Revision of ring theory

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1 CHAPTER 1 Revision of ring theory 1.1. Basic definitions and examples In this chapter we will revise and extend some of the results on rings that you have studied on previous courses. A ring is an algebraic object in which we can operate in a similar way as we do with integers. Inside the set Z of integers we can perform operations of addition, substraction and multiplication, but in general Z is not closed under division; for instance the quotient 3/2 is not an integer. Addition and multiplication have a series of well known properties; the abstraction of those properties is what constitutes the formal notion of ring. DEFINITION A ring is a nonempty set R together with two operations, a sum + and a product satisfying the following properties: Sum Product S1 Associativity: (a + b)+c = a +(b + c) for all a, b, c R, S2 Commutativity: a + b = b + a for all a, b R, S3 Zero: There is an element 0 R such that a +0=a =0+a for each a R, S4 Inverses: For each a R there is an element a R such that a +( a) =0. In what follows we will always write a b to denote a +( b). REMARK. These properties simply mean that (R, +) is an abelian group. P1 Associativity: a(bc) =(ab)c for all a, b, c R, P2 Unit: There is an element 1 R such that 1a = a1 =a for all a R, REMARK. This properties mean that (R, ) is a multiplicative monoid. P3 Distributivity: For all a, b, c R, a(b + c) = ab + ac, (a + b)c = ac + bc. Some rings satisfy an additional property for the product: P4 Commutativity: ab = ba for all a, b R. When this extra property is satisfied, we will say that R is a commutative ring. In this course, we will be mostly interested on commutative rings, although we will occasionally deal with some noncommutative examples. EXAMPLE R = {0} with the trivial operations. This is called the trivial ring, and is the only ring for which one has 1=0. In all what follows, we will assume our rings to be nontrivial, i.e. 0 = 1. EXAMPLE The integers Z with the usual addition and multiplication. EXAMPLE The fields of rational numbers Q, real numbers R or complex numbers C, or in general any field F. 1

2 2 1. REVISION OF RING THEORY EXAMPLE The rings of integers modulo n, Z/nZ (sometimes also denoted by Z n ) consisting of the set {a a Z} = {0, 1,...,n 1}, where a is the residue class of a modulo n, so a = {a = rn r Z}. Addition and multiplication are defined as a + b := a + b, ab := ab EXAMPLE Let R be any ring, and define R[x] to be the set of all polynomials a 0 + a 1 x + + a n x n where the coefficients a i are elements of the ring R. Then R[x] is a ring with addition and multiplication defined in the usual way (check as an exercise!). The ring R[x] is called the polynomial ring in one variable with coefficients in R. EXAMPLE The polynomial ring in n variables with coefficients in R, denoted by R[x 1,...,x n ] and defined inductively by R[x 1,...,x n ]:=R[x 1,...,x n 1 ][x n ]. EXAMPLE The ring M n (R) of n n matrices (a i,j ) i,j=1,...,n with coefficients a i,j in R, and the usual matrix addition and multiplication. EXAMPLE The power set ring. Let X be a set, and let P(X) ={Y Y X} the set of all subsets of X. On P(X) consider the operations: Y + Z := Y Z =(Y Z) \ (Y Z) the symmetric difference as addition, YZ := Y Z the intersection as a product. With this operations P(X) becomes a ring with zero element 0= and unit element 1=X. EXAMPLE Let V be a vector space, consider the set End(V ):={f : V V f is a linear map}, then End(V ) is a ring with pointwise addition (f +g)(v) :=f(v)+ g(v) and multiplication given by composition f g(v) :=f(g(v)), where the zero element is the constant map 0(v) =0and the unit element is the identity map Id(v) =v. REMARK. The ring of endomorphisms of a vector space is nothing but a matrix ring in disguise. We will state more precisely what we mean by this once we talk about ring isomorphisms. EXAMPLE Consider the set C(R) :={f : R R f continuous function} of real-valued continuous functions. The C(R) is a ring with pointwise addition (f +g)(x) := f(x)+g(x) and multiplication (fg)(x) :=f(x)+g(x). REMARK. One might wonder whether one could define a different ring structure on C(R) by replacing pointwise multiplication by composition as a product. Unlike it happened in the case of linear map, this operation does not turn C(R) into a ring. EXAMPLE Let X be a set and R be a ring. In a similar fashion to the previous example, the set X R of all R-valued maps f : X R on X becomes naturally a ring with pointwise addition and multiplication. EXAMPLE Quaternion algebras. Let F be a field (of characteristic different from 2), and let α, β F. The quaternion algebra α F β is defined as the set {a + bi + cj + dk a, b, c, d F} with standard sum and product defined by the rules ij = k = ji, i 2 = α, j 2 = β. If F is a subfield of the real numbers, α F β can also be described as the subring a + b α c β + d αβ of M 2 (C) consisting of matrices of the form c β d αβ a b, where α a, b, c, d F. EXAMPLE The ring of power series with real coefficients R[[x]] := a n x n a n R n. n 0 In general one can construct the ring of formal power series R[[x]] with coefficients in any ring R.

3 1.2. SUBRINGS, IDEALS AND QUOTIENT RINGS 3 EXAMPLE Let G be a group, R a ring, the group ring R[G] is defined as R[G] := x G a xx a x R x, a x =0for all x except a finite number = {f : G R f has a finite support}. The addition and product are defined (in the functional notation) as follows: (f + g)(x) :=f(x)+g(x), (fg)(x) :=(f g)(x) = y G f(y)g(y 1 x). The product in R[G] receives the name of convolution product of functions. REMARK. Note that in this case the convolution product actually defines a different ring structure in the set of functions f : G R than the pointwise product, so this example is actually different from Example ; for instance, if R is a commutative ring then G R (with the pointwise product) is also commutative, whereas the group ring R[G] will be noncommutative whenever G is Subrings, ideals and quotient rings Let R be a ring, we look at subsets of R which are in fact themselves rings in their own right when we restrict to them the sum and product of R. More precisely, we say that S is a subring of R if: (i) 1 R S, (ii) S is an additive subgroup of R. In other words, whenever a, b S, one has a b S. (iii) S is closed under the product of R, in other words, S is a multiplicative submonoid of R, i.e. for all a, b S one has ab S. We will write S R to denote that S is a subring of R. EXAMPLES (1) Z Q R C. (2) The trivial ring {0} is NOT a subring of Z, as 1 / {0}. (3) For any ring R, one has R R[x], where R consists of all constant polynomials in R[x]. (4) The rings of matrices M n (R) contain several interesting subrings. Some examples are The ring D n (R) of diagonal matrices. The ring U n (R) of upper-triangular matrices. The ring L n (R) of lower-triangular matrices. (5) If {S i } i I is a family of rings such that S i R for all i I, then i I S i R. (6) Let X R be a subset of a ring R, define [X] := {S R X S} the intersection of all subrings of R containing X. Then [X] is a subring of R, called the subring generated by X. EXERCISE Show that [X] can be identified with the set of all sums of the form ±x1 x n where x i X {1}. We move now to the key notion of ideal. Ideals are certain subsets of rings that play a similar role to that of normal subgroups in group theory, in the sense they allow us to build quotients of rings. Also, knowing the ideals of a ring in full detail often lead to a complete description of all the modules, so understanding ideals is a fundamental topic of this course. DEFINITION Let R be a commutative ring. A subset I R of R is said to be an ideal of R if it has the following properties: I1 Additive closure: I (R, +) is an additive subgroup of R, i.e. I = is nonempty and for all a, b I one has a b I. I2 Absorbency: For all r R and for all a I one has ra = ar I.

4 4 1. REVISION OF RING THEORY When I is an ideal of R we will write I R. REMARK In the previous definition we are assuming that R is commutative. For noncommutative rings one has to be more careful and distinguish among the notions of left, right and two-sided ideals. EXAMPLES (1) 0:={0} R the zero ideal, which is an ideal for every ring R. (2) RR the total ideal, also an ideal for every ring. The ideals I which are different from the total ideal are called proper ideals. (3) Let R be a ring, I,JR ideals, then I JR and I+J := {i + j i I, j J} R are ideals of R, respectively called the intersection and the sum of I and J. I + J I J I J I J is the greatest ideal (with respect to the inclusion ordering) contained in both I and J, whilst I + J is the least ideal containing both I and J. (4) If {I α } α A is a family of ideals of R, then α A I α R is also an ideal of R. (5) If I 1,...,I n are ideals of R, define I I n := {i i n i j I j }, then I I n R is also an ideal of R. (6) Let R be a ring, and a R an element of R, and define (a) =Ra := {ra r R}. Then (a) R is an ideal, called the principal ideal generated by a. More generally, if a 1,...,a n R are elements of R, then the set Ra Ra n = {r 1 a r n a n r i R} is an ideal of R, called the ideal generated by a 1,...,a n. (7) Particular examples of principal ideals are the trivial ideal 0 = (0) and the total ideal R = (1). If R = Z, the principal ideal generated by 2 is the set (2) = {2n n Z} of even numbers. Let R be a ring and I R an ideal. For each a R we define the coset of a with respect to I as the set a = a + I := {a + i i I}. Since I is an additive subgroup of R, and (R, +) is a commutative subgroup, I is normal in R, and thus the set R/I = {a a R} is an additive group with addition defined by a + b := a + b, i.e. (a + I)+(b + I) :=(a + b)+i. The absorbency property of an ideal also ensures that the product of cosets ab = ab is well defined, endowing the set R/I with a ring structure. In particular, the zero and unit elements of R/I are 0 and 1, respectively. EXAMPLES (1) R/0 =R, (2) R/R =0, (3) Z/(n) =Z/nZ = Z n Ring homomorphisms DEFINITION Let R and S be rings; a map f : R S is said to be a ring homomorphism (or ring morphism for short) if (1) f(1) = 1,

5 1.3. RING HOMOMORPHISMS 5 (2) f(a + b) =f(a)+f(b) for all a, b R, (3) f(ab) =f(a)f(b) for all a, b R. If f is injective we say it is a monomorphism, if it is surjective it is called an epimorphism, and if it is a bijection it is called an isomorphism. In this case we say that R is isomorphic to S and write R = S. DEFINITION Let f : R S be a ring morphism. We define the image of f as the set Im(f) :=f(r) ={f(r) r R} S, and the kernel of f as the set Ker(f) :=f 1 (0) = {r R f(r) =0} R. LEMMA Let R and S be rings, and f : R S be a ring homomorphism, then the following properties hold: (1) Im(f) S is a subring of S, (2) Ker(f) R is an ideal of R. PROOF. (1). 1 S = f(1 R ), and hence 1 S Im(f). If s 1,s 2 Im(f) then there are r 1,r 2 R such that f(r 1 )=s 1,f(r 2 )=s 2, but then using that f is a ring morphism one has s 1 s 2 = f(r 1 ) f(r 2 )=f(r 1 r 2 ) Im(f), s 1 s 2 = f(r 1 )f(r 2 )=f(r 1 r 2 ) Im(f), thus Im(f) S subring. (2). f(0 R )=0 S, so 0 R Ker(f), and thus Ker(f) =. Now, if a, b Ker(f), one has f(a b) =f(a) f(b) =0 0=0, hence a b Ker(f), so Ker(f) is an additive subgroup of R. Now, for any a Ker(f), and for any r R one has f(ra) =f(r)f(a) =f(r)0 = 0, thus ra Ker(f), and consequently Ker(f) R is an ideal of R. THEOREM (First isomorphism theorem). Let R and S be rings, and f : R S be a ring homomorphism, then the mapping f(r) r = r +Ker(f) provides an isomorphism R/ Ker(f) = Im(f). PROOF. Let a = a +Ker(f) R/ Ker(f) for a R. One has a = b a b Ker(f) f(a b) =0 f(a) =f(b), so the application π : R/ Ker(f) Im(f) given by π(a) :=f(a) is well defined and injective. Now, if b Im(f) then there is some a R such that b = f(a), and thus b = π(a); henceforth, π is surjective, and so it is a bijection. Now, π((1)) = f(1) = 1, and for any a, b R one has π(a b) =π(a b) =f(a b) =f(a) f(b) =π(a) π(b), π(ab) =π(ab) =f(ab) =f(a)f(b) =π(a)π(b), hence, π is a ring homomorphism, and since it is bijective we have a ring isomorphism R/ Ker(f) = Im(f). EXAMPLES (1) The identity map Id : R R is a ring isomorphism. If S R is a subring, the inclusion map i S : S R is a ring monomorphism.

6 6 1. REVISION OF RING THEORY (2) The complex conjugation map σ : C C defined by σ(z) =z is a ring isomorphism. (3) Let I R be an ideal of R, The canonical projection π I : R R/I is defined by π ( r) := r = r + I. It is easy to see that π I is a ring homomorphism, π I (a) =0 a =0 a + I = I a I, so Ker(π I )=I. For any a R/I we can write a = π I (a), thus π I is an epimorphism, and hence Im(π I )=R/I. The statement of the first isomorphism theorem in this particular case is just a tautology, saying that R/I is isomorphic to itself. (4) For any n 2, the ring Z n = Z/nZ is the image of the canonical projection π n : Z Z/(n) =Z n. (5) Let R S subring, and a S. We define the evaluation map e a : R[x] S by p(x) p(a), i.e. if p(x) = r 0 + r 1 x +, +r n x n then e a (p(x)) = r 0 + r 1 a +, +r n a n S. The evaluation map is a ring homomorphism for which we have Im(e a )={p(a) p R[x]} = ri a i r i R = R[a] S, Ker(e a )={p(x) R[x] p(a) =0}. Note that, if we assume that a R, one has p(x) Ker(e a ) p(a) =0 (x a) p(x) p(x) =(x a)r(x) In this case one gets p(x) ((x a)). Im(e a ) = R[x]/ Ker(e a )=R/((x a)) = R. LEMMA Let f : R S and g : S T be ring homomorphisms. Then the composition g f : R T is also a ring homomorphism. PROOF. The proof is immediate and follows from applying repeatedly that f and g are homomorphisms: (g f)(0) = g(f(0)) = g(0) = 0. (g f)(a + b) = g(f(a + b)) = g(f(a)+f(b)) = g(f(a)) + g(f(b)) = (g f)(a)+(g f)(b). (g f)(1) = g(f(1)) = g(1) = 1. (g f)(ab) = g(f(ab)) = g(f(a)f(b)) = g(f(a))g(f(b)) = (g f)(a) (g f)(b). THEOREM (Second isomorphism theorem). Let R be a ring, I R an ideal, S R a subring; then the following properties hold: (1) S + I R is a subring of R, (2) I S + I is an ideal of S + I, (3) S I S is an ideal of S, (4) There is a ring isomorphism PROOF. (1). Let s 1,s 2 S, i 1,i 2 I, then S + I I = S S I (s 1 + i 1 ) (s 2 + i 2 )=(s 1 s 2 )+(i 1 i 2 ) S + I, and thus S + I is an additive subgroup of R. Similarly, (s 1 + i 1 )(s 2 + i 2 )=s 1 s 2 +(s 1 i 2 + i 1 s 2 + i 1 i 2 ) S + I,

7 1.3. RING HOMOMORPHISMS 7 and 1 S S + I, so S + I is a submonoid of S, and hence S + I R is a subring. (2). The I is additively closed because it is a subgroup of R; moreover, since S + I R the absorbency property for I with respect to S + I is immediate. Hence, I S + I is an ideal of S + I. (3). As both I and S are additive subgroups of R, S I is also additively closed, and since it S I S, is is an additive subgroup of S. For the absorbency, let x S I, s S; since x I and s S R, absorbency property for I tells us that xs I. Also, as x S, s S and S is a subring, we get xs S, and thus sx S I. This proves that S I is an ideal of S. (4). Consider the map ϕ : S (S + I)/I given by ϕ(s) =s = s + I. Since ϕ can be seen as the composition of the inclusion S S + I with the canonical projection S + I (S + I)/I, and the composition of ring morphisms is a ring morphism by the previous lemma, we get the ϕ is a ring morphism. Now, for any s + i (S + I)/I, we have ϕ(s) =s, but since (s + i) s = i I we have s = s + i, and thus ϕ(s) =s + i. Hence ϕ is surjective, i.e. Im ϕ =(S + I)/I. Moreover, one has Ker ϕ = {s S ϕ(s) =0} = {s S s =0} = {s S s I} = S I, so applying the first isomorphism theorem to the morphism ϕ we obtain the result. THEOREM (Third isomorphism theorem). Let R be a ring, I J R ideals of R, then J/I is an ideal of R/I and moreover R/I J/I = R/J. PROOF. It is immediate to check that J/I is an ideal of R/I. consider the map ϕ : R/I R/J given by ϕ(x + I) :=x + J. Assume that x + I = x + I, i.e. x x I; as I J, one gets x x J, so x + J = x + J; and thus ϕ is well defined. The mapping ϕ is also obviously surjective. Now, for the kernel of ϕ we have Ker ϕ = {x + I R/I ϕ(x + I) =0} = {x + I R/I x + J} = {x + I R/I x J} = J/I, and the desired result follows from the first isomorphism theorem. COROLLARY (Correspondence theorem). Let R be a ring and I R an ideal of R; the map S S/I defines a correspondence between the set of subrings of R containing I and the set of subrings of R/I. Similarly, the map J J/I gives a correspondence between the set of ideals of R containing I and the set of ideals of R/I. REMARK Note that in principle we cannot state that different ideals of R containing I will give rise to different ideals in the quotient ring. The only bit we can be sure of is that all ideals of R/I will be of the form J/I for some J.