Note. From now on all rings are commutative rings with identity 1 6= 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 hSi = the smallest ideal of R that contains S We say that hSi is the ideal of R generated by the set S. 27.2 Note. We have hSi = fb1a1 + : : : bkak j ai 2 I; bi 2 R; k ≥ 0g 27.3 Definition. An ideal I C R is finitely generated if I = ha1; : : : ; ani for some a1; : : : ; an 2 R. An ideal I C R is a principal ideal if I = hai for some a 2 R. 27.4 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. 27.5 Proposition. The ring of integers Z is a PID. Proof. Let I C Z. If I = f0g then I = h0i, so I is a principal ideal. If I 6= f0g then let a be the smallest integer such that a > 0 and a 2 I. We will show that I = hai. 110 Since a 2 I we have hai ⊆ I. Conversely, if b 2 I then we have b = qa + r for some q; r 2 Z, 0 ≤ r ≤ a − 1. This gives r = b − qa, so r 2 I. Since a is the smallest positive element of I, this implies that r = 0. Therefore b = qa, and so b 2 hai. 27.6 Proposition. If F is a field then F is a PID. Proof. If I C F and 0 6= a 2 I then for every b 2 F we have b = (ba−1)a 2 I And so I = F. As a consequence the only ideals of F are f0g = h0i and F = h1i. 27.7 Proposition. If F is a field then the ring of polynomials F[x] is a PID. Proof. Let I 2 R. If I = f0g then I = h0i. Otherwise let 0 6= p(x) be a polynomial such that p(x) 2 I and deg p(x) ≤ deg q(x) for all q(x) 2 I − f0g. Check that I = hp(x)i. 27.8 Note. Z[x] is not a PID. E.g. the ideal h2; xi is not a principal ideal of Z[x] (check!). 27.9 Definition. A Euclidean domain is an integral domain R equipped with a function N : R − f0g −! N = f0; 1;::: g such that 111 1) N(ab) ≥ N(a) for all a; b 2 R − f0g 2) for any a; b 2 R, a 6= 0 there exist q; r 2 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. 27.10 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 2 F − f0g. 3) If F is a field then F[x] is a Euclidean domain with N(p(x)) = deg p(x) for p(x) 2 F[x], p(x) 6= 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) 2 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] := fa + bi 2 C j a; b 2 Zg Z[i] is a Euclidean domain with N(a + bi) = a2 + b2 = (a + bi)(a + bi) (exercise). 5) Define p p Z[ −5] := fa + b 5i j a; b 2 Zg p Exercise: Z[ −5] is not a Euclidean domain. 112 27.11 Theorem. If R is a Euclidean domain then R is a PID. Proof. Let ICR, I 6= f0g. Choose a 2 I such that a 6= 0 and that N(a) ≤ N(b) for all b 2 I − f0g. Check that hai = I. 27.12 Note.p It is not true that every PID is a Euclidean domain. Take e.g. 1 19 α = 2 + 2 i and let Z[α] = fa + bα j a; b 2 Zg 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), 34-38. (Note: this link requires a JSTOR access) 113 28 Prime ideals and maximal ideals 28.1 Definition. Let R be a ring. 1) An ideal I C R is a prime ideal if I 6= R and for any a; b 2 R we have ab 2 I iff either a 2 I or b 2 I 2) An ideal I C R is a maximal ideal if I 6= R and for any J C R such that I ⊆ J ⊆ R we have either J = I or J = R. 28.2 Examples. 1) The zero ideal f0g 2 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 C 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) hxi C Z[x] is prime ideal (check!) but it is not a maximal ideal: hxi ⊆ h2; xi C Z[x] 28.3 Proposition. Let I C 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. 28.4 Corollary. Every maximal ideal is a prime ideal. 114 Proof. If I C 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 28.3. 1) Assume that I C R is a prime ideal. For a + I; b + I 2 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 2 I. Since I is a prime ideal we get that either a 2 I (and so a + I = 0 + I) or b 2 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 C R is a maximal ideal. Let a + I 2 R=I, a + I 6= 0 + I. We want to show that there exists b + I 2 R=I such that (a + I)(b + I) = 1 + I Take the ideal J = hai + I. We have I ⊆ J ⊆ R and I 6= J (since a 62 I). Since I is a maximal ideal we must have J = R. In particular 1 2 J, so 1 = ab + c for some b 2 R, c 2 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 C R be an ideal such that I ⊆ J ⊆ R 115 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) = f0 + Ig or π(J) = R=I. Also, since I ⊆ J, we have π−1(π(J)) = J. If follows that either J = π−1(f0 + Ig) = 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 x2 + 1 2 R[x]. We have an epimorphism of rings f : R[x] −! C f(p(x)) = p(i) Check: Ker(f) = hx2 + 1i. By the First Isomorphism Theorem (26.13) 2 ∼ 2 we get R[x]=hx + 1i = C. Since C is a field this shows that hx + 1i is a maximal ideal of R[x]. 28.6 Note. For I;J C R define IJ := fa1b1 + ::: + akbk j ai 2 I; bi 2 J; k ≥ 0g Check: IJ is an ideal of R. 28.7 Proposition. Let I C R, I 6= R. The ideal I is a prime ideal iff for any ideals J1;J2 such that J1J2 ⊆ I we have either J1 ⊆ I or J2 ⊆ I. Proof. Exercise. 116 29 Zorn's Lemma and maximal ideals Goal: 29.1 Theorem. If R is a ring, I C R, and I 6= R then there exists a maximal ideal J C R such that I ⊆ J. 29.2 Definition. A partially ordered set (or poset) is a set S equipped with a binary relation ≤ satisfying 1) x ≤ x for all x 2 S (reflexivity) 2) if x ≤ y and y ≤ z then x ≤ z (transitivity) 3) if x ≤ y and y ≤ x then y = x (antisymmetry). 29.3 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. 29.4 Definition.