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Filters and Ultrafilters (continued) We would like to describe the maximal ideals of the ring R= Y k i∈I where k is a field and I is an index set. This is a special case of describing the maximal ideals of a ring of functions on a topological space. We can view R as the ring M ap(I, k) of all functions from I to k. For instance, I could be the unit interval [0, 1] on the real line, and k could be the real numbers. Later in the course we will try to determine the maximal ideals of particular subrings of M ap(I, k). 1 Here is one example: Theorem: (Hilbert’s Nullstellensatz) Suppose A is the ring of polynomials C[x1, · · · , xn], so that A ⊂ M ap(Cn, C). Then every maximal ideal M of A has the form Ma = {f ∈ A : f (a1, · · · , an) = 0} for some point a = (a1, · · · , an) of Cn. (The same is true if C is replaced by any algebraically closed field.) Q Returning to the case of R = i∈I k = M ap(I, k), we discussed last time how if I is infinite, there are maximal ideals M not of the form Mi0 = {f ∈ R : f (i0) = 0} for some i0 ∈ I. One can construct such an M by taking a maximal ideal containing the ideal of all f ∈ R which are zero at all but finitely many elements of I. This shows that M ap(I, k) doesn’t satisfy the Nullstellensatz when I is infinite. 2 To describe the maximal ideals of M ap(I, k), let’s recall the definition of filters and ultrafilters on the index set I. Def: Suppose I is a set. A filter for I is a collection J of subsets J of I with these properties: 1. If j ∈ J and J ⊂ J 0 ⊂ I then J 0 ∈ J . 2. If J1, J2 ∈ J then J1 ∩ J2 ∈ J . In other words, J is closed under taking oversets and intersections. Def: We say J is an ultrafilter if it is a maximal proper filter, in the sense that it is not the power set of I, and it is not properly contained in any filter for I other than the power set. 3 Example: Suppose i0 ∈ I. Let’s show that the collection Ji0 of all subsets J ⊂ I such that i0 ∈ J is an ultrafilter. Clearly Ji0 is closed under oversets and finite intersections, so its a filter. It doesn’t contain the emptyset, so it is a proper filter. Suppose Ji0 is properly contained in some other filter J 0. Then there would be an element J 0 ∈ J 0 such that i0 6∈ J 0. However, {i0} ∈ Ji0 ⊂ J . So by the axioms of filters, ∅ = {i0} ∩ J 0 ∈ J 0. which implies J 0 is the powerset. This implies J is an ultrafilter. Example: It’s a homework problem this week to use Zorn’s Lemma to show that every proper filter is contained in an ultrafilter. So for example, if I is infinite, then there is an ultrafilter which contains the filter of all complements of finite subsets of I. This ultrafilter can’t be of the form Ji0 4 Def: Suppose F is an filter for the set I. SupQ Q pose α = i∈I αi is an element of R = i∈I k. Let zer(α) = {i ∈ I : αi = 0}. Thus zer(α) is a subset of I. Define MF to be the set of all α ∈ R such that zer(α) ∈ F . Theorem: (See this week’s Homework) The set MF is an ideal. The map F → MF defines a bijection between the set of ultrafilters on I and the set of maximal ideals of R. Example: The filter of all complements of finite subsets of I corresponds the ideal of all elements of R which have only finitely many non-zero components. 5 Prime ideals Let R be a commutative ring. Def: A ideal P of R is a prime ideal if P 6= R, and whenever a, b ∈ R have the property that ab ∈ P , then one of a or b is in P . Comment: An equivalent definition is that R/P is an integral domain, i.e. that R/P is a commutative non-zero ring which has no zero divisors. Example: All maximal ideals M are prime, since R/M is a field. Example: The zero ideal {0} is prime if and only if R itself is an integral domain. This is so, for example, if R = Z. 6 Def: The spectrum Spec(R) of a commutative ring is the following topological space. 1. The elements of Spec(R) are the prime ideals of R. 2. The closed subsets of Spec(R) (which are the complements of the open subsets) have the form VA = {P ∈ Spec(R) : A ⊂ P } as A ranges over all ideals of R. ( This is called the Zariski topology.) 7 Example: Suppose R = Z. The prime ideals of Z are the zero ideal {0} together with the maximal ideals Zp as p ranges over all prime numbers. Each ideal A has the form Zn for some (possibly zero) integer n. One has V{0} = Spec(R) and for n 6= 0, VZn = {P = Zp : p|n} It follows that the closed sets of Spec(Z) are just the finite subsets of Spec(Z) − {{0}} together with all of Spec(Z). Picture: 8 One does have to check that the Zariski topology is a topology. For this one needs to show: 1. ∅ and Spec(R) are closed. This is so because ∅ = VR and Spec(R) = V{0}. 2. The intersection ∩A∈S VA is closed if S is any collection of ideals of R. This is true since ∩A∈S VA = VB when B is the R-ideal generated by all the A ∈ S. (A prime ideal P contains B if and only if it contains each A ∈ S.) 9 3. The union ∪A∈S VA is closed if S is any finite set of ideals of R. We claim ∪A∈S VA = VC when C = ∩A∈S A. In one direction, if P ∈ VA for some A ∈ S, then C ⊂ P so P ∈ VC . If P ∈ VC we want to show P ∈ VA for some A ∈ S. Suppose this is not the case. Then for each A ∈ S, it is not true that A ⊂ P . Choose for each A ∈ S an element pA ∈ A not in P . Then Y A∈S pA ∈ Y A⊂C A∈S but A∈S pA 6∈ P since P is prime. (Note that the product is finite.) This contradicts P ∈ VC . Q 10 Picture of Spec(C[t]): Picture of Spec(C[x, y]). We will later show that the prime ideals P of Spec(C[x, y]) are either (i) P = {0} (the generic point); (ii) P = C[x, y] · f for some irreducible polynomial f = f (x, y); (iii) maximal ideals, which by the Nullstellensatz have the form Ma = {h : h(a) = 0} for some a = (a1, a2) ∈ C. 11 Localization Idea: Generalize the construction of Q as the ring of quotients ab for which a, b ∈ Z and b 6= 0. Def: Suppose R is a commutative ring. A multiplicatively closed subset D of R is a subset which contains 1 and such that d1d2 ∈ D if d1, d2 ∈ D. Theorem: There is a commutative ring D−1R and a ring homomorphism φ : R → D−1R with the following properties: 1. φ(D) is contained in the group of units of D−1R. 2. The homomorphism φ is universal in the following sense. If φ0 : R → R0 is any other ring homomorphism for which φ0(D) ⊂ (R0)∗, there is a unique ring homomorphism h : D−1R → R0 such that φ0 = h ◦ φ. 12 Example: Suppose R = Z and D = Z − {0}. Then D−1R = Q and the inclusion φ : Z → Q have the right universal property. To construct h : Q → R0 given φ0 : R → R0, one sets a h( ) = φ0(a) · φ0(b)−1. b We would like to construct D−1R by formalizing the definition of ab for a ∈ R and d ∈ D. To do this we need: Lemma: There is an equivalence relation ˜ on the product set R × D defined by (r, d) ˜ (r0, d0) d”(rd0 − r0d) = 0 iff d” ∈ D for some Comment: To motivate this, one has formally d”(rd0 − r0d) = d”d0d r0 r − 0 d d ! So (r, d) ˜ (r0, d0) should correspond to the right side being 0. 13 Proof: Reflexivity and symmetry are easy, since R is a commutative ring. For transititivity, one has to show that if (r, d) ˜ (r 0, d0) and (r0, d0) ˜ (r 00, d00) then (r, d) ˜ (r00, d00). The principle in proving this sort of statement is to treat (r, d) as the formal fraction dr . One then takes an identity to be proved and puts the terms over a common denominator. In the present situation, one has formally that r” r − = d” d r0 r” − 0 d” d ! + r0 r − d0 d ! Putting the left and right hand sides over the common denominator dd0d” and then multiplying by dd0d” gives the identity d0(r”d − d”r) = d(r”d0 − d”r0) + d”(r0d − rd0) which holds because R is commutative. 14 The hypotheses that (r, d) ˜ (r 0, d0) and (r 0, d0) ˜ (r00, d00) imply that we can multiply the right hand side of d0(r”d − d”r) = d(r”d0 − d”r0) + d”(r0d − rd0) by d0 · d1 for some d0, d1 ∈ D to get 0. Since d0 · d1 · d ∈ D because D is multiplicatively closed, the left hand side of the resulting equality shows (r, d) ˜ (r00, d00). 15 Corollary: We can define D−1R to be the set of equivalence classes [(r, d)] of pairs (r, d) ∈ R × D with respect to the equivalence relation ˜ . This set becomes a ring if we define addition and multiplication by 1. [(r, d)] + [(r0, d0)] = [(rd0 + r 0d, dd0)]. 2. [(r, d)] ∗ [(r 0, d0)] = [(rr0, dd0)]. One defines φ : R → D−1R by φ(r) = [(r, 1)]. The proof of the Corollary consists of first checking that the addition and multiplication are well defined; this is necessary because we used representatives for equivalence classes in the definitions. Then one has to check + is an abelian group law, ∗ defines a monoid with multiplicative identity (1, 1), and the distributive laws hold. 16 To prove any of these facts, one treats [(r, d)] as dr and then puts terms over a common denominator in the identities to be proved. The check the universal property of D−1R, suppose φ0 : R → R0 is a ring homomorphism with φ(D) ⊂ (R0)∗. The ring homomorphism h : D−1R defined by h([(r, d)]) = φ(d)−1φ(r) has the property we want. In the following examples, we’ll write dr for the equivalence class [(r, d)] ∈ D−1R. Example: Suppose R an integral domain. Then D = R−{0} is multiplicatively closed. The ring D−1R is a field, since dr · dr = 1. Furthermore, φ : R → D−1R is injective. This is because r = 0 = 0 implies d(r · 1 − 1 · 0) = dr = 0 for 1 1 some d ∈ D, and d 6= 0 then forces r = 0 since R has no zero divisors. Once calls D−1R the fraction field of R. Subexample: The field of p-adic numbers Qp is the fraction field of Zp. 17 Example: Suppose P is a proper ideal R. The set D = R − P is multiplicatively closed if and only if P is a prime ideal. Let’s suppose this is the case. The ring D−1R is called the localization of R at P , and is denoted RP . Subexample 1: If R = Z, primes P ∈ Spec(Z) are either P = {0} or P = Zp = (p) for some prime p. When P = {0}, one gets D = Z − {0}, so R{0} is the fraction field Q of Z. Suppose now that P = (p) for some prime number p. We have D = Z − P = {n ∈ Z : p 6 |n}. So r Z(p) = D−1Z = { : r, n ∈ Z, p 6 |n} n is the subring of rationals which have denominators prime to p. There is an embedding Z(p) → Zp = the p − adic numbers. This is because each rational integer r ∈ Z not in Zp is a unit in Zp, since it can be inverted mod pn for all n. But Zp is much bigger than Z(p), e.g. its uncountable. 18 Subexample 2: Suppose R = C[x]. We now the prime ideals of R are either {0} or the ideal (x − α) for some α ∈ C. As in subexample 1, the localization R{0} is the fraction field C(x) of R. This is the field of all (x) rational functions fh(x) in one variable (where h(x) 6= 0). Suppose now that P = (x − α). Then R − P is the set of h(x) for which h(α) 6= 0. This means C[x](x−α) = { f (x) : f (x), h(x) ∈ C[x], h(α) 6= 0} h(x) Thus C[x](x−α) consists of those rational functions which take on well-defined values in some small open disc around α. This is where the term ”localization” comes from; one has functions defined locally near α. 19 Local Rings Def: A commutative ring R is called local if it has a unique maximal ideal mR . Proposition: A commutative ring R is a local ring if and only if R −R∗ is an ideal; in this case R − R∗ = mR is the unique maximal ideal of R. Proof: If R is local, and β ∈ R − mR , then Rβ can’t be a proper ideal of R, since this would imply Rβ ⊂ mR because proper ideals have to be contained in some maximal ideal. Hence Rβ = R, so β ∈ R∗ = R − mR . Conversely, no proper ideal of R can contain an element of R∗. So if R − R∗ is an ideal, this must be the unique maximal ideal of R, and R is local. 20 Example: Each field F is a local ring, since {0} is the unique proper ideal of F and hence the unique maximal ideal. Example: The ring Z is not a local ring, since it has many distinct maximal ideals Zp. Example: Suppose R is a commutative ring with prime ideal P . Let’s check that the localization RP is a local ring. Define an ideal of RP by r mRP = { ∈ RP : r ∈ P, d ∈ R − P }. d Using the fact that P is an ideal of R, one finds that mRP is an ideal of RP . We claim mRP 6= RP . Otherwise, r 1 = 1 d for some r ∈ P and d ∈ R − P , there is a d0 ∈ R − P such that d0 · (d − r) = 0. This would force d0d = d0r ∈ P , contradicting the fact that d0d 6∈ P since P is a prime ideal. 1= 21 We have r R − mRP = { ∈ RP : r, d ∈ R − P } d But such elements are units in RP , since dr ∈ RP if r, d ∈ R − P . Thus R − mRP ⊂ R∗. Because mRP is a proper ideal, R∗ ⊂ R − mRP . So R − mRP = R∗ and R is local. 22 Example: Let X be a topological space, and let F be the presheaf which over each open subset U ⊂ X has sections F (U ) the ring of real valued continuous functions on U . Suppose z is a point of X. The stalk Fz = lim F (U ) −→ z∈U is a local ring. To see why this is so, define m(Fz ) to be the kernel of the ring homomorphism Fz → R which sends each f ∈ Fz to its value at z. Then m(Fz ) is an Fz -ideal, and it is not all of Fz since it does not contain 1. If h is an element of Fz − m(Fz ), then h(z) 6= 0, so h is non-zero in some neighborhood U of z. This means 1/h is a continuous function on U , so 1/h ∈ Fz and h is a unit of Fz . Thus Fz is a local ring. 23