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Math 371 Lecture #21 §6.1: Ideals and Congruence, Part II §6.2: Quotients and Homomorphisms, Part I

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Math 371 Lecture #21 §6.1: Ideals and Congruence, Part II §6.2: Quotients and Homomorphisms, Part I Math 371 Lecture #21 x6.1: Ideals and Congruence, Part II x6.2: Quotients and Homomorphisms, Part I Why ideals? For a commutative ring R with identity, we have for a; b 2 R, that a j b if and only if (b) ⊆ (a), and u 2 R is a unit if and only if (u) = R. Having the established the notion of an ideal, we can now extend the notion of congruence to any ring with respect to any ideal. Definition. Let I be an ideal in a ring R. For a; b 2 R, we say a is congruent to b modulo I, written a ≡ b (mod I), provided that a − b 2 I. Examples. (a) For any integer k ≥ 1, we see that congruence modulo k in Z is the same thing as congruence modulo the principal ideal (k) = kZ of Z. (b) For a field F and any nonconstant polynomial p(x) in F [x], we see that congruence modulo p(x) in F [x] is the same thing as congruence modulo the principal ideal (p(x)) of F [x]. We know that the congruence in each of these examples is an equivalence relation, but is this the case for an arbitrary ideal of an arbitrary ring? Theorem 6.4. For any ideal I of a ring R, the relation of congruence modulo I is an equivalence relation. Proof. Reflexive: Since I is closed under subtraction, we have for any a 2 R that a − a = 0R 2 I; hence a ≡ a (mod I). Symmetric: For a; b 2 R, suppose that a ≡ b (mod I). Then a − b 2 I, so there is i 2 I such that a − b = i. Since 0R and i both belong to I, then −i = 0R − i 2 I; hence b − a = −i 2 I; so b ≡ a (mod I). Transitive: For a; b; c 2 R, suppose that a ≡ b (mod I) and b ≡ c (mod I). Then there are i; j 2 I such that a − b = i and b − c = j. Hence a − c = a − b + b − c = i + j 2 I, so that a ≡ c (mod I). Congruence modulo an ideal I respects the addition and multiplication of the ring. Theorem 6.5. For an ideal I in a ring R, if a ≡ b (mod I) and c ≡ d (mod I), then (1) a + c ≡ b + d (mod I), and (2) ac ≡ bd (mod I). Proof. (1) We have i; j 2 I such that a − b = i and c − d = j. Then (a + c) − (b + d) = (a − b) + (c − d) = i + j 2 I, so that a + c ≡ b + d (mod I). For the product we have ac − bd = ac − bc + bc − bd = (a − b)c + b(c − d) = ic + bj: Since the ideal I absorbs on the left and on the right, we have that ic and bj both belong to I. Hence ac − bd = ic + bj 2 I. For an ideal I in a ring R, the congruence class of a 2 R modulo I is fb 2 R : b ≡ a (mod I)g = fb 2 R : b − a 2 Ig = fb 2 R : b − a = i for some i 2 Ig = fb 2 R : b = a + i with i 2 Ig = fa + i : i 2 Ig: We use the notation a + I to denote the congruence class of a modulo I. Definition. A (left) coset of an ideal I in a ring R is a subset of the form a + I. Note: if R is commutative, then the right coset I + a = fi + a : i 2 Ig is the same as the left coset a + I, in which case we call a + I a coset of I. Example. We can think of the + in the coset a + I as an addition: for integers a; n with n ≥ 1, we have [a] = a + (n) = a + nZ = fa; a ± n; a ± 2n; : : : g. Theorem 6.6. For an ideal I in a ring R, and a; c 2 R, we have a ≡ c (mod I) if and only if a + I = c + I. Proof. Suppose a ≡ c (mod I). If b 2 a + I, then b ≡ a (mod I), so by transitivity b ≡ c (mod I), hence b 2 c + I. If b 2 c + I, then b ≡ c (mod I), so by symmetry and transitivity b ≡ a (mod I), hence b 2 a + I. Thus a + I = c + I. Now suppose that a + I = c + I. Then by reflexivity, a 2 a + I = c + I, so that a ≡ c (mod I). Corollary 6.7. For an ideal I in a ring R, two cosets of I are either disjoint or identical. Notation. For an ideal I in a ring R, the set of all cosets of I is denoted by R=I. We attempt to turn R=I into a ring by defining an addition and multiplication of cosets: for a; c 2 R, (a + I) + (c + I) = (a + c) + I; (a + I)(c + I) = ac + I: Since we are defining these operations via representatives of the cosets, there is a question of whether they are well-defined. Theorem 6.8. For an ideal I in a ring R, if a + I = b + I and c + I = d + I, then (a + c) + I = (b + d) + I and ac + I = bd + I: Proof. Suppose a + I = b + I and c + I = d + I. By Theorem 6.6, we know that a ≡ b (mod I), and hence that a − b 2 I. Similarly, we have that c − d 2 I. Since the ideal I is closed under addition, we obtain (a+c)−(b+d) = (a−b)+(c−d) 2 I, and so a + c ≡ b + d (mod I). By Theorem 6.6, we conclude that (a + c) + I = (b + d) + I. For the product we have ac − bd = ac − bc + bc − bd = (a − b)c + b(c − d): Since the ideal I absorbs multiplication on the left and on the right, we have ac − bd 2 I. Hence by Theorem 6.6, we conclude that ac + I = bd + I. Theorem 6.9. For an ideal I in a ring R, we have (1) R=I is a ring, (2) R=I is commutative if R is commutative, and (3) R=I has an identity if R has an identity. Proof. (1) One verifies that the eight axioms of a ring hold for the set of (left) cosets R=I with the addition and multiplication of cosets as defined. (2) If R is commutative, then for a; c 2 R, we have (a + I)(c + I) = ac + I = ca + I = (c + I)(a + I): (3) If R has a multiplicative identity 1R, then the coset 1R + I satisfies, for any a 2 R, (a + I)(1R + I) = a1R + I = a + I; (1R + I)(a + I) = 1Ra + I = a + I: Thus 1R + I is the multiplicative identity of R=I. Definition. For a ring R and an ideal I in R, the ring R=I is called the quotient ring (or factor ring) of R by I. One may say that we factor R by I to obtain the quotient ring R=I. a b Example. Let R = : a; b; c; d 2 , and I the subset of R consisting of matri- c d Z ces with even entries. The subset I is nonempty (it contains the zero matrix). The nonempty subset I is an ideal because it is closed under subtraction and it absorbs multiplication by elements of R on the both the left and the right. What is R=I? We investigate a coset for a randomly chosen matrix in R: 7 8 1 0 6 8 1 0 + I = + + I = + I: 5 −3 1 1 4 −4 1 1 This demonstrates that representatives of the distinct cosets of I in R are given by matrices with 0 or 1 as entries. So the quotient ring R=I consists of 24 = 16 elements, the distinct cosets of I in R..
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