Chapter 6, Ideals and quotient rings Ideals. Finally we are ready to study kernels and images of ring homomorphisms. We have seen two major examples in which congruence gave us ring homomorphisms: Z → Zn and F [x] → F [x]=(p(x)). We shall generalize this to congruence in arbitrary rings and then see that it brings us very close to a complete understanding of kernels and images of ring homomorphisms. Recall the definition of a ring. For congruence, we need a special subring that will behave like nZ or like p(x)F [x]={p(x)f(x)|f(x)∈F[x]}. Definition, p. 135. A subring I of a ring R is an ideal if whenever r ∈ R and a ∈ I, then ra ∈ I and ar ∈ I. If R is commutative, we only need to worry about multiplication on one side. More generally, one can speak of left ideals and right ideals and two-sided ideals.Our main interest is in the two-sided ideals; these turn out to give us the congruences we want. Before we look at examples, recall that to be a subring means I is closed under multi- plication and subtraction. Thus we get Theorem 6.1. A nonempty subset I in a ring R is an ideal iff it satisfies (1) if a; b ∈ I,thena−b∈I; (2) if r ∈ R; a ∈ I,thenra ∈ I and ar ∈ I. Note that if R has an identity, we can replace (1) by (10) if a; b ∈ I,thena+b∈I because (2) implies that, since −1 ∈ R, −b ∈ I whenever b ∈ I. Examples. 1. nZ = { kn | k ∈ Z } for any n ∈ Z is an ideal in Z.Ifn=0wegetthe zero ideal, an ideal of any ring R.Ifn=1wegetZ, the whole ring. Again, for any ring R, the whole ring is an ideal of R. 2. In Z6, the set I = { [2k] ∈ Z6 | k ∈ Z } is an ideal. 3. p(x)R[x]={p(x)f(x)|f(x)∈R[x]}is an ideal of R[x] for any commutative ring R with 1. 4. In Z[x], the set I = { f(x) ∈ Z[x] | f(0) ≡ 0 (mod n) } is an ideal for any n ≥ 2inZ. This generalizes an example on page 136 where n =2. 5. For R = C(R; R), fix any r ∈ R. The set I = { f ∈ R | f(r)=0}is an ideal. Note that it does not work to use a number other than 0. 1 2 ab 6. For R = M2(R), the set of first rows I = a; b ∈ R is a right ideal but not 00 a left ideal. (A similar thing is done for columns and left ideals in the book.) In particular, I is not a (two-sided) ideal. Check. Examples 1, 2 and 3 above were all of a special type which we can generalize. Theorem 6.2. Let R be a commutative ring with identity. Let c ∈ R.Theset I={rc | r ∈ R } is an ideal of R. Proof. Given two elements r1c and r2c in I,wehaver1c−r2c=(r1−r2)c∈I. For any a ∈ R, a(r1c)=(ar1)c ∈ I. Therefore I is an ideal. (We have implicitly used the fact that R is commutative so that multiplication on the right also works.) We call the ideal in Theorem 6.2 the principal ideal generated by c and denote it by (c)orbyRc. The ideal in Example 4 was not principal. To see this, note that n ≥ 2 and x both lie in I.IfIwere generated by some polynomial p(x), then both n and x must be multiples of p(x). But then n = p(x)q(x) implies that p(x)isaconstantc.Notethat c=6 1, for that would make I the whole ring, which it is not since 1 ∈= I.Nowwealso have x = cr(x)forsomer(x), which is impossible since c does not divide x (i.e., c has no inverse in Z[x] since the only units are 1). In fact it is easy to see that I is generated by the two elements n and x in the sense of the next theorem. Theorem 6.3. Let R be a commutative ring with identity. Let c1;c2;:::;cn ∈ R.Then the set I = {r1c1 + ···+rncn |r1;:::;rn ∈R} is an ideal of R. Proof. Homework; generalize the proof of Theorem 6.2. We call the ideal I of Theorem 6.3 the ideal generated by c1;:::;cn and denote it by (c1;c2;:::;cn). Comments. 1. If R does not have an identity, there is a complication in the definition since one wants the elements ci ∈ I (see exercise 31, p. 143). 2. If R is not commutative, one needs multiplication on both sides in the definition of I. 3. Ideals with finitely many generators are called finitely generated ideals. One has to work a bit to find ideals which are not finitely generated and we will avoid them in this course. One example is the ideal generated by all the indeterminates in the polynomial ring R[x1;x2;x3;:::] with infinitely many indeterminates. 3 In analogy to congruence in Z and F [x] we now will build a ring R=I for any ideal I in any ring R.Fora; b ∈ R,wesaya is congruent to b modulo I [and write a ≡ b (mod I)] if a − b ∈ I. Note that when I =(n)⊂Zis the principal ideal generated by n,thena−b∈I ⇐⇒ n | ( a − b ), so this is our old notion of congruence. As before, we require congruence to be an equivalence relation if it is going to work for us, so we check this. Theorem 6.4. Let I be an ideal of a ring R. Congruence modulo I is an equivalence relation. Proof. reflexive: a − a =0∈I since I is a subring. symmetric: Assume a ≡ b (mod I). Then a − b ∈ I.SinceIis a subring, its additive inverse, b − a is also in I,andsob≡a(mod I). transitive: Assume a ≡ b (mod I)andb≡c (mod I). Then a − b ∈ I and b − c ∈ I, hence the sum a − c =(a−b)+(b−c)∈I,soa≡c(mod I). We use this to show that arithmetic works \modulo I". Theorem 6.5. Let I be an ideal of a ring R.Ifa≡b(mod I) and c ≡ d (mod I),then (1) a + c ≡ b + d (mod I); (2) ac ≡ bd (mod I). Proof. (1) (a + c) − (b + d)=(a−b)+(c−d). Since a − b ∈ I and c − d ∈ I,sois (a+c)−(b+d), hence a + c ≡ b + d (mod I). (2) ac−bd = ac−bc+bc−bd =(a−b)c+b(c−d)∈Isince I is closed under multiplication on both sides. Therefore ac ≡ bd (mod I). Looking at this proof, we see that it is multiplication that fails if we have only a left or right ideal that is not 2-sided. The equivalence classes for this relation, are commonly called cosets. What do they look like? The congruence class of a modulo I is { b ∈ R | b ≡ a (mod I) } = { b ∈ R | b − a ∈ I } = { b ∈ R | b − a = i for some i ∈ I } = { b ∈ R | b = a + i for some i ∈ I } = { a + i | i ∈ I }: We denote this coset by a + I. As earlier, we have a ≡ b (mod I)iffa+I=b+I.The set of all cosets of I (congruence classes of R modulo I) will be denoted by R=I. 4 Selected problems from pp. 141{145. 13 (generalized). Let I be an ideal in a ring R with 1. I = R iff I contains a unit. Proof. (=⇒)IfI=R,then1∈Iis a unit in I. ( ⇐= ) Let u ∈ I be a unit. Then there exists v ∈ R with vu = 1. For any r ∈ R,weget r=r·1=r(vu)=(rv)u ∈ I. 14/35. A commutative ring R with 1 is a field iff its only two ideals are (0) and R. Proof. (=⇒) Any nonzero ideal I contains some nonzero element, which is a unit since R is a field. By #13, I = R. ( ⇐= ) Let 0 =6 a ∈ R and let I =(a). By hypothesis, I = R,soIcontains the identity 1. Therefore 1 = ra for some r ∈ R,sothatris the inverse of a. Therefore R is a field. 38. Every ideal I in Z is principal. Proof. Assume I =6 (0) (which is principal). Let c be the smallest positive element in I (exists by the well-ordering axiom). Then (c) ⊆ I. Conversely, let a ∈ I. By the division algorithm, we can write a = cq + r with 0 ≤ r<c. Then r = a − cq ∈ I. By our choice of c,wemusthaver= 0, as otherwise it is a smaller positive element of I. Therefore a ∈ (c), so I =(c) is principal. m 39. (a) S = { n | m; n ∈ Z;n odd } is a subring of Q. { m ∈ | } (b) I = n S m even is an ideal in S. (c) S=I has exactly two cosets. Proof. (a) Check closure under subtraction and multiplication. (b) Check closure under subtraction and multiplication by elements of S. m m m−n (c) If n ∈= I, then, since m and n are both odd, we see that n =1+ n ∈1+I.Sothe only cosets are I and 1 + I. 5 Quotient rings and homomorphisms. Theorem 6.5 gives the fact that addition and multiplication are well-defined on congru- ence classes. Translating this into the language of cosets gives Theorem 6.8.