Math 371 Lecture #20 x6.1: Ideals and Congruence, Part I We have seen congruence in the ring of integers Z and congruence in the ring of polyno- mials F [x]. Can we apply the notion of congruence to any ring? Examples. (a) In Z, the congruence a ≡ b (mod 6) means that a − b is a multiple of 6. Let I be the subset of Z that consists of all integer multiplies of 6: I = f6k : k 2 Zg = 6Z: Then a ≡ b (mod 6) means that a − b 2 I. We know that I = 6Z is a subring of Z, but it has a very important \absorption" property. For any k 2 Z and any t 2 I, does kt 2 I? Well, we have t = 6i for some i 2 Z, so that kt = k(6i) = 6(ki) where ki 2 Z, and hence kt 2 I. (b) In Q[x] congruence f(x) ≡ g(x) (mod x3 − 2) means that f(x) − g(x) is a polynomial multiple of x3 − 2. For the subring 3 I = fh(x)(x − 2) : h(x) 2 Q[x]g; of Q[x], we have that f(x) ≡ g(x) (mod x3 − 2) when f(x) − g(x) 2 I. Does I have the \absorption" property? Well, for k(x) 2 Q[x] and t(x) 2 I, we have k(x)t(x) = k(x)h(x)(x3 − 2) 2 I: These examples suggest that congruence in a ring might be obtained by subrings having the \absorption" property. We give these kind of absorbing rings a name. Definition. A subring I of a ring R is called an ideal if whenever r 2 R and a 2 I we have ra 2 I and ar 2 I. We include absorption on the right and on the left to accommodate noncommutative rings. Of course, if the ring R is commutative, the subring I is commutative, so that ar = ra and only one absorption condition needs to be checked. Not every subring I of a ring R is an ideal. Example. The subring Z of Q is not an ideal because for 3 2 Z and 1=2 2 Q, we have 3(1=2) 62 Z. Sometimes for noncommutative rings, the \left" absorption ra 2 I or the \right" absorp- tion ar 2 I may hold while the other does not. Example. The subset a b I = : a; b 2 0 0 Q of M(Q) is a subring that absorbs on the right, a b r s ar + bt as + bu = 2 I 0 0 t u 0 0 but not on the left, r s a b ra rb = 62 I: t u 0 0 ta tb We say that subring I is a right ideal but not a two-sided ideal. There is a simple way to determine when a nonempty subset of a ring is an ideal. Theorem 6.1. A nonempty subset I of a ring R is an ideal if and only if (1) for a; b 2 I we have a − b 2 I, and (2) for r 2 R and a 2 I, we have ra 2 I and ar 2 I. Proof. An ideal I of R satisfies the two conditions. Suppose that a nonempty subset I of R satisfies the two conditions. By condition (2), the set I absorbs on the left and the right, so it remains to show that I is a subring. For this we need to show that I is closed under subtraction (which is condition (1)), and closed under multiplication which follows from condition (2) because I ⊂ R so that a; b 2 I implies ab 2 I. Thus I is an ideal of R. Example. For an integer k ≥ 2, the subring kZ of Z is an ideal by Theorem 6.1. The ideal kZ has a special form. Theorem 6.2. Let R be a commutative ring with identity. For any c 2 R, the set I = frc : r 2 Rg is an ideal of R. Proof. For r1c; r2c 2 I, we have r1c − r2c = (r1 − r2)c 2 I because r1 − r2 2 R; this is condition (1) of Theorem 6.1. For r 2 R and sc 2 I, we have r(sc) = (rs)c 2 I because rs 2 R; by commutativity of R we also have cr 2 I; this is condition (2) of Theorem 6.1. Thus I is an ideal of R. We give the ideals described in Theorem 6.1 a special name. Definition. An ideal of the form frc : r 2 Rg for some c 2 R is called the principal ideal generated by c, and is denoted by (c). Example. For a commutative ring F with identity and p(x) 2 F [x], the ideal (p(x)) (no- tice that we have already been using this notation) is a principal ideal of F [x] generated by p(x). If F is a field, then F [x]=(p(x)) is the commutative ring with identity obtained from F [x] by forming congruence classes modulo the principal ideal generated by p(x). Not every ideal I in a commutative ring R with identity needs be a principal ideal generated by one element of R. n Example. Is the subset I of polynomials f(x) = anx + ··· + a1x + a0 in Z[x] for which 3 j a0 and 3 j a1 is a principal ideal of Z[x]? n k For f(x) = anx + ··· + a1x + a0 and g(x) = bkx + ··· + b1x + b0 in I, we have that the coefficients of x and 1 in f(x) + g(x) are a1 + b1 and a0 + b0, which are both divisible by 3 since a0; a1; b0; b1 are. l For h(x) = clx +···+c1x+c0 2 Z[x], the coefficients of x and 1 in f(x)h(x) are a1c0 +a0c1 and a0c0 which are divisible by 3 because a0; a1 are divisible by 3. By Theorem 6.1, the set I is an ideal of Z[x]. If I were a principal ideal generated by a single element of Z[x], then there would be p(x) 2 Z[x] such that I = (p(x)). This would mean that f(x) = 3 2 I would be a nonzero polynomial multiple of p(x). This requires that that either p(x) = ±1 or p(x) = ±3. If p(x) = ±1, then 2x would be a polynomial multiple of p(x), and hence 2x 2 I, a contradiction. So p(x) = ±3. But g(x) = x2 2 I is not a polynomial multiple of p(x) = ±3 in Z[x]. Thus there is no p(x) 2 Z[x] such that I = (p(x)). There is generalization of a principal ideal that accounts for this ideal and many other ideals. Theorem 6.3. Let R be a commutative ring with identity, and let c1; c2; : : : ; cn 2 R. Then the set I = fr1c1 + r2c2 + ··· + rncn : ri 2 Rg is an ideal in R. Idea of Proof. Apply Theorem 6.1. Definition. The ideal I in Theorem 6.3 is called an ideal generated by the finitely many generators c1; c2; ··· ; cn, or a finitely generated ideal. n Example. The ideal I of Z[x] of polynomials f(x) = anx + ··· + a1x + a0 where 3 j a0 2 and 3 j a1 is a finitely generated ideal of Z[x] with generators c1 = x and c2 = 3: 2 I = fr1(x)x + r2(x)3 : r1(x); r2(x) 2 Z[x]g:.