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Introduction to Commutative Algebra

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Introduction to Commutative Algebra MATH 603: INTRODUCTION TO COMMUTATIVE ALGEBRA THOMAS J. HAINES 1. Lecture 1 1.1. What is this course about? The foundations of differential geometry (= study of manifolds) rely on analysis in several variables as \local machinery": many global theorems about manifolds are reduced down to statements about what hap- pens in a local neighborhood, and then anaylsis is brought in to solve the local problem. Analogously, algebraic geometry uses commutative algebraic as its \local ma- chinery". Our goal is to study commutative algebra and some topics in algebraic geometry in a parallel manner. For a (somewhat) complete list of topics we plan to cover, see the course syllabus on the course web-page. 1.2. References. See the course syllabus for a list of books you might want to consult. There is no required text, as these lecture notes should serve as a text. They will be written up \in real time" as the course progresses. Of course, I will be grateful if you point out any typos you find. In addition, each time you are the first person to point out any real mathematical inaccuracy (not a typo), I will pay you 10 dollars!! 1.3. Conventions. Unless otherwise indicated in specific instances, all rings in this course are commutative with identity element, denoted by 1 or sometimes by e. We will assume familiarity with the notions of homomorphism, ideal, kernels, quotients, modules, etc. (at least). We will use Zorn's lemma (which is equivalent to the axiom of choice): Let S; ≤ be any non-empty partially ordered set. A chain T in S is a subset T ⊆ S such that x; y 2 T implies x ≤ y or y ≤ x holds. If S; ≤ is such that every chain has an upper bound in S (an element s 2 S with t ≤ s for all t 2 T ), then S contains at least one maximal element. 1.4. Correspondence between ideals and homomorphisms. We call any sur- jective homomorphism A ! B a quotient. We say the quotients f1 : A ! B1 and f2 : A ! B2 are equivalent if there exists a ring isomorphism φ : B1 f! B2 satisfying φ ◦ f1 = f2. The terminology is justified because any surjective homomorphism f : A ! B is clearly equivalent to the canonical quotient A ! A=ker(f). Proposition 1.4.1. (1) There is an order-preserving correspondence fideals I ⊆ Ag ! fequivalence classes of quotients A ! Bg: The correspondence sends an ideal I to the equivalence class of the canonical quotient A ! A=I, and the quotient f : A ! B to the ideal ker(f) ⊆ A. Date: Fall 2005. 1 2 THOMAS J. HAINES (2) Fix an ideal I ⊂ A. There is an order-preserving correspondence fideals J ⊆ A containing Ig ! fideals of A=Ig; given by: send an ideal J ⊃ I to its image J in A=I, and send an ideal J 0 ⊆ A=I to its pre-image under the canonical map A ! A=I. 1.5. Prime and maximal ideals. A domain is a ring A with the property: 1 6= 0 and if x; y 2 A and xy = 0, then x = 0 or y = 0. Examples are the integers Z, and any ring of polynomial functions over a field. An ideal p ⊂ A is prime if it is proper (p 6= A) and xy 2 p implies x 2 p or y 2 p. Thus, p is prime if and only if A=p is a domain. An ideal m ⊂ A is maximal if m 6= A and there is no ideal I satisfying m ( I ( A. Equivalently, m is maximal if and only A=m is a field. To see this, check that any ring R having only (0) and R as ideals is a field. Now m is maximal if and only if A=m has no ideals other than (0) and A=m (Prop. 1.4.1), so the result follows on taking R = A=m in the previous statement. In particular, every maximal ideal is prime. Proposition 1.5.1. Maximal ideals exist in any ring A with 1 6= 0. Proof. This is a standard application of Zorn's lemma. Let S be the set of all proper ideals in A, ordered by inclusion. Let T = fIαgα2A be a chain of proper ideals. Then the union [α2AIα is an ideal which is an upper bound of T in S. Hence by Zorn's lemma S has maximal elements, and this is what we claimed. Let us define Spec(A) to be the set of all prime ideals of A, and Specm(A) to be the subset consisting of all maximal ideals. These are some of the main objects of study in this course. The nomenclature \spectrum" comes from functional analysis, and will be explained later on. Also, pretty soon we will give the set Spec(A) the structure of a topological space and discuss the foundations of algebraic geometry... 1.6. Operations of contraction and extension. Fix a homomorphism φ : A ! B. For an ideal I ⊆ A define its extension Ie ⊆ B to the ideal generated by the e image φ(I); equivalently, I = \J J where J ⊆ B ranges over all ideals containing the set φ(I). Dually, for an ideal J ⊆ B define the contraction J c := φ−1(J), an ideal in A. Note that J prime ) J c prime, so contraction gives a map of sets Spec(B) ! Spec(A). On the other hand, contraction does not preserve maximality: consider the con- traction of J = (0) under the inclusion Z ,! Q. Therefore, a homomorphism φ : A ! B does not always induce a map of sets Specm(B) ! Specm(A). As we will see later on, there is a natural situation where φ does induce a map Specm(B) ! Specm(A): this happens if A; B happen to be finitely generated al- gebras over a field. This is quite important and is a consequence of Hilbert's Nullstellensatz, one of the first important theorems we will cover. 1.7. Nilradical. Define the nilradical of A by rad(A) := ff 2 A j f n = 0; for some n ≥ 1g: Check that rad(A) really is an ideal. Elements f satisfying the condition f n = 0 for some n ≥ 1 are called nilpotent. MATH 603: INTRODUCTION TO COMMUTATIVE ALGEBRA 3 Counterexample: For a non-commutative ring, it is no longer always true that 0 1 0 0 the sum of two nilpotent elements is nilpotent. The elements and , 0 0 1 0 0 1 in the ring M (R) over a ring R with 1 6= 0, are nilpotent, but their sum is 2 1 0 not. Lemma 1.7.1. \ rad(A) = p: p2Spec(A) Proof. The inclusion ⊆ is clear from the definition of prime ideal. For the reverse inclusion, suppose f 2 A is not nilpotent, i.e., suppose f n 6= 0 for every n ≥ 1. Let Σ = fI j f n 2= I; 8n ≥ 1g. This set is non-empty (it contains the ideal I = (0)) and this set has a maximal element (Zorn). Call it p. We claim that p is prime (and this is enough to prove ⊇). If not, choose x; y2 = p such that xy 2 p. Since p + (x) ) p and p + (y) ) p, we have f n 2 p + (x) and f m 2 p + (y) for some n+m positive integers n; m. But then f 2 p, a contradiction. n 1.8. Radical of anp ideal. Define rp(I) = ff 2 A j f 2 I; for some n ≥ 1g. Often, we denote r(I) = I. Check that I is an ideal. Note that rad(A) = p(0). Also, it is easy to check the following fact: p \ (1.8.1) I = p: p⊇I Here p ranges over prime ideals containing I. T 1.9. Jacobson radical. Define the ideal radm(A) = m: m2Specm(A) Proposition 1.9.1. radm(A) = fx 2 A j 1 − xy is a unit for all y 2 Ag: Proof. ⊆: Say x 2 radm(A). If y is such that 1 − xy is not a unit, then 1 − xy 2 m, for some maximal ideal m. But then 1 2 m, which is nonsense. ⊇: If x2 = m for some m, then (x) + m = A. But then 1 = z + xy, for some z 2 m and y 2 A. So 1 − xy is not a unit. Exercise 1.9.2. Prove the following statements. (i) r(r(I)) = r(I); (ii) rad(A=rad(A)) = 0; (iii) radm(A=radm(A)) = 0. We call an ideal I radical if r(I) = I. So, (i) shows that the radical ideals are precisely those of the form r(I), for some ideal I. There exist ideals which are not radical. Consider (X2) ⊂ C[X], and note that r(X2) = (X). Both rad(A) and radm(A) have some meaning in algebraic geometry, which we will return to shortly. Also, we will see that radical ideals play an important role too. 4 THOMAS J. HAINES 1.10. Modules. Let M be an abelian group. Then the ring of group endomor- phisms of M, denoted End(M), is a ring (in general non-commutative). Giving M the structure of an A-module is precisely the same thing as giving a ring homomor- phism A ! End(M): We have correspondences as in Prop. 1.4.1 fsubmodules N ⊆ Mg ! fquotients M ! M 0g and fsubmodules N 0 ⊆ M containing Ng ! fsubmodules of M=Ng: Also, we have the fundamental isomorphisms of A-modules N + N 0 N 0 (i) If N; N 0 ⊆ M, then =∼ ; N N \ N 0 M=N 0 (ii) If N 0 ⊆ N ⊆ M, then =∼ M=N.
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