Mathematics and Algorithms for Computer Algebra Part 1 c 1992 Dr Francis J. Wright – CBPF, Rio de Janeiro July 9, 2003 2a: Introduction to Abstract Algebra (continued) 1 Ideals and quotient rings 1.1 Quotient algebras Homomorphic images provide simpler models of algebras, and quotient al- gebras provide a technique for constructing homomorphic images. The quo- tienting is by a congruence relation. Definition 1 A congruence relation E on an Ω-algebra A is an equivalence relation on A such that for any operation ω ∈ Ω of arity n ai E bi for i = 1, . , n ⇒ ω(a1, . , an) E ω(b1, . , bn). That is, if the operands of any operator are equivalent then so are the corresponding values. Example: Equivalence mod m on Z generalizes immediately to congru- ence mod m on an arbitrary commutative ring R, thus: a ≡m b ⇐⇒ a − b = km ⇐⇒ m | (a − b), a, b, k, m ∈ R. Theorem 1 (Quotient algebras) Let A be an Ω-algebra and E a congru- ence relation on A. Then: 1. the quotient set A/E is an Ω-algebra if ω ∈ Ω is defined on A/E by ω([a1],..., [an]) = [ω(a1, . , an)] (i.e. ω : A/E → A/E maps equivalence classes to equivalence classes); 1 2. A/E is a homomorphic image of A under the natural map ν : a 7→ [a]. Hence we have a source of homomorphic images. But do all homomorphic images of algebras have this form? Essentially, yes, as follows. Lemma 2 Let φ : A → A0 be a morphism of Ω-algebras. Then the kernel relation Eφ, defined by a Eφ b ⇐⇒ φ(a) = φ(b), is a congruence relation on A. Theorem 3 (Universal Isomorphism Theorem) Let A and A0 be Ω- algebras, with A0 a homomorphic image of A under φ : A → A0. Then 0 ∼ 0 A = A/Eφ, where ψ :[a] 7→ φ(a) is an isomorphism A/Eφ → A , thus φ A - A0 ¨* ¨¨ ν : ¨ a 7→ [a] ¨ ψ :[a] 7→ φ(a) ? ¨ ¨¨ A/Eφ [This is essentially the Decomposition Theorem for Functions applied to algebras instead of sets.] 1.2 Ideals Let R be a ring. Then all homomorphic images of R are given up to isomor- phism by quotient rings R/E for some congruence relation E on R. Ideals provide congruence relations. Definition 2 An ideal I in a ring R is a nonempty subset of R such that: 1. a, b ∈ I ⇒ a − b ∈ I; 2. a ∈ I, r ∈ R ⇒ ar, ra ∈ I. Requirement (1) makes I an additive subgroup of R. However: Remark A proper ideal I ⊂ R is not a subring because it does not contain 1R. If 1R ∈ I then I = R. 2 The equivalence relation EI induced by I is defined by a EI b ⇐⇒ a − b ∈ I. This is also written as a ≡ b (mod I) (“a is congruent to b mod I”). 1.2.1 Principal ideals If R is a commutative ring and m ∈ R then (m) = {km | k ∈ R} is the smallest ideal of R containing m, called the principal ideal generated by m (principal because it is generated by a single element). Examples: (2) ⊂ Z is the principal ideal of even integers, 63 1; (x) ⊂ R[x] is the principal ideal of all polynomials containing x as a factor, i.e. with zero constant term, 63 1. The equivalence relation a E(m) b ⇐⇒ a − b ∈ (m) induced by (m) is the same as congruence mod m, ≡m, defined above, and is normally written “mod m” rather than “mod (m)”. 1.2.2 Ideal generators More generally, let a1, a2, . , an ∈ R, a commutative ring. Then Pn (a1, a2, . , an) = { i=1riai | ri ∈ R} n is the smallest ideal of R containing {ai}i=1, called the ideal generated by n {ai}i=1. (It is principal iff ∃b ∈ R such that (a1, a2, . , an) = (b).) Theorem 4 Let R be a ring and I an ideal in R. Then the equivalence relation EI induced by I (a EI b ⇐⇒ a − b ∈ I) is a congruence relation on R. Hence ideals generate homomorphic images via their equivalence relations. 1.3 Quotient rings First we need a notion from group theory. Definition 3 For any element a of a ring R regarded as an additive group and any additive subgroup I, the (left) coset of I by a is defined to be a + I = {a + i | i ∈ I}. 3 Since an additive group is commutative, the right coset I + a = a + I. Denote by R/I (“R mod I”) the set of all cosets of I by elements of R: R/I = {a + I | a ∈ R}. Because the coset a + I is the same as the equivalence class [a]EI , R/I and R/EI are just different notations for the same set. Theorem 5 (Quotient rings) Let I be an ideal in a ring R. Then 1. R/I = R/EI = {a + I | a ∈ R}, the set of all additive cosets of I in R, is a quotient ring under the following “mod I” operations: (a + I) + (b + I) = (a + b) + I, −(a + I) = −a + I, 0R/I = 0R + I = I, (a + I)(b + I) = ab + I, 1R/I = 1R + I; 2. R/I is a homomorphic image of R under the natural map ν : a 7→ a+I; 3. R/I is a ring (commutative if R is). Let us consider some examples. 1.3.1 The quotient ring Z/(m) This is identical to Z/≡m, and each element a + (m) has a unique represen- tative rm(a) in Zm. 1.3.2 The quotient ring R[x]/(m(x)) Let m(x) be a polynomial in R[x], R commutative (and assume deg m(x) > 0). Then (m(x)) is an ideal in R[x], and by the quotient ring theorem R[x]/(m(x)) = {a(x) + (m(x)) | a(x) ∈ R[x]} is a ring – the “quotient ring of polynomials mod m(x)” – and a homomor- phic image of R[x] under the (natural) map a(x) 7→ a(x) + (m(x)). 4 Moreover, R[x]/(m(x)) and R[x]/≡m(x) are the same sets and the coset a(x) + (m(x)) ∈ R[x]/(m(x)) is the equivalence class [a(x)] ∈ R[x]/≡m(x). Each coset or equivalence class has a unique representative having degree < deg m(x), namely rm(x)(a(x)), because:- Theorem 6 1. For any a(x) + (m(x)) ∈ R[x]/(m(x)), a(x) + (m(x)) = rm(x)(a(x) + (m(x))). 2. If deg a(x), deg b(x) < deg m(x) then a(x) 6= b(x) ⇒ a(x) + (m(x)) 6= b(x) + (m(x)). Corollary 7 If the number of elements in the set R, denoted |R|, is k and deg m(x) = n, then |R[x]/(m(x))| = kn (whereas |R[x]| is infinite because there is no limit on the degrees of the polynomials in R[x]). 1.3.3 The quotient ring R[[x]]/(xm) For fixed positive m, m a(x) ≡xm b(x) ⇐⇒ a(x) − b(x) ∈ (x ) ⇐⇒ ai = bi (i = 0, . , m − 1) and each element has a unique polynomial representative having degree < m. 1.4 Principal ideal domains A principal ideal domain (PID) is an integral domain D in which every ideal I is principal, i.e. I = (a) = {ra | r ∈ D} for some a ∈ D. Examples: Z, F [x] and F [[x]] are PIDs, but F [x, y] is not, because (x, y) is not principal. More generally: Theorem 8 A Euclidean domain is a PID. 5 Proof Let D be a Euclidean domain with degree function d, and let I be an ideal in D. If I = {0} then I is the principal ideal (0), so assume I 6= {0}. Let m 6= 0 be an element of I having minimum degree; then we claim that I = (m). (m) ⊆ I trivially by the definition of ideal. To prove that I ⊆ (m), perform a Euclidean division of any a ∈ I by m to give a = mq + r, d(r) < d(m) or r = 0. Then a, m ∈ I ⇒ r = a − mq ∈ I. But d(r) 6< d(m) because m has minimum degree by assumption, so r = 0 ⇒ a = mq ∈ (m). 2 Generally in an integral domain D the generator of a principal ideal is determined only up to associates, i.e. (a) = (b) ⇐⇒ a ∼ b, because a ∈ (m) ⇐⇒ au ∈ (m) if u is a unit in D from the definition of ideal. Hence Z is a PID in which every ideal has the form (m) where m ≥ 0 and F [x] is a PID in which every ideal 6= (0) has the form (m(x)) where m(x)) is monic, because both are Euclidean domains. Although F [[x]] is not a Euclidean domain is can be proved to be a PID in a similar way to the proof for Euclidean domains, in which every ideal 6= (0) has the form (xm) for some m ≥ 0. Because any homomorphic image of a ring R must be isomorphic to R modulo an ideal, and a PID has only one class of ideals, the conclusion is that the only non-trivial homomorphic image of Z is Zm, of F [x] is F [x]m(x) (polynomials mod m(x)) and of F [[x]] is F [[x]]xm (truncated power series under truncated arithmetic) – hence the importance of those image rings. 1.5 Simple rings Any ring R has two improper ideals, the zero ideal (0) = {0} and the unit ideal (1) = R, with corresponding improper homomorphic images R/(0) =∼ R and R/(1) =∼ {0} (and their isomorphs).