring is a prime ideal.

Maximal ideals are de¬ned again as in the commutative setting just as maximal

elements in the (non-empty) set of ideals. By Zorn™s lemma there exist maximal ideals.

Claim. If M is a maximal ideal then it is a prime ideal.

Proof. Take I and J ideals of R which are not contained in M . Then by the maximality

of M we get (I + M ) = R and (J + M ) = R hence,

R · R = R = (I + M )(J + M ) = I · J + M · J + I · M + M · M .

If we assume I · J ⊆ M then R ⊆ M which is a contradiction. Hence I · J M . This

shows M is prime.

By this result we see that every ring has prime ideals.

In the commutative case if we approach the theory of ideals from the point of view of

modules over R we obtain an equivalent description. This is not true anymore in the