Math 371 Lecture #20 §6.1: Ideals and Congruence, Part I

We have seen congruence in the ring of integers Z and congruence in the ring of polynomials 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 = {6k : k ∈ Z} = 6Z.

Then a ≡ b (mod 6) means that a − b ∈ I. We know that I = 6Z is a subring of Z, but it has a very important “absorption” property. For any k ∈ Z and any t ∈ I, does kt ∈ I? Well, we have t = 6i for some i ∈ Z, so that kt = k(6i) = 6(ki) where ki ∈ Z, and hence kt ∈ 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 = {h(x)(x − 2) : h(x) ∈ Q[x]}, of Q[x], we have that f(x) ≡ g(x) (mod x3 − 2) when f(x) − g(x) ∈ I. Does I have the “absorption” property? Well, for k(x) ∈ Q[x] and t(x) ∈ I, we have k(x)t(x) = k(x)h(x)(x3 − 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 ∈ R and a ∈ I we have ra ∈ I and ar ∈ 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 ∈ Z and 1/2 ∈ Q, we have 3(1/2) 6∈ Z. Sometimes for noncommutative rings, the “left” absorption ra ∈ I or the “right” absorption ar ∈ I may hold while the other does not. Example. The subset a b I = : a, b ∈ 0 0 Q of M(Q) is a subring that absorbs on the right, a b r s ar + bt as + bu = ∈ I 0 0 t u 0 0 but not on the left, r s a b ra rb = 6∈ 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 ∈ I we have a − b ∈ I, and (2) for r ∈ R and a ∈ I, we have ra ∈ I and ar ∈ 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 ∈ I implies ab ∈ 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 ∈ R, the set

I = {rc : r ∈ R} is an ideal of R.

Proof. For r1c, r2c ∈ I, we have r1c − r2c = (r1 − r2)c ∈ I because r1 − r2 ∈ R; this is condition (1) of Theorem 6.1. For r ∈ R and sc ∈ I, we have r(sc) = (rs)c ∈ I because rs ∈ R; by commutativity of R we also have cr ∈ 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 {rc : r ∈ R} for some c ∈ 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) ∈ 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 | a0 and 3 | 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 ∈ 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) ∈ Z[x] such that I = (p(x)). This would mean that f(x) = 3 ∈ 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 ∈ I, a contradiction. So p(x) = ±3. But g(x) = x2 ∈ I is not a polynomial multiple of p(x) = ±3 in Z[x]. Thus there is no p(x) ∈ 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 ∈ R. Then the set I = {r1c1 + r2c2 + ··· + rncn : ri ∈ R} 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 | a0 2 and 3 | a1 is a finitely generated ideal of Z[x] with generators c1 = x and c2 = 3:

2 I = {r1(x)x + r2(x)3 : r1(x), r2(x) ∈ Z[x]}.