MATH 8020 CHAPTER 1: COMMUTATIVE RINGS

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1 MATH 8020 CHAPTER 1: COMMUTATIVE RINGS PETE L. CLARK Contents 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 Additional exercises Fixing terminology. 1. Commutative rings We are interested in studying properties of commutative rings with unity. 1 So let us begin by clarifying this terminology. 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 1 In particular, all rings considered in these notes are meant to be commutative. Any references to possibly non-commutative rings are vestiges of an older version of these notes. If you find them, please point them out to be so they can be removed. 1

2 2 PETE L. CLARK 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 working always in the category of commutative rings with unity. In a sense which will be made precise shortly, this means that the identity 1 is regarded as a part of the structure of a ring, and must therefore be preserved by all homomorphisms of rings. 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 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 (commutative!) ring. An element x R is said to be a unit if there exists an element y such that xy = 1. Exercise 1.1: a) Show that if x is a unit, the element y with xy = 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. 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 rin g 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.

3 MATH 8020 CHAPTER 1: COMMUTATIVE RINGS 3 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). (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 that TFAE: (i) f is a bijection. (ii) f is an isomorphism. Remark: 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 that for any ring S there exists a unique homomorphism from the ring Z of integers to S. (One says that Z is the initial object in the category of rings.)

4 4 PETE L. CLARK 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. 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 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 X.X. 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.

5 MATH 8020 CHAPTER 1: COMMUTATIVE RINGS 5 Exercise 1.6: Let P, Q be two sets of prime numbers. Show TFAE: (i) Z P = ZQ. (ii) Z P = Z Q. (iii) P = Q. Exercise 1.7: Show that every subring of Q is of the form Z P for some P. Note well that 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 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 ] Ideals and quotient rings. Let f : R S be a homomorphism of rings, and put I = f 1 (0) = {x R f(x) = 0}. Then, since f is in particular a homomorphism of commutative groups (R, +) (S, +), I is a subgroup of (R, +). Moreover, it enjoys the following property: (I) For all i I and y R, iy I. Indeed, 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 (I). Theorem 1. Let R be a ring, and let I be a subgroup of (R, +). TFAE: (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.

6 6 PETE L. CLARK 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 (I) holds. 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. 2 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 that TFAE: (i) I R. (ii) I = R. Exercise 1.9: a) Let R be a 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 that TFAE: (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 1.10: In contemporary undergraduate algebra texts, there is a trend to 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?) Exercise 1.11: a) The intersection of an arbitrary family of ideals in a ring is an ideal. b) Let {I i } be a set of ideals in the 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 2. For a subset S of a commutative ring R, S is an ideal, the intersection of all ideals containing S. 2 If this is not known and/or obvious to the reader, these notes will probably be too brisk.

7 MATH 8020 CHAPTER 1: COMMUTATIVE RINGS 7 Exercise 1.12: Prove Proposition 2. When S is a subset of R such that I = S, we say that S is a set of generators for I. Of course the same ideal will in general have (infinitely) many sets of generators. Recall that an ideal I is principal if it can be generated by a single element. For any element 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 has a finite generating set, of course. 3 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, as it happens, most or all of the rings you are likely to meet in your undergraduate algebra courses 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. Exercise 1.13: 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 an arbitrary set of ideals of I. Give an explicit description of the ideal I i. Remark: 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 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) carries each ideal J to the smallest ideal containing both J and I. 3 Nevertheless the definition is too important to omit!

8 8 PETE L. CLARK The correspondence theorem will be a constant companion in our study of rings and ideals. As is common, we will often use the map Ψ to identify the sets I(R/I) and I I (R). Exercise 1.14: Let I be an ideal of R and {J i } be a set of ideals of R. that Φ preserves suprema and Ψ preserves infima: Φ( J i ) = Φ(J i ) and Ψ( J i ) = Ψ(J i ). Show 1.4. The monoid of ideals of R. Let R be a commutative ring, and let I and J be ideals of 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 1.15: 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 1.16: 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 1.17: 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 1.18: Let R be a PID. Show that I (R) is cancellative. 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.

9 MATH 8020 CHAPTER 1: COMMUTATIVE RINGS 9 Exercise 1.19: Show that for a commutative monoid M, TFAE: (i) M is cancellative. (ii) There exists a commutative group G and an injective monoid homomorphism ι : M G. Exercise 1.20: 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 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 1.21: 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. Then the pushforward and pullback of ideals were studied in detail in Theorem 3. In particular, in this case the pullback map f : Ĩ(S) Ĩ(R) is an injection, and 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 I of R, f f I = I + 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

10 10 PETE L. CLARK composite maps f f and/or f f? In these notes, the most satisfying and important answers will come for localizations and integral extensions 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 1.22: For an ideal I in a ring R, show that TFAE: (i) I is maximal. (ii) R/I is a field. Exercise 1.23: For an ideal I in a ring R, show that TFAE: (i) I is prime. (ii) R/p is an integral domain. Exercise 1.24: Deduce from the previous two exercises that any maximal ideal is prime. Exercise 1.25: 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 X.X: Show that (I : J) is indeed an ideal of R 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 1.26: 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 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.

11 MATH 8020 CHAPTER 1: COMMUTATIVE RINGS 11 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 1.27: 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. 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. We will see later that 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. It is easy to see that 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 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 }.

12 12 PETE L. CLARK 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. Remark: 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 ). 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 (coming relatively soon) 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 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 1.28: Prove Proposition 5. Exercise 1.29: Generalize the preceding discussion to decompositions into a finite number of factors: R = R 1 R n. However, there is no straightforward extension to ideals in infinite direct products. Some preliminary exploration of this issue occurs in Exercise 1.37 below. 7 We will see later that there is a topological space Spec R associated to every ring, such that Spec R is disconnected in the usual topological sense iff R can be written as a nontrivial product of rings

13 MATH 8020 CHAPTER 1: COMMUTATIVE RINGS Additional exercises. Exercise 1.30: A free R-module is an R-module M of the form i I R, i.e., a possibly infinite direct sum of copies of R itself. An R-module M is torsionfree if for all x R \ {0} and all m M \ {0} we have xm 0. a) Note that R is always a free R-module. Show that R is torsionfree as an R- module iff R is a domain. b) Show that the following are equivalent: (i) R is a domain. (ii) Every free R-module is torsionfree. 8 c) Show that if R is a field, every R-module is free. d) Suppose I is a nonzero, proper ideal of R. Show that the R-module R/I is neither free nor torsionfree. e) Conclude that a ring R such that every R-module is free is a field, and that a domain R such that every R-module is torsionfree is a field. Exercise 1.31: Let k be a field, X a set, and let k X be the set of all functions f : X k. a) Show that k X becomes a ring under the operations of pointwise addition and multiplication. What are the additive and multiplicative identity elements? b) Show that k X is an integral domain iff k X is connected iff #X = 1. (N.B.: When X is the empty set, k X is the zero ring, which by wise convention is decreed not to be a domain.) c) Suppose X is finite. Show that every ideal of k X is principal. d)* Suppose X is infinite. Is every ideal of k X principal? e) Show that k X has no nonzero nilpotent elements. (Recall an element x of a ring R is nilpotent if x n = 0 for some positive integer n. Rings with no nonzero nilpotent elements are reduced.) f) Show that any subring of a reduced ring is reduced, and conclude that a nonreduced ring like C[t]/(t 2 ) cannot be a subring of any ring k X. Exercise 1.32: Let R be a ring which contains a field k as a subring. a) Let m be any maximal ideal of R. Show that the quotient ring R/m is a field extension of k, and in particular that it has the same characteristic as k (i.e., either 0 or a prime number). b) Let R be a ring which admits two maximal ideals m 1, m 2 such that the fields R/m 1 and R/m 2. c) Deduce that the ring Z of integers is not a ring of k-valued functions on any set, i.e., is not a subring of k X for any field k, despite the fact that Z is reduced. Comment: Problems 4) and 5) give two different obtructions for a ring R to be realizable as a ring of k-valued functions on a set X. As I mentioned in class, nevertheless the philosophy is that every ring will be viewed as a ring of functions on its prime spectrum Spec R, and as I implied this requires relaxing somewhat the notion of a function. To be more accurate, this relaxed notion of function will fix the obstruction raised in Problem 5). It does not fix the problem of nilpotent 8 Accordingly, the notion of torsionfree modules is much more useful over domains. So far as I know, in this course we will only be interested in torsionfree modules over domains.

14 14 PETE L. CLARK elements, which will rather be ignored: for instance, we will see that the elements 0 and t of the ring C[t]/(t 2 ) induce the same function. I apologize if I was misleading on this account. Exercise 1.33 (The Zariski Topology): Let R be a ring. We endow the set Spec R of prime ideals of R with a topology by specifying the closed sets: for any ideal I of R, put V (I) = {p Spec R I p.} In this problem you will verify that the family V (I) of subsets of Spec R satisfies the axioms necessary to be the closed sets for a unique topology. a) Show that V (R) =. b) Show that V ((0))) = Spec R. c) Show that V (I 1 ) V (I 2 ) = V (I 1 I 2 ). d) Let {I j } j J be any family of ideals. Let I = I j be the ideal generated by the I j s, i.e., the least ideal containing every I j. Show that V (I) = j J V (I j). Exercise 1.34 (First Steps With the Zariski Topology): let R be a ring. a) Suppose that we tried to enlarge the topology by adding, for each subset S of R, the closed set V (S) = {p Spec R S p}. Show that this does not change the topology, i.e., each V (S) is already closed in the Zariski topology. b) For each f R, define U f = {p Spec R f / p}. Thus U f = Spec R \ V ((f)) and each U f is open. Show that in fact {U f } f R forms a base for the Zariski topology. c) Let p Spec R. Show that the singleton subset {p} of Spec R is closed iff p is a maximal ideal. d) Deduce from part b) that for any integral domain R that is not a field, the space Spec R is not a separated (a.k.a. T 1 space). e) Let S be a finite set. Determine the Zariski topology on k S. f) Show that for any ring R, the Zariski topology on Spec R is quasi-compact. (Feel free to look this up if you like.) Exercise 1.35 (Functoriality of the spectrum): let f : R S be a homomorphism of rings. a) Let q be a prime ideal of S. Show that f (q) = {x R f(x) q} is a prime ideal of R. b) Via part a), we have defined a map f : Spec S Spec R, q f (q). Show that in fact f is continuous with respect to the Zariski topologies on Spec R and Spec S. c) Upshot: R Spec R is a contravariant functor from the category of commutative rings to the category of topological spaces. Exercise 1.36 (The Maximal Spectrum): For any ring R, let MaxSpec(R) be the set of maximal ideals of R, so MaxSpec(R) Spec(R). Give it the subspace topology. a) Show that MaxSpec(R) is a separated space, but that it need not be Hausdorff. b) Show that MaxSpec(R) need not be a closed subset of Spec R. c) Show that, nevertheless, MaxSpec(R) is quasi-compact. d) Give an example of a homomorphism of rings f : R S and a maximal ideal m of S such that f (m) is not a maximal ideal of R. Thus MaxSpec does not have the good functorial properties of Spec.

15 MATH 8020 CHAPTER 1: COMMUTATIVE RINGS 15 Advertisement: one of the results of Melvin Hochster s thesis (1967) is that every separated quasi-compact space is homeomorphic to MaxSpec R for some ring R. In particular, this holds for every compact (i.e., Hausdorff!) space. But the compact case is much more classical, going back at least to Gelfand in the 1930 s. We will see the classical proof later on in the course. Exercise 1.37: Let X be an infinite set, and consider the ring k X. a) For each x X, define m x to be the set of all functions f : X k with f(x) = 0. Show that m x is a maximal ideal of k S and that k s /m s = k. b) By part a), we have an embedding S MaxSpec k S. Show that the induced topology on S is discrete. c) Deduce that there must be other maximal ideals of k X besides the m x s. Can you find any?? Exercise 1.38: A commutative group (G, +) is said to be divisible if for all x G and n Z +, there exists y G with ny = x. G is said to be uniquely divisible if for all x, n as above, the element y is unique. a) Show that the additive group (Q, +) is divisible. b) Show that any direct sum or direct product of divisible groups is divisible. Show that any quotient of a divisible group is divisible. In particular, Q/Z is a divisible group. c) Show that a divisible group if uniquely divisible iff it is torsionfree. d) Show that any uniquely divisible group (G, +) admits the canonical structure of a Q-vector space, so is isomorphic to a direct sum of copies of Q. e) Show that no nontrivial free abelian group is divisible. In particular, (Q, +) is a torsionfree but nonfree Z-module. f) Show that a subgroup of a divisible group need not be divisible. Remark: Later we will show that every torsionfree abelian group can be realized as a subgroup of the additive group of a Q-vector space. Indeed this will be a nice application of the notion of a flat module.