Commutative Algebra. Pete L. Clark

Transcription

1 Commutative Algebra Pete L. Clark

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3 Contents Introduction 9 1. What is Commutative Algebra? 9 2. Why study Commutative Algebra? 9 3. What distinguishes this text More on the contents Acknowledgments 12 Chapter 1. Commutative rings Fixing terminology Adjoining elements Ideals and quotient rings The monoid of ideals of R Pushing and pulling ideals Maximal and prime ideals Products of rings A cheatsheet 24 Chapter 2. Galois Connections The basic formalism Lattice properties Examples of Antitone Galois Connections Antitone Galois Connections Decorticated: Relations Isotone Galois Connections Examples of Isotone Galois Connections 33 Chapter 3. Modules Basic definitions Finitely presented modules Torsion and torsionfree modules Tensor and Hom Projective modules Injective modules Flat modules Nakayama s Lemma Ordinal Filtrations and Applications Tor and Ext More on flat modules Faithful flatness 85 Chapter 4. First Properties of Ideals in a Commutative Ring 89 3

4 4 CONTENTS 1. Introducing maximal and prime ideals Radicals Comaximal ideals Local rings The Prime Ideal Principle of Lam and Reyes Minimal Primes An application to unit groups 102 Chapter 5. Examples of Rings Rings of numbers Rings of continuous functions Rings of holomorphic functions Kapovich s Theorem Polynomial rings Semigroup algebras 118 Chapter 6. Swan s Theorem Introduction to (topological) vector bundles Swan s Theorem Proof of Swan s Theorem Applications of Swan s Theorem Stably free modules 131 Chapter 7. Localization Definition and first properties Pushing and pulling via a localization map The fibers of a morphism Commutativity of localization and passage to a quotient Localization at a prime ideal Localization of modules Local properties Local characterization of finitely generated projective modules 151 Chapter 8. Noetherian rings Chain conditions on partially ordered sets Chain conditions on modules Semisimple modules and rings Normal Series The Krull-Schmidt Theorem Some important terminology Introducing Noetherian rings Theorems of Eakin-Nagata, Formanek and Jothilingam The Bass-Papp Theorem Artinian rings: structure theory The Hilbert Basis Theorem The Krull Intersection Theorem Krull s Principal Ideal Theorem The Dimension Theorem, following [BMRH] The Artin-Tate Lemma 181

5 CONTENTS 5 Chapter 9. Boolean rings First Properties Boolean Algebras Ideal Theory in Boolean Rings The Stone Representation Theorem Boolean Spaces Stone Duality 191 Chapter 10. Associated Primes and Primary Decomposition Associated Primes The support of a module Primary Ideals Primary Decomposition, Lasker and Noether Irredundant primary decompositions Uniqueness properties of primary decomposition Applications in dimension zero Applications in dimension one 204 Chapter 11. Nullstellensätze Zariski s Lemma Hilbert s Nullstellensatz The Real Nullstellensatz The Combinatorial Nullstellensatz The Finite Field Nullstellensatz Terjanian s Homogeneous p-nullstellensatz 215 Chapter 12. Goldman domains and Hilbert-Jacobson rings Goldman domains Hilbert rings Jacobson Rings Hilbert-Jacobson Rings Application: Zero-Dimensional Ideals in Polynomial Rings 227 Chapter 13. Spec R as a topological space The Prime Spectrum Properties of the spectrum: quasi-compactness Properties of the spectrum: connectedness Properties of the spectrum: separation and specialization Irreducible spaces Noetherian spaces Krull Dimension of Topological Spaces Jacobson spaces Hochster s Theorem Rank functions revisited The Forster-Swan Theorem 246 Chapter 14. Integral Extensions First properties of integral extensions Integral closure of domains Spectral properties of integral extensions 253

6 6 CONTENTS 4. Integrally closed domains The Noether Normalization Theorem Some Classical Invariant Theory Galois extensions of integrally closed domains Almost Integral Extensions 264 Chapter 15. Factorization Kaplansky s Theorem (II) Atomic domains, ACCP EL-domains GCD-domains GCDs versus LCMs Polynomial rings over UFDs Application: the Schönemann-Eisenstein Criterion Application: Determination of Spec R[t] for a PID R Power series rings over UFDs Nagata s Criterion 280 Chapter 16. Principal rings and Bézout domains Principal ideal domains Some structure theory of principal rings Euclidean functions and Euclidean rings Bézout domains 291 Chapter 17. Valuation rings Basic theory Ordered abelian groups Connections with integral closure Another proof of Zariski s Lemma Discrete valuation rings 302 Chapter 18. Normalization theorems The First Normalization Theorem The Second Normalization Theorem The Krull-Akizuki Theorem 307 Chapter 19. The Picard Group and the Divisor Class Group Fractional ideals The Ideal Closure Invertible fractional ideals and the Picard group Divisorial ideals and the Divisor Class Group 316 Chapter 20. Dedekind domains Characterization in terms of invertibility of ideals Ideal factorization in Dedekind domains Local characterization of Dedekind domains Factorization into primes implies Dedekind Generation of ideals in Dedekind domains Finitely generated modules over a Dedekind domain Injective Modules 326

7 CONTENTS 7 Chapter 21. Prüfer domains Characterizations of Prüfer Domains Butts s Criterion for a Dedekind Domain Modules over a Prüfer domain 334 Chapter 22. One Dimensional Noetherian Domains Residually Finite Domains Cohen-Kaplansky domains Rings of Finite Rank 344 Chapter 23. Structure of overrings Introducing overrings Overrings of Dedekind domains Elasticity in Replete Dedekind Domains Overrings of Prüfer Domains Kaplansky s Theorem (III) Every commutative group is a class group 359 Bibliography 363

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9 Introduction 1. What is Commutative Algebra? Commutative algebra is the study of commutative rings and attendant structures, especially ideals and modules. This is the only possible short answer I can think of, but it is not completely satisfying. We might as well say that Hamlet, Prince of Denmark is about a fictional royal family in late medieval Denmark and especially about the title (crown) prince, whose father (i.e., the King) has recently died and whose father s brother has married his mother (i.e., the Queen). Informative, but not the whole story! 2. Why study Commutative Algebra? What are the purely mathematical reasons for studying any subject of pure mathematics? I can think of two: I. Commutative algebra is a necessary and/or useful prerequisite for the study of other fields of mathematics in which we are interested. II. We find commutative algebra to be intrinsically interesting and we want to learn more. Perhaps we even wish to discover new results in this area. Most beginning students of commutative algebra can relate to the first reason: they need, or are told they need, to learn some commutative algebra for their study of other subjects. Indeed, commutative algebra has come to occupy a remarkably central role in modern pure mathematics, perhaps second only to category theory in its ubiquitousness, but in a different way. Category theory provides a common language and builds bridges between different areas of mathematics: it is something like a circulatory system. Commutative algebra provides core results and structures that other results and structures draw upon are overlayed upon: it is something like a skeleton. The branch of mathematics which most of all draws upon commutative algebra for its structural integrity is algebraic geometry, the study of geometric properties of manifolds and singular spaces which arise as solution sets to systems of polynomial equations. There is a hard lesson here: in the 19th century algebraic geometry split off from complex function theory and differential geometry as its own discipline and then burgeoned dramatically at the turn of the century and the years thereafter. But by 1920 or so the practitioners of the subject had found their way into territory in which purely geometric reasoning led to serious errors. In particular 9

10 10 INTRODUCTION they had been making arguments about how algebraic varieties behave generically, but they lacked the technology to even give a precise meaning to the term. Thus the subject eventually became invertebrate and began to collapse under its own weight. Then (starting in about 1930) came a heroic shoring up process in which the foundations of the subject were recast with commutative algebraic methods at the core. This was done several times over, in different ways, by Zariski, Weil, Serre and Grothendieck, among others. For the last 60 years it has been impossible to deeply study algebraic geometry without knowing commutative algebra a lot of commutative algebra. (More than is contained in these notes!) The other branch of mathematics which draws upon commutative algebra in an essential way is algebraic number theory. One sees this from the beginning in that the Fundamental Theorem of Arithmetic is the assertion that the ring Z is a unique factorization domain (UFD), a basic commutative algebraic concept. Moreover number theory was one of the historical sources of the subject. Notably the concept of Dedekind domain came from Dedekind s number-theoretic investigations. At the student level, algebraic number theory does not embrace commutative algebra as early or as thoroughly as algebraic geometry. This seems to me to be a pedagogical mistake: although one can do a good amount of algebraic number theory without explicit reliance on commutative algebra, this seems to come at the expense of not properly explaining what is going on. A modicum of commutative algebra greatly enriches the study of algebraic number theory: it clarifies it, generalizes it and (I believe) makes it more interesting. The interplay among number theory, algebraic geometry and commutative algebra flows in all directions. What Grothendieck did in the 1960s (with important contributions from Chevalley, Serre and others) was to create a single field of mathematics that encompassed commutative algebra, classical algebraic geometry and algebraic number theory: the theory of schemes. As a result, most contemporary number theorists are also partly commutative algebraists and partly algebraic geometers: we call this cosmopolitan take on the subject arithmetic geometry. There are other areas of mathematics that draw upon commutative algebra in important ways. To mention some which will show up in later in these notes: Differential topology. General topology. Invariant theory. Order theory. 3. What distinguishes this text The most straightforward raison d être for a commutative algebra text would be to provide a foundation for the subjects of algebraic geometry, arithmetic geometry and algebraic number theory. The bad news is that this task even, restricted to providing foundations for the single, seminal text of Hartshorne [Ha] is dauntingly large. The good news is that this has nevertheless been achieved some time ago by David Eisenbud (a leading contemporary expert on the interface of commutative algebra and algebraic geometry) in his text [Ei]. This work is highly recommended.

11 4. MORE ON THE CONTENTS 11 It is 797 pages long, so contains enough material for many courses in the subject. It would be folly to try to improve upon (and terrible drudgery to successfully imitate) Eisenbud s work here, and I certainly have not tried. The other standard commutative algebra texts are those by Atiyah-Macdonald [AM] and by Matsumura [M]. Any reader who is halfway serious in her study of commutative algebra should also have access to all of [AM], [M] and [Ei] and consult them frequently. While the current text does not rely on them in the logical sense, I am at best a part time commutative algebraist, and much of what I know comes from these texts. (Reading only derivative sources is never the best idea.) On the other hand, precisely because there are three standard excellent texts I have allowed my choice of topics to be, in places, much less standard. The topics covered in Atiyah-Macdonald s text in particular have become a de facto standard for a first course in commutative algebra. Here are the chapter titles from [AM]: 1. Rings and Ideals 2. Modules 3. Rings and Modules of Fractions 4. Primary Decomposition 5. Integral Dependence and Valuations 6. Chain Conditions 7. Noetherian Rings 8. Artin Rings 9. Discrete Valuation Rings and Dedekind Domains 10. Completions 11. Dimension Theory. The entire text is 126 pages, and a distinctive feature is that a substantial portion of the text is devoted to exercises, making [AM] one of the most amenable to student study graduate level mathematics texts I have ever seen. The exercises are especially attractive: some of them are easy, some of them are very challenging, and they treat both core topics and interesting side attractions. Much of the present text covers the material of the first nine chapters of [AM] but with a much more leisurely, detailed exposition. Many exercises in [AM] appear as proved results here. To give a specific example, Boolean rings appear in Exercises 1.11, 1.23, 1.24, 1.25, 3.28 of [AM], in which in particular proofs of the Stone Representation Theorem and Stone Duality Theorems are sketched. In this text 9 is devoted to Boolean rings, including proofs of these two results. The failure to cover completions and basic dimension theory in this text would be unforgivable were it not the case that [AM] covers it so nicely. Let us also compare to [M]. Here we treat the material of the first 12 sections of [M] except 8 (Completion and the Artin-Rees Lemma) as well as some material from 20 (UFDs). This is less than one third of the material covered in [M]. 4. More on the contents As mentioned above, one of the distinguishing features of this text and one of the things which makes it much lengthier compared to the portions of [AM] and [M] that cover the same material is that we digress to include many applications to other parts of mathematics. At one point I had the idea that every section of core material should be followed by a section giving applications. This conceit was not really feasible, but there are still some entire sections devoted to applications (generally characterized by making contact with topics outside of commutative algebra by their relative independence from the rest of the text):

12 12 INTRODUCTION 2 on Galois connections. 6 on vector bundles and Swan s Theorem. 9 on Boolean rings, Boolean spaces and Stone Duality. As for significant parts of sections, we have: 5.2 on rings of continuous functions. 5.3 on rings of holomorphic functions on the real Nullstellensatz 11.4 on the combinatorial Nullstellensatz 11.5 on the finite field Nullstellensatz 11.6 on Terjanian s Nullstellensatz 13.6 on Hochster s Theorem 14.6 on invariant theory and the Shephard-Todd-Chevalley Theorem 5. Acknowledgments Thanks to Pablo Barenbaum, Max Bender, Martin Brandenburg, John Doyle, Georges Elencwajg, Ernest Guico, Emil Jerabek, Jason Joseph, Keenan Kidwell, David Krumm, Allan Lacy, Casey LaRue, Stacy Musgrave, Hans Parshall, Alon Regev, Tomasz Rzepecki, Jacob Schlather, Jack Schmidt, Mariano Suárez-Álvarez, James Taylor, Peter Tamaroff, Matthé van der Lee and Lori Watson for catching errors 1 and making other useful suggestions. 1 Of which many, many remain: your name could go here!

13 CHAPTER 1 Commutative rings 1. Fixing terminology We are interested in studying properties of commutative rings with unity. By a general algebra R, we mean a triple (R, +, ) where R is a set endowed with a binary operation + : R R R called addition and a binary operation : R R R called multiplication satisfying the following: (CG) (R, +) is a commutative group, (D) For all a, b, c R, (a + b) c = a c + b c, a (b + c) = a b + a c. For at least fifty years, there has been agreement that in order for an algebra to be a ring, it must satisfy the additional axiom of associativity of multiplication: (AM) For all a, b, c R, a (b c) = (a b) c. A general algebra which satisfies (AM) will be called simply an algebra. A similar convention that is prevalent in the literature is the use of the term nonassociative algebra to mean what we have called a general algebra: i.e., a not necessarily associative algebra. A ring R is said to be with unity if there exists a multiplicative identity, i.e., an element e of R such that for all a R we have e a = a e = a. If e and e are two such elements, then e = e e = e. In other words, if a unity exists, it is unique, and we will denote it by 1. A ring R is commutative if for all x, y R, x y = y x. In these notes we will be (almost) always working in the category of commutative rings with unity. In a sense which will shortly be made precise, this means that the identity 1 is regarded as part of the structure of a ring and must therefore be preserved by all homomorphisms. Probably it would be more natural to study the class of possibly non-commutative rings with unity, since, as we will see, many of the fundamental constructions of rings give rise, in general, to non-commutative rings. But if the restriction to commutative rings (with unity!) is an artifice, it is a very useful one, since two of the most fundamental notions in the theory, that of ideal and module, become 13

14 14 1. COMMUTATIVE RINGS significantly different and more complicated in the non-commutative case. It is nevertheless true that many individual results have simple analogues in the noncommutative case. But it does not seem necessary to carry along the extra generality of non-commutative rings; rather, when one is interested in the non-commutative case, one can simply remark Proposition X.Y holds for (left) R-modules over a noncommutative ring R. Notation: Generally we shall abbreviate x y to xy. Moreover, we usually do not use different symbols to denote the operations of addition and multiplication in different rings: it will be seen that this leads to simplicity rather than confusion. Group of units: Let R be a ring with unity. An element x R is said to be a unit if there exists an element y such that xy = yx = 1. convention on exercises: Throughout the exercises, a ring means a commutative ring unless explicit mention is made to the contrary. Some but not all of the results in the exercises still hold for non-commutative rings, and it is left to the interested reader to explore this. Exercise 1.1. a) Show that if x is a unit, the element y with xy = yx = 1 is unique, denoted x 1. b) Show that if x is a unit, so is x 1. c) Show that, for all x, y R, xy is a unit x and y are both units. d) Deduce that the units form a commutative group, denoted R, under multiplication. Remark 1. For elements x, y in a non-commutative ring R, if x and y are units so is xy, but the converse need not hold. (Thus Exercise 1.1c) is an instance of a result in which commutativity is essential.) Nevertheless this is enough to deduce that in any ring the units R form a group...which is not necessarily commutative. Example (Zero ring): Our rings come with two distinguished elements, the additive identity 0 and the multiplicative identity 1. Suppose that 0 = 1. Then for x R, x = 1 x = 0 x, whereas in any ring 0 x = (0 + 0) x = 0 x + 0 x, so 0 x = 0. In other words, if 0 = 1, then this is the only element in the ring. It is clear that for any one element set R = {0}, = 0 0 = 0 endows R with the structure of a ring. We call this ring the zero ring. The zero ring exhibits some strange behavior, such that it must be explicitly excluded in many results. For instance, the zero element is a unit in the zero ring, which is obviously not the case in any nonzero ring. A nonzero ring in which every nonzero element is a unit is called a division ring. A commutative division ring is called a field. Let R and S be rings (with unity). of sets which satisfies A homomorphism f : R S is a map (HOM1) For all x, y R, f(x + y) = f(x) + f(y).

15 1. FIXING TERMINOLOGY 15 (HOM2) For all x, y R, f(xy) = f(x)f(y). (HOM3) f(1) = 1. Note that (HOM1) implies f(0) = f(0 + 0) = f(0) + f(0), so f(0) = 0. Thus we do not need to explcitly include f(0) = 0 in the definition of a group homomorphism. For the multiplicative identity however, this argument only shows that if f(1) is a unit, then f(1) = 1. Therefore, if we did not require (HOM3), then for instance the map f : R R, f(x) = 0 for all x, would be a homomorphism, and we do not want this. Exercise 1.2. Suppose R and S are rings, and let f : R S be a map satisfying (HOM1) and (HOM2). Show that f is a homomorphism of rings (i.e., satisfies also f(1) = 1) iff f(1) S. A homomorphism f : R S is an isomorphism if there exists a homomorphism g : S R such that: for all x R, g(f(x)) = x; and for all y S, f(g(y)) = y. Exercise 1.3. Let f : R S be a homomorphism of rings. Show the following are equivalent: (i) f is a bijection. (ii) f is an isomorphism. In many algebra texts, an isomorphism of rings (or groups, etc.) is defined to be a bijective homomorphism, but this gives the wrong idea of what an isomorphism should be in other mathematical contexts (e.g. for topological spaces). Rather, having defined the notion of a morphism of any kind, one defines isomorphism in the way we have above. Exercise 1.4. a) Suppose R and S are both rings on a set containing exactly one element. Show that there is a unique ring isomorphism from R to S. (This is a triviality, but explains why are we able to speak of the zero ring, rather than simply the zero ring associated to one element set. We will therefore denote the zero ring just by 0.) b) Show that any ring R admits a unique homomorphism to the zero ring. One says that the zero ring is the final object in the category of rings. Exercise 1.5. Show: for a not-necessarily-commutative-ring S there exists a unique homomorphism from the ring Z of integers to S. (Thus Z is the initial object in the category of not-necessarily-commutative-rings. It follows immediately that it is also the initial object in the category of rings.) A subring R of a ring S is a subset R of S such that (SR1) 1 R. (SR2) For all r, s R, r + s R, r s R, and rs R. Here (SR2) expresses that the subset R is an algebra under the operations of addition and multiplication defined on S. Working, as we are, with rings with unity, we have to be a bit more careful: in the presence of (SR2) but not (SR1) it is possible that R either does not have a multiplicative identity or, more subtly, that it has a multiplicative identity which is not the element 1 S.

16 16 1. COMMUTATIVE RINGS An example of the first phenomenon is S = Z, R = 2Z. An example of the second is S = Z, R = 0. A more interesting example is S = Z Z i.e., the set of all ordered pairs (x, y), x, y Z with (x 1, y 1 ) + (x 2, y 2 ) = (x 1 + x 2, y 1 + y 2 ), (x 1, y 1 ) (x 2, y 2 ) = (x 1 x 2, y 1 y 2 ) and R = {(0, y) y Z}. Then with the induced addition and multiplication from S, R is isomorphic to the ring Z and the element (0, 1) serves as a multiplicative identity on R which is different from the (always unique) multiplicative identity 1 S = (1, 1), so according to our conventions R is not a subring of S. Notice that if R is a subring of S, the inclusion map R S is an injective homomorphism of rings. Conversely, if ι : R S is an injective ring homomorphism, then R = ι(r) and ι(r) is a subring of S, so essentially we may use ι to view R as a subring of S. The only proviso here is that this certainly depends on ι: in general there may be other injective homomorphisms ι : R S which realize R as a different subset of S, hence a different subring. 2. Adjoining elements Let ι : R S be an injective ring homomorphism. As above, let us use ι to view R as a subring of S; we also say that S is an extension ring of R and write S/R for this (note: this has nothing to do with cosets or quotients!) We wish now to consider rings T such that R T S; such a ring T might be called a subextension of S/R or an intermediate ring. Let X = {x i } be a subset of S. Then the partially ordered set of all subrings of T containing R and X contains a bottom element, given (as usual!) by taking the intersection of all of its elements. (This partially ordered set is nonempty, since S is in it.) We call this the ring obtained by adjoining the elements of X to R. In the commutative case, we denote this ring by R[{x i }], for reasons that will become more clear when we discuss polynomial rings in 5.4. Example: Take R = Z, S = C. Then Z[i] = Z[ 1] is the smallest subring of C containing (Z and) 1. Example: Take R = Z, S = Q, let P be any set of prime numbers, and put X = { 1 p } p P. Then there is a subring Z P := Z[{ 1 p } p P] of Q. Exercise 1.6. Let P, Q be two sets of prime numbers. Show the following are equivalent: (i) Z P = ZQ. (ii) Z P = Z Q. (iii) P = Q. Exercise 1.7. Show: every subring of Q is of the form Z P for some P. The adjunction process R R[X] is defined only relative to some extension ring S of R, although the notation hides this. In fact, one of the recurrent themes of the subject is the expression of the adjunction process in a way which depends only on R itself. In the first example, this is achieved by identifying 1 with its minimal polynomial t and replacing Z[ 1] with the quotient ring Z[t]/(t 2 + 1). The

17 3. IDEALS AND QUOTIENT RINGS 17 second example will eventually be turned around: we will be able to give an independent definition of Z P as a certain ring of fractions formed from Z and then Q will be the ring of fractions obtained by taking P to be the set of all prime numbers. Nevertheless, the existence of such turnabouts should not cause us to forget that adjunction is relative to an extension; indeed forgetting this can lead to serious trouble. For instance, if 3 2 is the unique real cube root of 2 and ζ3 is a primitive cube root of unity, then the three complex numbers with cube 2 are z 1 = 3 2, z 2 = 3 2ζ 3 and z 3 = 3 2ζ3. 2 Each of the rings Q[z 1 ], Q[z 2 ], Q[z 3 ] is isomorphic to the ring Q[t]/(t 3 2), so all three are isomorphic to each other. But they are not the same ring: on the one hand Q[z 1 ] is contained in R and the other two are not. More seriously Q[z 1, z 2, z 3 ] = Q[ 3 2, ζ 3 ], which strictly contains any one of Q[z 1 ], Q[z 2 ] and Q[z 3 ]. 3. Ideals and quotient rings Let f : R S be a homomorphism of rings, and put I = f 1 (0) = {x R f(x) = 0}. In particular f is a homomorphism of commutative groups (R, +) (S, +), I is a subgroup of (R, +). Moreover, it enjoys both of the following properties: (LI) For all j I and x R, xj I. (RI) For all i I and y R, iy I. Indeed, f(xj) = f(x)f(j) = f(x) 0 = 0 = 0 f(y) = f(i)f(y) = f(iy). In general, let R be a ring. An ideal is a subset I R which is a subgroup of (R, +) (in particular, 0 I) and which satisfies (LI) and (RI). Theorem 1.1. Let R be a ring, and let I be a subgroup of (R, +). The following are equivalent: (i) I is an ideal of R. (ii) There exists a ring structure on the quotient group R/I making the additive homomorphism R R/I into a homomorphism of rings. When these conditions hold, the ring structure on R/I in (ii) is unique, and R/I is called the quotient of R by the ideal I. Proof. Consider the group homomorphism q : R R/I. If we wish R/I to be a ring in such a way so that q is a ring homomorphism, we need (x + I)(y + I) = q(x)q(y) = q(xy) = (xy + I). This shows that there is only one possible ring structure, and the only question is whether it is well-defined. For this we need that for all i, j I, (x+i)(y +j) xy = xj +iy +ij I. Evidently this holds for all x, y, i, j iff (LI) and (RI) both hold. Remark: If R is commutative, then of course there is no difference between (LI) and (RI). For a non-commutative ring R, an additive subgroup I satisfying condition (LI) but not necessarily (RI) (resp. (RI) but not necessarily (LI)) is called a left ideal (resp. a right ideal). Often one says two-sided ideal to emphasize that (LI) and (RI) both hold. Much of the additional complexity of the non-commutative

18 18 1. COMMUTATIVE RINGS theory comes from the need to distinguish between left, right and two-sided ideals. We do not wish to discuss such complexities here, so henceforth in this section we assume (except in exercises, when indicated) that our rings are commutative. Example: In R = Z, for any integer n, consider the subset (n) = nz = {nx x Z} of all multiples of n. This is easily seen to be an ideal. 1 The quotient Z/nZ is the ring of integers modulo n. An ideal I R is called proper. Exercise 1.8. Let R be a ring and I an ideal of R. Show the following are equivalent: (i) I R. (ii) I = R. Exercise 1.9. a) Let R be a commutative ring. Show that R is a field iff R has exactly two ideals, 0 and R. b) Let R be a not necessarily commutative ring. Show the following are equivalent: (i) The only one-sided ideals of R are 0 and R. (ii) R is a division ring. c) For a field k and an integer n > 1, show that the matrix ring M n (k) has no two-sided ideals but is not a division ring. Exercise Some contemporary undergraduate algebra texts define the finite ring Z/nZ in a different and apparently simpler way: put Z n = {0, 1,..., n 1}. For any integer x, there is a unique integer k such that x kn Z n. Define a function mod n : Z Z n by mod n(x) := x kn. We then define + and on Z n by x + y := mod n(x + y), xy = mod n(xy). Thus we have avoided any mention of ideals, equivalence classes, quotients, etc. Is this actually simpler? (Hint: how do we know that Z n satisfies the ring axioms?) For any commutative ring R and any element y R, the subset (y) = yr = {xy x R} is an ideal of R. Such ideals are called principal. A principal ideal ring is a commutative ring in which each ideal is principal. Exercise a) The intersection of any family of (left, right or two-sided) ideals in a not-necessarily-commutative-ring is a (left, right or two-sided) ideal. b) Let {I i } be a set of ideals in the commutative ring R. Show that i I i has the following property: for any ideal J of R such that J I i for all i, J i I. Let R be a ring and S a subset of R. There is then a smallest ideal of R containing S, namely I i, where I i are all the ideals of R containing S. We call this the ideal generated by S. This is a top-down description; as usual, there is a complementary bottom-up description which is not quite as clean but often more useful. Namely, put S := { r i s i r i R, s i S} i.e., the set of all finite sums of an element of R times an element of S. Proposition 1.2. For a subset S of a commutative ring R, S is an ideal, the intersection of all ideals containing S. 1 If this is not known and/or obvious to the reader, these notes will probably be too brisk.

19 3. IDEALS AND QUOTIENT RINGS 19 Exercise Prove Proposition 1.2. When S is a subset of R such that I = S, we say S is a set of generators for I. In general the same ideal will have many (most often infinitely many) sets of generators. An ideal I is principal if it can be generated by a single element. In any ring, the zero ideal 0 = 0 and the entire ring R = 1 are principal. For x R, we tend to denote the principal ideal generated by x as either Rx or (x) rather than x. An ideal I is finitely generated if...it admits a finite set of generators. 2 Stop and think for a moment: do you know an example of an ideal which is not finitely generated? You may well find that you do not. It turns out that there is a very large class of rings including most or all of the rings you are likely to meet in undergraduate algebra for which every ideal is finitely generated. A ring R in which every ideal is finitely generated is called Noetherian. This is probably the single most important class of rings, as we will come to appreciate slowly but surely over the course of these notes. Exercise Let R be a ring. a) For ideals I and J of R, define I + J = {i + j i I, j J}. Show that I + J = I J is the smallest ideal containing both I and J. b) Extend part a) to any finite number of ideals I 1,..., I n. c) Suppose {I i } is a set of ideals of I. Give an explicit description of the ideal I i. The preceding considerations show that the collection of all ideals of a commutative ring R, partially ordered by inclusion, form a complete lattice. If I is an ideal in the ring R, then there is a correspondence between ideals J of R containing I and ideals of the quotient ring R/I, exactly as in the case of a normal subgroup of a group: Theorem 1.3. (Correspondence theorem) Let I be an ideal of a ring R, and denote the quotient map R R/I by q. Let I(R) be the lattice of ideals of R, I I (R) be the sublattice of ideals containing I and I(R/I) the lattice of ideals of the quotient ring R/I. Define maps Φ : I(R) I(R/I), J (I + J)/I, Ψ : I(R/I) I(R), J q 1 (J). Then Ψ Φ(J) = I + J and Φ Ψ(J) = J. In particular Ψ induces an isomorphism of lattices from I(R/I) to I I (R). Proof. For all the abstraction, the proof is almost trivial. For J I(R), we check that Ψ(Φ(J)) = Ψ(J +I (mod I)) = {x R x+i J +I} = J +I I I (R). Similarly, for J I(R/I), we have Φ(Ψ(J)) = J. Remark: In fancier language, the pair (Φ, Ψ) give an isotone Galois connection between the partially ordered sets I(R) and I(R/I). The associated closure operator Φ Ψ on I(R/I) is the identity, whereas the closure operator Ψ Φ on I(R) 2 Well, obviously. Nevertheless this definition is so critically important that it would have been a disservice to omit it.

20 20 1. COMMUTATIVE RINGS carries each ideal J to the smallest ideal containing both J and I. 3 The Correspondence Theorem will be our constant companion. As is common, we will often use the map Ψ to identify the sets I(R/I) and I I (R). Exercise Let I be an ideal of R and {J i } be a set of ideals of R. Show that Φ preserves suprema and Ψ preserves infima: and Φ( J i ) = Φ(J i ) Ψ( J i ) = Ψ(J i ). 4. The monoid of ideals of R Let I and J be ideals of the ring R. The product ideal IJ is the least ideal containing all elements of the form xy for x I and y J. (It is easy to see that IJ = { x i y i x i I, y i J} is precisely the set of all finite sums of such products.) Recall that we have written I(R) for the lattice of all ideals of R. Then (I, J) IJ gives a binary operation on I(R), the ideal product. Exercise Show that I(R) under the ideal product is a commutative monoid, with identity element R and absorbing element the (0) ideal of R. 4 If you are given a commutative monoid M, then invariably the property you are hoping it has is cancellation: for all x, y, z M, xz = yz = x = y. 5 For example, if R is a ring, then the set R of nonzero elements of R is cancellative iff R is a domain. Note that 0 is an absorbing element of (R, ), which we have removed in order to have any chance at cancellativity. Exercise a) Let M be a cancellative monoid of cardinality greater than one. Show that M does not have any absorbing elements. b) Let R be a ring which is not the zero ring. Show that the monoid I(R) is not cancellative. In light of the previous exercise, for a domain R we define I (R) to be the monoid of nonzero ideals of R under multiplication. Warning: Just because R is a domain, I (R) need not be cancellative! Exercise Let R = Z[ 3], and let p 2 = 1 + 3, 1 3 (i.e., the ideal generated by these two elements. a) Show that #R/(2) = 4 and R/p 2 = Z/2Z. b) Show that p 2 2 = p 2 (2). c) Conclude that I (R) is not cancellative. Exercise Let R be a PID. Show that I (R) is cancellative. Exercise Show: for a commutative monoid M, the following are equivalent: (i) M is cancellative. 3 This point of view will be explored in more detail in 2. 4 An element z of a monoid M is called absorbing if for all x M, zx = xz = z. 5 Well, obviously this is an exaggeration, but you would be surprised how often it is true.

21 5. PUSHING AND PULLING IDEALS 21 (ii) There exists a commutative group G and an injective monoid homomorphism ι : M G. Exercise Let M be a commutative monoid. A group completion of M consists of a commutative group G(M) and a monoid homomorphism F : M G(M) which is universal for monoid homomorphisms into a commutative group. That is, for any commutative group G and monoid homomorphism f : M G, there exists a unique homomorphism of groups q : G G(M) such that F = q f. a) Show that any two group completions are isomorphic. b) Show that any commutative monoid has a group completion. c) Show that a commutative monoid injects into its group completion iff it is cancellative. 5. Pushing and pulling ideals Let f : R S be a homomorphism of commutative rings. We can use f to transport ideals from R to S and also to transport ideals from S to R. More precisely, for I an ideal of R, consider f(i) as a subset of S. Exercise Give an example to show that f(i) need not be an ideal of S. Nevertheless we can consider the ideal it generates: we define f (I) = f(i), and we call f (I) the pushforward of I to S. Similarly, let J be an ideal of S, and consider its complete preimage in R, i.e., f 1 (J) = {x R f(x) J}. As you are probably already aware, preimages have much nicer algebraic properties than direct images, and indeed f 1 (J) is necessarily an ideal of R. We denote it by f (J) and call it the pullback of J to R. Example: Suppose that I is an ideal of R, S = R/I and f : R R/I is the quotient map. In this case, pushforwards and pullbacks were studied in detail in Theorem 1.3. In this case f : I(S) I(R) is an injection, which allows us to view the lattice of ideals of S as a sublattice of the lattice of ideals of R. Moreover we have a push-pull formula: for all ideals J of R, f f J = J + I and also a pull-push formula: for all ideals J of R/I, f f J = J. These formulas are extremely useful at all points in the study of ring theory. More generally, whenever one meets a homomorphism f : R S of rings (or better, a certain class of homomorphisms), it is fruitful to ask about properties of f and f : in particular, is f necessarily injective, or surjective? Can we identify the composite maps f f and/or f f? In these notes, the most satisfying and important answers will come for localizations and integral extensions.

22 22 1. COMMUTATIVE RINGS 6. Maximal and prime ideals An ideal m of R is maximal if it is proper and there is no proper ideal of R strictly containing m. An ideal p of R is prime if for all x, y R, xy p implies x p or y p or both. Exercise For an ideal I in a ring R, show: the following are equivalent: (i) I is maximal. (ii) R/I is a field. Exercise For an ideal I in a ring R, show: the following are equivalent: (i) I is prime. (ii) R/p is a domain. Exercise Show: maximal ideals are prime. Exercise Let f : R S be a homomorphism of rings. a) Let I be a prime ideal of R. Show that f I need not be a prime ideal of S. b) Let J be a prime ideal of S. Show that f J is a prime ideal of R. c) Let J be a maximal ideal of S. Show that f J need not be maximal in R. If I and J are ideals of a ring R, we define the colon ideal 6 (I : J) = {x R xj I}. Exercise Show: (I : J) is indeed an ideal of R. 7. Products of rings Let R 1 and R 2 be rings. The Cartesian product R 1 R 2 has the structure of a ring with componentwise addition and multiplication: (r 1, r 2 ) + (s 1, s 2 ) := (r 1 + s 1, r 2 + s 2 ). (r 1, r 2 ) (s 1, s 2 ) := (r 1 s 1, r 2 s 2 ). Exercise a) Show that R 1 R 2 is commutative iff both R 1 and R 2 are commutative. b) R 1 R 2 has an identity iff both R 1 and R 2 do, in which case e := (e 1, e 2 ) is the identity of R 1 R 2. As for any Cartesian product, R 1 R 2 comes equipped with its projections π 1 : R 1 R 2 R 1, (r 1, r 2 ) r 1 π 2 : R 1 R 2 R 2, (r 1, r 2 ) r 2. The Cartesian product X 1 X 2 of sets X 1 and X 2 satisfies the following universal property: for any set Z and any maps f 1 : Z X 1, f 2 : Z X 2, there exists a unique map f : Z X 1 X 2 such that f 1 = π 1 f, f 2 = π 2 f. The Cartesian product R 1 R 2 satisfies the analogous universal property in the category of rings. Exercise For rings R 1, R 2, S and ring homomorphisms f i : S R i, there exists a unique homomorphism of rings f : S R 1 R 2 such that f i = π i f. 6 The terminology is unpleasant and is generally avoided as much as possible. One should think of (I : J) as being something like the ideal quotient I/J (which of course has no formal meaning). Its uses will gradually become clear.

23 7. PRODUCTS OF RINGS 23 So the Cartesian product of R 1 and R 2 is also the product in the categorical sense. As with sets, we can equally well take the Cartesian product over an arbitrary indexed family of rings: if {R i } i I is a family of rings, their Cartesian product i I R i becomes a ring under coordinatewise addition and multiplication, and satisfies the universal property of the product. Details are left to the reader. It is natural to ask whether the category of rings has a direct sum as well. In other words, given rings R 1 and R 2 we are looking for a ring R together with ring homomorphisms ι i : R i R such that for any ring S and homomorphisms f i : R i S, there exists a unique homomorphism f : R S such that f i = f ι i. We recall that in the category of abelian groups, the Cartesian product group G 1 G 2 also the categorical direct sum, with ι 1 : g (g, 0) and ι 2 : g (0, g). Since each ring has in particular the structure of an abelian group, it is natural to wonder whether the same might hold true for rings. However, the map ι 1 : R 1 R 1 R 2 does not preserve the multiplicative identity (unless R 2 = 0), so is not a homomorphism of rings when identities are present. Moreover, even in the category of algebras, in order to satisfy the universal property on the underlying additive subgroups, the homomorphism f is uniquely determined to be (r 1, r 2 ) f 1 (r 1 ) + f 2 (r 2 ), and it is easily checked that this generally does not preserve the product. (The category of rings does have direct sums in the categorical sense: the categorical direct sum of R 1 and R 2 is given by the tensor product R 1 Z R 2.) Now returning to the case of commutative rings, let us consider the ideal structure of the product R = R 1 R 2. If I 1 is an ideal of R 1, then I 1 {0} = {(i, 0) i I} is an ideal of the product; moreover the quotient R/I 1 is isomorphic to R 1 /I 1 R 2. Similarly, if I 2 is an ideal, {0} I 2 is an ideal of R 2. Finally, if I 1 is an ideal of R 1 and I 2 is an ideal of R 2, then I 1 I 2 := {(i 1, i 2 ) i 1 I 1, i 2 I 2 } is an ideal of R. In fact we have already found all the ideals of the product ring: Proposition 1.4. Let R 1 and R 2 be commutative rings, and let I be an ideal of R := R 1 R 2. Put I 1 := {r 1 R 1 r 2 R 2 (r 1, r 2 ) I}, I 2 := {r 2 R 2 r 1 R 1 (r 1, r 2 ) I}. Then I = I 1 I 2 = {(i 1, i 2 ) i 1 I 1, i 2 I 2 }. Proof. Observe first that I 1 {0} and {0} I 2 are ideals of R contained in I. Indeed, if i 1 I 1, then (i 1, r 2 ) I for some r 2 and then (i 1, 0) = (i 1, r 2 ) (1, 0), and similarly for I 2. Therefore Conversely, if (x, y) I, then I 1 I 2 = (I 1 {0}) + ({0} I 2 ) I. (x, y) = (x, 0)(1, 0) + (0, y)(0, 1) I 1 I 2. Another way to express the result is that, corresponding to a decomposition R = R 1 R 2, we get a decomposition I(R) = I(R 1 ) I(R 2 ).

24 24 1. COMMUTATIVE RINGS Let us call a commutative ring R disconnected if there exists nonzero rings R 1, R 2 such that R = R 1 R 2, and connected otherwise. 7 If R is disconnected, then choosing such an isomorphism ϕ, we may put I 1 = ϕ 1 (R 1 {0}) and I 2 = ϕ 1 ({0} R 2 ). Evidently I 1 and I 2 are ideals of R such that I 1 I 2 = {0} and I 1 I 2 = R. Conversely, if in a ring R we can find a pair of ideals I 1, I 2 with these properties then it will follow from the Chinese Remainder Theorem (Theorem 4.19) that the natural map Φ : R R/I 2 R/I 1, r (r + I 2, r + I 1 ) is an isomorphism. Now Φ restricted to I 1 induces an isomorphism of groups onto R/I 2 (and similarly with the roles of I 1 and I 2 reversed). We therefore have a distinguished element of I 1, e 1 := Φ 1 (1). This element e 1 is an identity for the multiplication on R restricted to I 1 ; in particular e 2 1 = e 1 ; such an element is called an idempotent. In any ring the elements 0 and 1 are idempotents, called trivial; since e 1 = Φ 1 (1, 0) and not the preimage of (0, 0) or of (1, 1) e 1 is a nontrivial idempotent. Thus a nontrivial decomposition of a ring implies the presence of nontrivial idempotents. The converse is also true: Proposition 1.5. Suppose R is a ring and e is a nontrivial idempotent element of R: e 2 = e but e 0, 1. Put I 1 = Re and I 2 = R(1 e). Then I 1 and I 2 are ideals of R such that I 1 I 2 = 0 and R = I 1 + I 2, and therefore R = R/I 1 R/I 2 is a nontrivial decomposition of R. Exercise Prove Proposition 1.5. Exercise Generalize the preceding discussion to decompositions into a finite number of factors: R = R 1 R n. 8. A cheatsheet Let R be a commutative ring. Here are some terms that we will analyze in loving detail later, but would like to be able to mention in passing whenever necessary. R is a domain if xy = 0 = x = 0 or y = 0. An ideal p of R is prime if the quotient ring R/p is a domain. Equivalently, p is an ideal such that xy p = x p or y p. An ideal m of R is maximal if it is proper i.e., not R itself and not strictly contained in any larger proper ideal. Equivalently, m is an ideal such that the quotient ring R/m is a field. R is Noetherian if it satisfies any of the following equivalent conditions: 8 (i) For any nonempty set S of ideals of R, there exists I S which is not properly contained in any J S. (ii) There is no infinite sequence of ideals I 1 I 2... in R. (iii) Every ideal of R is finitely generated. 7 We will see later that there is a topological space Spec R associated to every ring, and Spec R is a disconnected topological space iff R can be written as a nontrivial product of rings 8 See Theorem 8.25 for a proof of their equivalence.

25 8. A CHEATSHEET 25 R is Artinian (or sometimes, an Artin ring) if the partially ordered set of ideals of R satisfies the descending chain condition: there is no infinite sequence of ideals I 1 I If I and J are ideals of a ring R, we define the colon ideal 9 (I : J) is also an ideal. (I : J) = {x R xj I}. Let R S be an inclusion of rings. We say that s S is integral over R if there are a 0,..., a n 1 R such that s n + a n 1 s n a 1 s + a 0 = 0. We say that S is integral over R if every element of S is integral over R. This is the appropriate generalization to rings of the notion of an algebraic field extension. We will study integral elements and extensions, um, extensively in 14, but there is one easy result that we will need earlier, so we give it now. Proposition 1.6. Let R S be an integral extension of domains. If S is a field then R is a field. Proof. Let α R. Then α 1 is integral over R: there are a i R such that Multiplying through by α n 1 gives α n = a n 1 α n a 1 α 1 + a 0. α 1 = a n 1 + a n 2 α a 1 α n 2 + a 0 α n 1 R. 9 The terminology is unpleasant and is generally avoided as much as possible. One should think of (I : J) as being something like the ideal quotient I/J (which of course has no formal meaning). Its uses will gradually become clear.

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27 CHAPTER 2 Galois Connections 1. The basic formalism Let (X, ) be a partially ordered set. We denote by X the order dual of X: it has the same underlying set as X but the inverse order relation: x y y x. Let (X, ) and (Y, ) be partially ordered sets. A map f : X Y is isotone (or order-preserving) if for all x 1, x 2 X, x 1 x 2 = f(x 1 ) f(x 2 ); f is antitone (or order-reversing) if for all x 1, x 2 X, x 1 x 2 = f(x 1 ) f(x 2 ). Exercise 2.1. Let X, Y, Z be partially ordered sets, and let f : X Y, g : Y Z be functions. Show: a) If f and g are isotone, then g f is isotone. b) If f and g are antitone, then g f is isotone. c) If one of f and g is isotone and the other is antitone, then g f is antitone. Let (X, ) and (Y, ) be partially ordered sets. An antitone Galois connection between X and Y is a pair of maps Φ : X Y and Ψ : Y X such that: (GC1) Φ and Ψ are both antitone maps, and (GC2) For all x X and all y Y, x Ψ(y) y Φ(x). There is a pleasant symmetry in the definition: if (Φ, Ψ) is a Galois connection between X and Y, then (Ψ, Φ) is a Galois connection between Y and X. If (X, ) is a partially ordered set, then a mapping f : X X is called a closure operator if it satisfies all of the following properties: (C1) For all x X, x f(x). (C2) For all x 1, x 2 X, x 1 x 2 = f(x 1 ) f(x 2 ). (C3) For all x X, f(f(x)) = f(x). Proposition 2.1. The mapping Ψ Φ is a closure operator on (X, ) and the mapping Φ Ψ is a closure operator on (Y, ). Proof. By symmetry, it is enough to consider the mapping x Ψ(Φ(x)) on X. If x 1 x 2, then since both Φ and Ψ are antitone, we have Φ(x 1 ) Φ(x 2 ) and thus Ψ(Φ(x 1 )) Ψ(Φ(x 1 )): (C2). For x X, Φ(x) Φ(x), and by (GC2) this implies x Ψ(Φ(x)): (C1). Finally, for x X, applying (C1) to the element Ψ(Φ(x)) of X gives Ψ(Φ(x)) Ψ(Φ(Ψ(Φ(x)))). 27