MATH 250B: COMMUTATIVE ALGEBRA 11 4. Lecture 4 Visualizing rings We describe several ways of visualizing rings. The simplest is just to draw a point for each element of the ring.p For example for Z[i] we get a square lattice, and for the Eisenstein integers Z[(1 + −3)=2] we get a hexagonal lattice. As an application we will show that Z[i] has unique factorization. Recall that a ring is called Euclidean if it has division with remainder: given a, b 6= 0 we can find q, r with a = qb + r and jrj < jbj, where jxj lies in some well ordered set (such as the non-negative integers). Any Euclidean ring has unique factorization (and is even a principal ideal domain): copy the proof for Z. The key point is to show that if P is irreducible (not a non-trivial product) then it is prime (pjab implies pja or pjb. This is in Euclid book 7, though Euclid could not states that Z is a UFD as multiplication was done geometrically so it was hard to multiply more than 3 numbers. So we want to show Z[i] is Euclidean. We let jXj be the usual absolute value of a complex number. We have to show that given a=b we can find q with ja=b−qj < j1j. It is enough to show that the complex plane is covered by open unit radius disks with centers the Gaussian integers,p which is geometricallyp obvious. The same argument works for Z[ −2] but fails for Z[ −3], where the argument just fails: the plane is covered by closed unit disks but not by open ones.p In factp this ring does NOT have unique factorization: for example,p 2×2 = (1+ −3)(1− −3). It has some non-principal ideals, such as (2; 1 + −3. Note that the non-principal ideals are a different shape from the principal ideals. It is contained in its integral closure the Eisenstein integers, which do have unique factorization. Euclidean rings are rather rare even among unique factorization domains. For example, if R is a UFD then so is R[x], but k[x; y] is not a Euclidean ring. It is not even a PID, as (x; y) is not principal. (Remark. A Euclidean ring R has the property that for a non-unit a of minimal norm, every non-zero element of R=(pa) is represented by a unit, so in particular R=(a) is a field. In particular Z[(1 + −19=2)] is not Euclidean, even though it is an imaginary quadratic number field with unique factorization.) Moreover not many rings can be embedded inside Cn (though this does apply to all algebraic number fields, where is is a very important tool.) The second method of representing rings geometrically is to draw a point for each basis element. Here we assume that the ring is a free module over some nice ring such as a field. For example k[x; y] can be represented by a 2-dimensional quadrant of a 2-dimensional lattice. This is widely used in combinatorics. As examples, we can easily see lots of ideals of finite codimension, and can see that the ring k[x; xy; xy2; ::::] is not Noetherian. (In particular subrings of Noetherian rings need not be Noetherian.) Example 4.1. What is the dimension of the vector space k[x; y]=(x3; x2y2; y5)? Drawing a picture makes it obvious that the dimension is 12. (Can do this using points; some people prefer boxes.) Example 4.2. We can also see from this that there are large numbers of finite dimensional algebras over a field generated by 2 nilpotent elements. This is related to the problem of classifying pairs of commuting (nilpotent) matrices, which is in some sense a "wild" problem. 12 RICHARD BORCHERDS Example 4.3. We see that k[x2; xy; y2] is a free module of rank 2 over k[x2; y2] with a basis 1; xy, so is isomorphic to k[u; v; w]=(v2 − uw). Geometric meaning: if we quotient a plane by (x; y) ! (−x; −y) we get a cone. The main disadvantage is that this representation only works well for rather special classes of rings and ideals, such as polynomial rings and monomial ideals. The third method of representing rings geometrically is to represent them by their spectrum, the set of prime ideals with the Zariski topology. This works for all rings and is by far the most important representation, though seems somewhat mysterious at first. As motivation, we will explain the connection between the spectrum of a ring and the electromagnetic spectrum. The spectrum of light given off by stars or elements has certain lines in it, and Hilbert thought that these might be given by the eigenvalues of some self adjoint operator (this later turned out to be correct). So the eigenvalues of an operator are called its spectrum. If A is a self adjoint operator on a finite dimensional complex vector space, we can look at the commutative ring C[A], a finite dimensional vector space over C. Then the eigenvalues of A are exactly the homomorphisms from this ring to C, which can be identified with the maximal ideals of R[A]. So the set of maximal ideals of a ring is called its maximal spectrum. The ring R[A] can be identified with the ring of (continuous) complex functions on the spectrum (a finite discrete space). Now suppose that X is any compact Hausdorff space and R is the ring of contin- uous complex functions on X. (This is a commutative C* algebra.) We can recover the space X from R as follows. The points of X correspond to the maximal ideals of X (the maximal ideal of a point is the ideal of functions vanishing there). The topology can be recovered as follows: take any ideal I of R, and let Z(I) be the set of maximal ideals containing I. These give the closed sets of X. (Different ideals can give the same closed set.) We can now do the same for any commutative ring R. We define its maximal spectrum to be the set of maximal ideals, with the Zariski topology (the closed sets are the sets Z(I)). There is a problem. A continuous map of spaces from X to Y gives a homomor- phism of their rings of continuous functions (in the opposite direction. We would like this to be some sort of anti-equivalence of categories, so homomorphisms of rings should give continuous functions between their maximal spectrums. However it does not in general (though it does for the case of C∗ algebras). The problem is that the inverse of a maximal ideal need not be maximal: look at Z ! Q for example. Let us analyze the reason for this problem. An ideal is maximal if and only if the quotient is a field. The fact that the inverse image of a maximal ideal need not be maximal corresponds to the fact that a subring of a field need not be a field: in general it is just an integral domain. Now we see how to fix this: instead of using maximal ideals, use the ideals such that R=I is an integral domain. These are called prime ideals: equivalently they have the property that if xy 2 I then x 2 I or y 2 I, corresponding to the fact that an integral domain is a ring such that if xy = 0 then x = 0 or y = 0. The name "prime ideal" is a little unfortunate because they do not quite correspond to primes. If p is a prime element of a ring (meaning that if pjxy then pjx or pjy and p is not 0 or a unit) then (p) is indeed MATH 250B: COMMUTATIVE ALGEBRA 13 a prime ideal. However 0 is a prime ideal in any integral domain even though 0 is not usually considered to be a prime. (By convention, integral domains and fields have more than 1 element, so R is not a prime or maximal ideal)..