DEDEKIND DOMAINS AND THE IDEAL CLASS GROUP BENJAMIN BOYAJIAN Abstract. In this paper, we will define a discrete valuation ring and Dedekind domain, and will develop a notion of the ideal group and ideal class group of a Dedekind domain. We will also prove that the ring of algebraic integers in an imaginary quadratic number field is a Dedekind domain, and will prove some additional theorems about the ideal class group for this special case. Contents 1. Factorization in Commutative Rings 1 2. Localization and Integrality 4 3. Discrete Valuation Rings 4 4. Dedekind Domains 6 5. The Ideal Class Group 7 Acknowledgments 12 References 12 1. Factorization in Commutative Rings First, we will provide some background material on factorization, which will be important for the rest of the paper. We assume that the reader is acquainted with basic ring theory, and proceed with a few important definitions. Throughout this paper, R will denote a commutative ring, 0 will denote the additive identity of R, and 1 will denote the multiplicative identity of R. Definition 1.1. An element u 2 R is a unit if it has a multiplicative inverse. Definition 1.2. For two elements a; b 2 R, we say that a divides b if there exists an element c 2 R such that ac = b. We write a j b if a divides b and a - b if a does not divide b. Definition 1.3. Two elements a; b 2 R are associates if there exists a unit u 2 R such that au = b. Remark 1.4. If a; b 2 R are associates, then a j b and b j a. Definition 1.5. If a; b 2 R, we say that a is a proper divisor of b if a is not a unit, and there exists c 2 R such that ac = b, where c is not a unit. Definition 1.6. An element a 2 R that is nonzero and not a unit is irreducible if there do not exist elements b; c 2 R that are not units such that a = bc. Date: August 26, 2011. 1 2 BENJAMIN BOYAJIAN Definition 1.7. An element p 2 R that is nonzero and not a unit is prime if for any elements a; b 2 R such that p j ab, either p j a or p j b. p Example 1.8. In Z[ −5], the element 2p is irreducible,p but not prime.p To prove 2 that it is irreducible, note that for all a + b −p5 2 Z[ −5], we have ja + b −5j = 2 2 a + 5b 2 Z. Suppose the only elements of Z[ −5] thatp have absolutep value 1 are ±1, which are units. Supposep there is an element a + b −5 2 Z[ −5] that is a 2 2 2 2 2 2 proper divisor of 2. Then ja+b −5j = a +5b divides j2pj = 4, thusp a +5b = 2, which isp impossible. Also,p 2 is not prime because 2 j (1 + −5)(1 − −5) = 6, but 2 - 1 + −5 and 2 - 1 − −5. Definition 1.9. An ideal I ⊆ R is principal if I = far j r 2 Rg for some a 2 R. We say that I is the principal ideal generated by a, and we write I = (a). Remark 1.10. If a; b 2 R are associates, then (a) = (b). Similarly, u 2 R is a unit if and only if (u) = (1) = R. Definition 1.11. A commutative ring R is an integral domain if for any elements a; b 2 R such that ab = 0, either a = 0 or b = 0. Definition 1.12. An ideal P ⊆ R is prime if for any elements a; b 2 R such that ab 2 P , either a 2 P or b 2 P . Remark 1.13. A principal ideal (p) ⊆ R is prime if and only if p 2 R is prime. This follows because saying that p j a is equivalent to saying that a 2 (p). Proposition 1.14. An ideal P ⊆ R is prime if and only if the quotient ring R=P is an integral domain. Proof. This follows because a 2 P if and only if a = 0, where a is the equivalence class of a in R=P . Definition 1.15. An ideal M ⊆ R is maximal if the only ideals that contain M are M and R. Proposition 1.16. An ideal M ⊆ R is maximal if and only if the quotient ring R=M is a field. Proof. By the Correspondence Theorem (see Artin Chapter 10, Proposition 4.3), the ideals of R that contain M are in bijective correspondence with the ideals of R=M. If M is a maximal ideal, then there only two ideals of R=M, thus R=M is a field. Conversely, if R=M is a field, then there are only two ideals of R that contain M, thus M is a maximal ideal. Corollary 1.17. A maximal ideal is a prime ideal. Proof. Let M ⊂ R be a maximal ideal. Then the quotient ring R=M is a field, thus it is an integral domain. By (1.12), M is a prime ideal. Proposition 1.18. If an element p 2 R is prime and R is an integral domain, then p is irreducible. Proof. Suppose there exist elements a; b 2 R that are not units such that ab = p. Then p j ab, but p - a. This is because a = cp implies p = ab = bcp, which implies bc = 1 by the Cancellation Law. Similarly, p - b, which contradicts the definition of prime. DEDEKIND DOMAINS AND THE IDEAL CLASS GROUP 3 Definition 1.19. An integral domain R is an principal ideal domain if every ideal of R is principal. Definition 1.20. n ring R is Noetherian if it does not contain an infinite increasing chain of ideals I1 ⊂ I2 ⊂ :::. Definition 1.21. A ring R has the ascending chain property if it does not contain an infinite increasing chain of principal ideals (a1) ⊂ (a2) ⊂ :::. Proposition 1.22. An integral domain R has the ascending chain property if and only if the process of factoring any element into irreducible elements terminates after a finite number of steps. Proof. Suppose that R contains an infinite increasing chain of principal ideals (a1) ⊂ (a2) ⊂ :::. Then (an) 6= R for all n 2 N, thus each an is not a unit. Also, (an−1) ⊂ (an), so an is a proper divisor of an−1. So then we can write a1 = b1a2 = b1b2a3 = :::, where the bi are not units, and the factoring process does not terminate. Conversely, if the process of factoring does not terminate, then we have a1 = b1a2 = b1b2a3 = ::: where the bi are not units, thus (a1) ⊂ (a2) ⊂ ::: is a strictly increasing chain of principal ideals. Definition 1.23. An integral domain R is a unique factorization domain if every element a 2 R can be factored a = a1 : : : an into irreducible elements, and this factorization is unique up to associates. p Example 1.24. The ring Z[ −5]p is not a uniquep factorization domain. To see this, observe that 6 = 2 · 3 = (1 + −5)(1 − −5)p has two distinctp factorizations into irreducible elements. The proofs that 2; 3; (1+ −5); (1− −5) are irreducible follow in the same way as Example 1.8. Proposition 1.25. Let R be an integral domain with the ascending chain property. Then R is a unique factorization domain if and only if every irreducible element is prime. Proof. Suppose that R is a unique factorization domain, and let a 2 R be an irreducible element. If that a j bc for some b; c 2 R, then bc can be factored bc = a1 : : : an. This factorization is unique, so ai is an associate of a for some 1 ≤ i ≤ n. Then either b or c must contain an associate of a in its unique factorization. Conversely, suppose that R is not a unique factorization domain, and then there exists a 2 R that has two distinct factorizations a = b1 : : : bn = c1 : : : cm. Then there exists 1 ≤ i ≤ n such that bi is not an associate of cj for any 1 ≤ j ≤ m. Then b j a, but b - c1 : : : cm−1 and b - cm. Proposition 1.26. In a principal ideal domain, every irreducible element is prime. Proof. Let R be a principal ideal domain, and let a 2 R be an irreducible element. Then the principal ideal (a) ⊂ R is a maximal ideal. To prove this, suppose that there exists an intermediate principal ideal (a) ⊂ (b) ⊂ R. Then b is a proper divisor of a, but b is not a unit, which contradicts the assumption that a is irreducible. Then the quotient ring R=(a) is a field, thus it is an integral domain. Then (a) is a prime ideal, thus a is prime. Theorem 1.27. A principal ideal domain is a unique factorization domain. 4 BENJAMIN BOYAJIAN Proof. Let R be a principal ideal domain, and then every irreducible element is prime. So we only need to prove the existence of factorizations for elements in R, which is equivalent to showing that R contains no infinite increasing chain of principal ideals. Suppose that (a1) ⊂ (a2) ⊂ ::: is such a chain, and then the union [n2N(an) is an ideal, thus is it generated by some b 2 R. Then b is contained in one of the ideals (an), thus (b) ⊆ (an).