27 Principal ideal domains and Euclidean rings

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1 Note. From now on all rings are commutative rings with identity 1 0 unless stated otherwise. 27 Principal ideal domains and Euclidean rings 27.1 Definition. If R is a ring and S is a subset of R then denote S = the smallest ideal of R that contains S We say that S is the ideal of R generated by the set S Note. We have S = {b 1 a b k a k a i I, b i R, k 0} 27.3 Definition. An ideal I R is finitely generated if I = a 1,..., a n for some a 1,..., a n R. An ideal I R is a principal ideal if I = a for some a R Definition. A ring R is a principal ideal domain (PID) if it is an integral domain (25.5) such that every ideal of R is a principal ideal Proposition. The ring of integers Z is a PID. Proof. Let I Z. If I = {0} then I = 0, so I is a principal ideal. If I {0} then let a be the smallest integer such that a > 0 and a I. We will show that I = a. 110

2 Since a I we have a I. Conversely, if b I then we have b = qa + r for some q, r Z, 0 r a 1. This gives r = b qa, so r I. Since a is the smallest positive element of I, this implies that r = 0. Therefore b = qa, and so b a Proposition. If F is a field then F is a PID. Proof. If I F and 0 a I then for every b F we have b = (ba 1 )a I And so I = F. F = 1. As a consequence the only ideals of F are {0} = 0 and 27.7 Proposition. If F is a field then the ring of polynomials F[x] is a PID. Proof. Let I R. If I = {0} then I = 0. Otherwise let 0 p(x) be a polynomial such that p(x) I and deg p(x) deg q(x) for all q(x) I {0}. Check that I = p(x) Note. Z[x] is not a PID. E.g. the ideal 2, x is not a principal ideal of Z[x] (check!) Definition. A Euclidean domain is an integral domain R equipped with a function N : R {0} N = {0, 1,... } such that 111

3 1) N(ab) N(a) for all a, b R {0} 2) for any a, b R, a 0 there exist q, r R such that b = qa + r and either r = 0 or N(r) < N(a). The function N is called the norm function on R Examples. 1) Z is a Euclidean domain with the norm function given by the absolute value. 2) Any field F is a Euclidean domain with N(a) = 0 for all a F {0}. 3) If F is a field then F[x] is a Euclidean domain with N(p(x)) = deg p(x) for p(x) F[x], p(x) 0. Note: Z[x] is not a Euclidean domain with N(p(x)) = deg p(x). E. g. there are no q(x), r(x) Z[x] such that either r(x) = 0 or deg r(x) < 1 and that x = 2q(x) + r(x) 4) The ring of Gaussian integers is the subring Z[i] C given by Z[i] := {a + bi C a, b Z} Z[i] is a Euclidean domain with N(a + bi) = a 2 + b 2 = (a + bi)(a + bi) (exercise). 5) Define Z[ 5] := {a + b 5i a, b Z} Exercise: Z[ 5] is not a Euclidean domain. 112

4 27.11 Theorem. If R is a Euclidean domain then R is a PID. Proof. Let I R, I {0}. Choose a I such that a 0 and that N(a) N(b) for all b I {0}. Check that a = I Note. It is not true that every PID is a Euclidean domain. Take e.g. α = i and let Z[α] = {a + bα a, b Z} Then Z[α] is a PID, but it is not a Euclidean domain. See J. C. Wilson, A principal ideal ring that is not a euclidean ring, Mathematics Magazine, 46 (1) (1973), (Note: this link requires a JSTOR access) 113

5 28 Prime ideals and maximal ideals 28.1 Definition. Let R be a ring. 1) An ideal I R is a prime ideal if I R and for any a, b R we have ab I iff either a I or b I 2) An ideal I R is a maximal ideal if I R and for any J R such that I J R we have either J = I or J = R Examples. 1) The zero ideal {0} R is a prime ideal iff R is an integral domain, and it is a maximal ideal iff R is a field. 2) Recall that if I Z then I = nz for some n 0. Check: (nz is a prime ideal) iff (nz is a maximal ideal) iff (n is a prime number) 3) x Z[x] is prime ideal (check!) but it is not a maximal ideal: x 2, x Z[x] 28.3 Proposition. Let I R 1) The ideal I is a prime ideal iff R/I is an integral domain. 2) The ideal I is a maximal ideal iff R/I is a field Corollary. Every maximal ideal is a prime ideal. 114

6 Proof. If I R is a maximal ideal then R/I is a field. In particular R/I is an integral domain, and so I is a prime ideal. Proof of Proposition ) Assume that I R is a prime ideal. For a + I, b + I R/I we have (a + I)(b + I) = ab + I Thus if (a + I)(b + I) = 0 + I then ab + I = 0 + I i.e. ab I. Since I is a prime ideal we get that either a I (and so a + I = 0 + I) or b I (and so b + I = 0 + I). Therefore R/I is an integral domain. The other implication follows from a similar argument. 2) Assume that I R is a maximal ideal. Let a + I R/I, a + I 0 + I. We want to show that there exists b + I R/I such that (a + I)(b + I) = 1 + I Take the ideal J = a + I. We have I J R and I J (since a I). Since I is a maximal ideal we must have J = R. In particular 1 J, so 1 = ab + c for some b R, c I. This gives 1 + I = (ab + c) + I = ab + I = (a + I)(b + I) Conversely, assume that R/I is a field, and let J R be an ideal such that I J R 115

7 We will show that either J = I or J = R. Take the canonical epimorphism π : R R/I. Since π(j) is an ideal of R/I (check!) and R/I is a field we have either π(j) = {0 + I} or π(j) = R/I. Also, since I J, we have π 1 (π(j)) = J. If follows that either J = π 1 ({0 + I}) = I or J = π 1 (R/I) = R 28.5 Examples. 1) Since an ideal nz of Z is prime (and maximal) iff n is a prime number therefore Z/nZ is an integral domain (and in fact a field) iff n is a prime number. 2) Take x R[x]. We have an epimorphism of rings f : R[x] C f(p(x)) = p(i) Check: Ker(f) = x By the First Isomorphism Theorem (26.13) we get R[x]/ x = C. Since C is a field this shows that x is a maximal ideal of R[x] Note. For I, J R define Check: IJ is an ideal of R. IJ := {a 1 b a k b k a i I, b i J, k 0} 28.7 Proposition. Let I R, I R. The ideal I is a prime ideal iff for any ideals J 1, J 2 such that J 1 J 2 I we have either J 1 I or J 2 I. Proof. Exercise. 116

8 29 Zorn s Lemma and maximal ideals Goal: 29.1 Theorem. If R is a ring, I R, and I R then there exists a maximal ideal J R such that I J Definition. A partially ordered set (or poset) is a set S equipped with a binary relation satisfying 1) x x for all x S (reflexivity) 2) if x y and y z then x z (transitivity) 3) if x y and y x then y = x (antisymmetry) Example. If A is a set and S is the set of all subsets of A then S is a poset with ordering given by the inclusion of subsets Definition. A linearly ordered set is a poset (S, ) such that for any x, y S we have either x y or y x Definition. If (S, ) is a poset then an element x S is a maximal element of S if we have x y only for y = x Example. Let S be the set of all proper subsets of a set A ordered with respect to the inclusion. For every a A the set A {a} is a maximal element of S. 117

9 29.7 Note. If (S, ) is a poset and T S then T is also a poset with ordering inherited from S Definition. Let (S, ) is a poset and let T S. An upper bound of T is an element x S such that y x for all y T Definition. If (S, ) is a poset. A chain in S is a subset T S such that T is linearly ordered Zorn s Lemma. If (S, ) is a non-empty poset such that every chain in S has an upper bound in S then S contains a maximal element. Proof of Theorem Let I R be an ideal of R, and let S be the set of all ideals J R such that I J and J R. Notice that S since I S. The set S is a poset ordered with respect to inclusion of ideals. We will show that every chain in S has an upper bound in S. Let T = {J i } i A be a chain in S. Check: J T := i A J i is an ideal of R. Moreover, I J T. Finally, we have J T R. Indeed, otherwise 1 J T, and so 1 J i for some i A. This would give J i = R, which contradicts our assumptions. It follows that J T S. Since J i J T for all i A, we obtain that J T is an upper bound of the chain T. By Zorn s Lemma (29.10) there is a maximal element J S. This means, in particular, that J is an ideal of R such I J. Moreover, let K R be any ideal such that K R and J K. We have I J K which means that K S. Maximality of J is S implies then that J = K. This shows that J is a maximal ideal of R. 118

10 29.11 Corollary. Every ring contains a maximal ideal. Proof. If R is a ring then by Theorem 29.1 there exists a maximal ideal I R such that I contains the zero ideal {0} R. 119

11 30 Unique factorization domains Motivation: 30.1 Fundamental Theorem of Arithmetic. If n Z, n > 1 then n = p 1 p 2... p k where p 1,..., p k are primes. Moreover, this decomposition is unique up to reordering of factors. Goal. Extend this to other rings Definition. Let R be an integral domain. An element a R is irreducible if a 0, a is not a unit and if a = bc for some b, c R then either b or c is a unit Examples. 1) n Z is irreducible iff n = ±p where p is a prime number. 2) A field has no irreducible elements. 3) Take p(x) R[x], p(x) = x Then p(x) is irreducible in R[x]. 4) Take p(x) C[x], p(x) = x Then p(x) is not irreducible in C[x]: p(x) = (x i)(x + i) 30.4 Note. If a R is an irreducible element and u R is a unit then ua is irreducible. 120

12 30.5 Definition. Let R be an integral domain. Elements a, b R are associates if a = ub for some unit u R. We write: a b Examples. 1) If m, n Z then m n iff m = ±n. 2) Check: units in R[x] are non-zero polynomials of degree 0. It follows that if p(x), q(x) R[x] then f(x) g(x) iff f(x) = ag(x) for some a R {0} Definition. A unique factorization domain (UFD) is an integral domain R that satisfies the following conditions: 1) if a R is a non-zero, non-unit element then a = b 1... b k for some irreducible elements b 1,..., b k R 2) if b 1,..., b k, c 1,..., c l are irreducible elements such that b 1... b k = c 1... c l then k = l and for some permutation σ : {1,..., k} {1,..., k} we have b 1 c σ(1),..., b k c σ(k) Examples. 1) Z is a UFD by the Fundamental Theorem of Arithmetic (30.1). 2) If F is a field then F is a UFD since all non-zero elements of F are units. 121

13 Recall. Z[ 5] is the subring of C given by Z[ 5] = {a + b 5i a, b Z} 30.9 Proposition. Z[ 5] is not a UFD. Proof. For a + b 5i Z[ 5] define N(a + b 5i) = (a + b 5i)(a + b 5i) = a 2 + 5b 2 N Notice that 1) N(α) = 1 iff α = ±1 2) N(α) = 0 iff α = 0 3) N(α) 3 for all α Z[ 5] 4) N(αβ) = N(α)N(β) Observation 1. The only units in Z[ 5] are 1 and 1. As a consequence for any α, β we have α β iff α = ±β. Indeed, if α Z[ 5] is a unit then Therefore N(α) = 1, and so α = ±1. N(α)N(α 1 ) = N(αα 1 ) = N(1) = 1 Observation 2. If α Z[ 5] is an element such that N(α) = 9 then α is irreducible. Indeed, if α = ββ then N(β)N(β ) = N(α) = 9 122

14 Therefore N(β) must be either 1 (and so β is a unit), 3 (impossible), or 9 (and then N(β ) = 1, i.e. β is a unit). Take 9 Z[ 5]. We have 3 3 = 9 = (2 + 5i)(2 5i) By Observation 2 the elements 3, 2 + 5i, 2 5i are irreducible in Z[ 5]. On the other hand by Observation 1 we obtain 3 (2 + 5i), 3 (2 5i) As a consequence 9 does not have a unique factorization in Z[ 5] 123

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