Commutative Algebra Background September 18, 2007 1 Spectra and Localization. 1.1 Prime Spectra of rings. Rings, unless otherwise signaled, will be commutative rings with unit element, and homomorphism of rings will be assumed to preserve the unit element. A ring denoted k will be understood to be a field, and k[X, Y, . .] the (commutative) ring of polynomials in the independent variables X, Y, . with coefficients in the field k. The standard list of notation for the rings of rational integers, rational numbers, reals, and complexes, will also be understood: Z, Q, R, C. If A is a ring, Spec A, its prime spectrum (for short: its spectrum) is—as a set—the set of prime ideals of A. Discuss: Spectrum versus Max Spectrum, with examples of Banach algebras. If f : A → B is a ring-homomorphism, the pullback, i.e., full inverse image, of any prime ideal of B under the homomorphism f is a prime ideal of A. This gives us a map on spectra Spec B → Spec A that we will denote by Spec(f) (or, sometimes, by just f again) allowing us to consider Spec as a contravariant functor from the category of commutative rings to the category of sets. If f : A → B is a surjective ring-homomorphism, then Spec(f) : Spec B → Spec A is an injection. 1.2 The “Universal solution” to the problem of inverting a given set of elements in a ring. Let A be a ring, and S ⊂ A any subset. Consider the “problem” of finding ring homomorphisms A → A0 such that the set S gets sent to a set of units in A0. The “universal solution” to this problem (which exists, and is therefore easily seen to be unique up to unique isomorphism) is a ring 1 −1 homomorphism ιS : A → S A which sends S to units, and has the property that any “solution” 0 −1 −1 A → A is obtained by composition of ιS with a unique homomorphism A → S A. The ring S A is called the localization of A with respect to the set S. Since the problem of inverting a set S is equivalent to the problem of inverting the multiplicative system in A generated by S, we may—when it is convenient for us to do so—suppose that the set S we wish to invert contains the identity element of A and is closed under multiplication. Exercise 1. If A is a ring and S ⊂ A a subset, let {Xs}s∈S be a collection of independent variables; −1 show that S A can be given as the quotient of the polynomial ring A[Xs; s ∈ S] by the ideal generated by the polynomials s · Xs − 1 for s ∈ S. Exercise 2. If A is a ring and S ⊂ A a multiplicative system, show that any element in the kernel of A → S−1A is also in the kernel of A → {f}−1A for some element f ∈ S. Exercise 3. If A is a ring and S ⊂ A a multiplicative system, show that the kernel of A → S−1A consist precisely in the elements of A which are annihilated by some element of S. (Hint: Let a ∈ A be in that kernel. By the previous exercise we may replace S by (the multiplicative system generated by) a single element s and our element a will “still” be in the kernel. But now we’re home since S−1A = A[X]/(Xs − 1) so we get that a = g(X) · (Xs − 1) is a equation in A[X] for some polynomial g(X) ∈ A[X]. Then compute.) This operation A 7→ S−1A is indeed a localization operation for on the level of spectra we have that the map induced by ιS, i.e., −1 Spec(ιS) : Spec S A → Spec A is injective, and its image is the complement of {P ∈ Spec A | P ∩ S 6= the empty set}. Discuss the difficulties of extending this localization procedure to noncommutative rings (or more general set-ups) these difficulties suggesting the scope of geometry. Here are some important special cases of localization: • Let A be an integral domain, and suppose that the multiplicative system S ⊂ A doesn’t contain 0 ∈ A. In this case, if K is the field of fractions of A (K being the localization of A relative to the multiplicative system A − {0}), then S−1A = {a/s ∈ K | a ∈ A, s ∈ S}; • Suppose the subset S is generated by finitely many elements s1, s2, . , sm of A. In this case Q −1 −1 put s = i si and you can take S A to be the ring A[1/s] := A[s]/(st−1) with A → S A the −1 natural homomorphism A → A[1/s]. Denoting X := Spec A and Xf := Spec S A we have that Xs ⊂ X is the open subset consisting of the set of all points of Spec A that correspond to prime ideals P in A that do not contain the element s; visually: “the complement of the locus of zeroes of s.” 2 • For a prime ideal P ⊂ A take S = A − P . Then S is a multiplicative system. We usually −1 denote S A as AP (the localization of A at P ). We have that AP is a local ring, with maximal ideal equal to PAP ⊂ AP , and with residue field AP /P AP = the field of fractions of the integral domain A/P . The set of prime ideals of AP pull back to the set of prime ideals of A contained in P . Define the Zariski topology on spectra as follows: If A is a ring, the open sets on Spec A is given by the complements of the injective mappings Spec(φ) : Spec B,→ Spec A where φ runs through all surjective ring-homomorphisms with domain A. A base for the open sets is given by the images −1 2 of the mappings Spec(Sf A) → Spec A for f ∈ A, where Sf = {1, f, f ,...} is the multiplicative set generated by f. For any homomorphism f of rings, the induced morphism Spec f is a continuous map of spectra. 1.3 Some examples • Regarding the spectra of local rings: – If A = k is a field, the zero ideal is the only prime ideal, so Spec A consists of one point. – If A is a local ring, its unique maximal ideal corresponds to the unique closed point in Spec A. – If A is a local ring with nilpotent maximal ideal (e.g., Z/p2Z, or any other artinian local ring) then Spec A consists of one point. • More generally, – if A is a ring and N ⊂ A is its nilradical, consider the associated reduced ring to A, denoted Ared := A/N . The natural projection A −→ Ared induces a homeomorphism on spectra, but wait ...1; – if A is noetherian and we let P1,P2,...,Pν denote the set of its minimal primes (i.e., those that are connected to a primary decomposition of the zero ideal in A) then Spec A is the union of the closed subsets Spec A/Pj for j = 1, 2, . , ν. – if A = B × C is a product then Spec A is the disjoint union of Spec B and Spec C, and conversely; 1.4 Partitions of unity. The last example of the previous subsection gives us a way of decomposing spectra into a disjoint union of smaller spectra (essentially equivalent to giving orthogonal idempotent decompositions of the unit element). We will now consider something much more general. Let A be a ring. An equation 1 = f1 + f2 + ... + fν 1We will be considering these spectra along with their natural local ring structure. 3 with fi ∈ A will be called a partition of unity attached to which we have a homomorphism of rings, ν Y −1 A −→ {fi} A, i=1 and therefore an open mapping on the level of spectra, ν ν G −1 Y −1 Spec({fi} A) = Spec ( {fi} A) −→ Spec A. i=1 i=1 −1 The collection of open subsets Ui := Spec {fi} A ⊂ Spec A for i = 1, 2, . , ν, form a cover of Spec A. (Proof: If P is a prime ideal of A there is some index i such that fi ∈/ P which strange as it sounds means exactly that the point P of Spec A is in the image of −1 Spec {fi} A.) Exercise 4. If A is an integral domain and K its field of fractions, and we are given a partition of unity for A as above, show that ν \ −1 A = ({fi} A) ⊂ K. i=1 • Discrete Valuation Rings. A ring is a DVR if it is a principal ideal domain and has precisely one nonzero prime ideal, so Spec A consists of two points, one closed point corresponding to the maximal ideal, the other open point corresponding to the zero ideal. If A is a DVR then its nonzero prime ideal, m(A) ⊂ A is visibly its only maximal ideal; so A is a local ring, and A/m(A) is its residue field. Any element not contained in m(A) is a unit in A; i.e., A∗ = A − m(A). Any generator z of the ideal m(A) is called a uniformizer, the uniformizers being the irreducible elements of A. Fix a uniformizer z. For any nonzero element α ∈ A there is a maximal integer (≥ 0) n such that α is divisible by zn in A, and for such an n we have that α = uzn for a unit u ∈ A∗ (proof: A is a PID so a UFD and has only one irreducible element up to multiplication by a unit, namely z).