# 3.1 The Definition and Some Basic Properties. We identify the natural class of integral domains in which unique factorization of ideals is possible.

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1 Chapter 3 Dedekind Domains 3.1 The Definition and Some Basic Properties We identify the natural class of integral domains in which unique factorization of ideals is possible Definition A Dedekind domain is an integral domain A satisfying the following three conditions: (1) A is a Noetherian ring; (2) A is integrally closed; (3) Every nonzero prime ideal of A is maximal. A principal ideal domain satisfies all three conditions, and is therefore a Dedekind domain. We are going to show that in the AKLB setup, if A is a Dedekind domain, then so is B, a result that provides many more examples and already suggests that Dedekind domains are important in algebraic number theory Proposition In the AKLB setup, B is integrally closed, regardless of A. If A is an integrally closed Noetherian ring, then B is also a Noetherian ring, as well as a finitely generated A-module. Proof. By (1.1.6), B is integrally closed in L, which is the fraction field of B by (2.2.8). Therefore B is integrally closed. If A is integrally closed, then by (2.3.8), B is a submodule of a free A-module M of rank n. IfA is Noetherian, then M, which is isomorphic to the direct sum of n copies of A, is a Noetherian A-module, hence so is the submodule B. An ideal of B is, in particular, an A-submodule of B, hence is finitely generated over A and therefore over B. It follows that B is a Noetherian ring Theorem In the AKLB setup, if A is a Dedekind domain, then so is B. In particular, the ring of algebraic integers in a number field is a Dedekind domain. 1

2 2 CHAPTER 3. DEDEKIND DOMAINS Proof. In view of (3.1.2), it suffices to show that every nonzero prime ideal Q of B is maximal. Choose any nonzero element x of Q. Since x B, x satisfies a polynomial equation x m + a m 1 x m a 1 x + a 0 =0 with the a i A. If we take the positive integer m as small as possible, then a 0 0by minimality of m. Solving for a 0, we see that a 0 Bx A Q A, so the prime ideal P = Q A is nonzero, hence maximal by hypothesis. By Section 1.1, Problem 6, Q is maximal. Problems For Section 3.1 This problem set will give the proof of a result to be used later. Let P 1,P 2,...,P s,s 2, be ideals in a ring R, with P 1 and P 2 not necessarily prime, but P 3,...,P s prime (if s 3). Let I be any ideal of R. The idea is that if we can avoid the P j individually, in other words, for each j we can find an element in I but not in P j, then we can avoid all the P j simultaneously, that is, we can find a single element in I that is in none of the P j. The usual statement is the contrapositive of this assertion. Prime Avoidance Lemma With I and the P i as above, if I s i=1 P i, then for some i we have I P i. 1. Suppose that the result is false. Show that without loss of generality, we can assume the existence of elements a i I with a i P i but a i / P 1 P i 1 P i+1 P s. 2. Prove the result for s =2. 3. Now assume s > 2, and observe that a 1 a 2 a s 1 P 1 P s 1, but a s / P 1 P s 1. Let a =(a 1 a s 1 )+a s, which does not belong to P 1 P s 1, else a s would belong to this set. Show that a I and a/ P 1 P s, contradicting the hypothesis. 3.2 Fractional Ideals Our goal is to establish unique factorization of ideals in a Dedekind domain, and to do this we will need to generalize the notion of ideal. First, some preliminaries Products of Ideals Recall that if I 1,...,I n are ideals, then their product I 1 I n is the set of all finite sums i a 1ia 2i a ni, where a ki I k,k=1,...,n. It follows from the definition that I 1 I n is an ideal contained in each I j. Moreover, if a prime ideal P contains a product I 1 I n of ideals, then P contains I j for some j.

3 3.2. FRACTIONAL IDEALS Proposition If I is a nonzero ideal of the Noetherian integral domain R, then I contains a product of nonzero prime ideals. Proof. Assume the contrary. If S is the collection of all nonzero ideals that do not contain a product of nonzero prime ideals, then, as R is Noetherian, S has a maximal element J, and J cannot be prime because it belongs to S. Thus there are elements a, b R such that a/ J, b / J, and ab J. By maximality of J, the ideals J + Ra and J + Rb each contain a product of nonzero prime ideals, hence so does (J +Ra)(J +Rb) J +Rab = J. This is a contradiction. (Notice that we must use the fact that a product of nonzero ideals is nonzero, and this is where the hypothesis that R is an integral domain comes in.) Corollary If I is an ideal of the Noetherian ring R (not necessarily an integral domain), then I contains a product of prime ideals. Proof. Repeat the proof of (3.2.2), with the word nonzero deleted. Ideals in the ring of integers are of the form nz, the set of multiples of n. A set of the form (3/2)Z is not an ideal because it is not a subset of Z, yet it behaves in a similar manner. The set is closed under addition and multiplication by an integer, and it becomes an ideal of Z if we simply multiply all the elements by 2. It will be profitable to study sets of this type Definitions Let R be an integral domain with fraction field K, and let I be an R-submodule of K. We say that I is a fractional ideal of R if ri R for some nonzero r R. We call r a denominator of I. An ordinary ideal of R is a fractional ideal (take r = 1), and will often be referred to as an integral ideal Lemma (i) If I is a finitely generated R-submodule of K, then I is a fractional ideal. (ii) If R is Noetherian and I is a fractional ideal of R, then I is a finitely generated R-submodule of K. (iii) If I and J are fractional ideals with denominators r and s respectively, then I J, I+J and IJ are fractional ideals with respective denominators r (or s), rs and rs. [The product of fractional ideals is defined exactly as in (3.2.1).] Proof. (i) If x 1 = a 1 /b 1,...,x n = a n /b n generate I and b = b 1 b n, then bi R. (ii) If ri R, then I r 1 R.AsanR-module, r 1 R is isomorphic to R and is therefore Noetherian. Consequently, I is finitely generated. (iii) It follows from the definition (3.2.4) that the intersection, sum and product of fractional ideals are fractional ideals. The assertions about denominators are proved by noting that r(i J) ri R, rs(i + J) ri + sj R, and rsij =(ri)(sj) R.

4 4 CHAPTER 3. DEDEKIND DOMAINS The product of two nonzero fractional ideals is a nonzero fractional ideal, and the multiplication is associative because multiplication in R is associative. There is an identity element, namely R, since RI I =1I RI. We will show that if R is a Dedekind domain, then every nonzero fractional ideal has a multiplicative inverse, so the nonzero fractional ideals form a group Lemma Let I be a nonzero prime ideal of the Dedekind domain R, and let J be the set of all elements x K such that xi R. Then R J. Proof. Since RI R, it follows that R is a subset of J. Pick a nonzero element a I, so that I contains the principal ideal Ra. Let n be the smallest positive integer such that Ra contains a product P 1 P n of n nonzero prime ideals. Since R is Noetherian, there is such an n by (3.2.2), and by (3.2.1), I contains one of the P i,sayp 1. But in a Dedekind domain, every nonzero prime ideal is maximal, so I = P 1. Assuming n 2, set I 1 = P 2 P n, so that Ra I 1 by minimality of n. Choose b I 1 with b/ Ra. Now II 1 = P 1 P n Ra, in particular, Ib Ra, hence Iba 1 R. (Note that a has an inverse in K but not necessarily in R.) Thus ba 1 J, but ba 1 / R, for if so, b Ra, contradicting the choice of b. The case n = 1 must be handled separately. In this case, P 1 = I Ra P 1,so I = Ra. Thus Ra is a proper ideal, and we can choose b R with b / Ra. Then ba 1 / R, but ba 1 I = ba 1 Ra = br R, soba 1 J. We now prove that in (3.2.6), J is the inverse of I Proposition Let I be a nonzero prime ideal of the Dedekind domain R, and let J = {x K : xi R}. Then J is a fractional ideal and IJ = R. Proof. If r is a nonzero element of I and x J, then rx R, sorj R and J is a fractional ideal. Now IJ R by definition of J, soij is an integral ideal. Using (3.2.6), we have I = IR IJ R, and maximality of I implies that either IJ = I or IJ = R. In the latter case, we are finished, so assume IJ = I. If x J, then xi IJ = I, and by induction, x n I I for all n =1, 2,... Let r be any nonzero element of I. Then rx n x n I I R, sor[x] is a fractional ideal. Since R is Noetherian, part (ii) of (3.2.5) implies that R[x] is a finitely generated R-submodule of K. By (1.1.2), x is integral over R. But R, a Dedekind domain, is integrally closed, so x R. Therefore J R, contradicting (3.2.6). The following basic property of Dedekind domains can be proved directly from the definition, without waiting for the unique factorization of ideals Theorem If R is a Dedekind domain, then R is a UFD if and only if R is a PID. Proof. Recall from basic algebra that a (commutative) ring R is a PID iff R is a UFD and every nonzero prime ideal of R is maximal.

5 3.3. UNIQUE FACTORIZATION OF IDEALS 5 Problems For Section If I and J are relatively prime ideals (I + J = R), show that IJ = I J. More generally, if I 1,...,I n are relatively prime in pairs, show that I 1 I n = n i=1 I i. 2. Let P 1 and P 2 be relatively prime ideals in the ring R. Show that P r 1 and P s 2 are relatively prime for arbitrary positive integers r and s. 3. Let R be an integral domain with fraction field K. IfK is a fractional ideal of R, show that R = K. 3.3 Unique Factorization of Ideals In the previous section, we inverted nonzero prime ideals in a Dedekind domain. We now extend this result to nonzero fractional ideals Theorem If I is a nonzero fractional ideal of the Dedekind domain R, then I can be factored uniquely as P n1 1 P n2 2 Pr nr, where the n i are integers. Consequently, the nonzero fractional ideals form a group under multiplication. Proof. First consider the existence of such a factorization. Without loss of generality, we can restrict to integral ideals. [Note that if r 0 and ri R, then I =(rr) 1 (ri).] By convention, we regard R as the product of the empty collection of prime ideals, so let S be the set of all nonzero proper ideals of R that cannot be factored in the given form, with all n i positive integers. (This trick will yield the useful result that the factorization of integral ideals only involves positive exponents.) Since R is Noetherian, S, if nonempty, has a maximal element I 0, which is contained in a maximal ideal I. By (3.2.7), I has an inverse fractional ideal J. Thus by (3.2.6) and (3.2.7), I 0 = I 0 R I 0 J IJ = R. Therefore I 0 J is an integral ideal, and we claim that I 0 I 0 J. For if I 0 = I 0 J, then the last paragraph of the proof of (3.2.7) can be reproduced with I replaced by I 0 to reach a contradiction. By maximality of I 0, I 0 J is a product of prime ideals, say I 0 J = P 1 P r (with repetition allowed). Multiply both sides by the prime ideal I to conclude that I 0 is a product of prime ideals, contradicting I 0 S.ThusSmust be empty, and the existence of the desired factorization is established. To prove uniqueness, suppose that we have two prime factorizations P n1 1 Pr nr = Q t1 1 Qts s where again we may assume without loss of generality that all exponents are positive. (If P n appears, multiply both sides by P n.) Now P 1 contains the product of the P ni i, so by (3.2.1), P 1 contains Q j for some j. By maximality of Q j, P 1 = Q j, and we may renumber so that P 1 = Q 1. Multiply by the inverse of P 1 (a fractional ideal, but there is no problem), and continue inductively to complete the proof.

6 6 CHAPTER 3. DEDEKIND DOMAINS Corollary A nonzero fractional ideal I is an integral ideal if and only if all exponents in the prime factorization of I are nonnegative. Proof. The only if part was noted in the proof of (3.3.1). The if part follows because a power of an integral ideal is still an integral ideal Corollary Denote by n P (I) the exponent of the prime ideal P in the factorization of I. (If P does not appear, take n P (I) = 0.) If I 1 and I 2 are nonzero fractional ideals, then I 1 I 2 if and only if for every prime ideal P of R, n P (I 1 ) n P (I 2 ). Proof. We have I 2 I 1 iff I 2 I1 1 R, and by (3.3.2), this happens iff for every P, n P (I 2 ) n P (I 1 ) Definition Let I 1 and I 2 be nonzero integral ideals. We say that I 1 divides I 2 if I 2 = JI 1 for some integral ideal J. Just as with integers, an equivalent statement is that each prime factor of I 1 is a factor of I Corollary If I 1 and I 2 are nonzero integral ideals, then I 1 divides I 2 if and only if I 1 I 2. In other words, for these ideals, DIVIDES MEANS CONTAINS. Proof. By (3.3.4), I 1 divides I 2 iff n P (I 1 ) n P (I 2 ) for every prime ideal P. By (3.3.3), this is equivalent to I 1 I GCD s and LCM s As a nice application of the principle that divides means contains, we can use the prime factorization of ideals in a Dedekind domain to compute the greatest common divisor and least common multiple of two nonzero ideals I and J, exactly as with integers. The greatest common divisor is the smallest ideal containing both I and J, that is, I + J. The least common multiple is the largest ideal contained in both I and J, which is I J. A Dedekind domain comes close to being a principal ideal domain in the sense that every nonzero integral ideal, in fact every nonzero fractional ideal, divides some principal ideal.

7 3.4. SOME ARITHMETIC IN DEDEKIND DOMAINS Proposition Let I be a nonzero fractional ideal of the Dedekind domain R. Then there is a nonzero integral ideal J such that IJ is a principal ideal of R. Proof. By (3.3.1), there is a nonzero fractional ideal I such that II = R. By definition of fractional ideal, there is a nonzero element r R such that ri is an integral ideal. If J = ri, then IJ = Rr, a principal ideal of R. Problems For Section 3.3 By (2.3.11), the ring B of algebraic integers in Q( 5) is Z[ 5]. In Problems 1-3, we will show that Z[ 5] is not a unique factorization domain by considering the factorization (1 + 5)(1 5) = By computing norms, verify that all four of the above factors are irreducible. 2. Show that the only units of B are ±1. 3. Show that no factor on one side of the above equation is an associate of a factor on the other side, so unique factorization fails. 4. Show that the ring of algebraic integers in Q( 17) is not a unique factorization domain. 5. In Z[ 5] and Z 17], the only algebraic integers of norm 1 are ±1. Show that this property does not hold for the algebraic integers in Q( 3). 3.4 Some Arithmetic in Dedekind Domains Unique factorization of ideals in a Dedekind domain permits calculations that are analogous to familiar manipulations involving ordinary integers. In this section, we illustrate some of the ideas. Let P 1,...,P n be distinct nonzero prime ideals of the Dedekind domain R, and let J = P 1 P n. Let Q i be the product of the P j with P i omitted, that is, Q i = P 1 P i 1 P i+1 P n. (If n = 1, we take Q 1 = R.) If I is any nonzero ideal of R, then by unique factorization, IQ i IJ. For each i =1,...,n, choose an element a i belonging to IQ i but not to IJ, and let a = n i=1 a i Lemma The element a belongs to I, but for each i, a / IP i. (In particular, a 0.) Proof. Since each a i belongs to IQ i I, we have a I. Now a i cannot belong to IP i, for if so, a i IP i IQ i, which is the least common multiple of IP i and IQ i [see (3.3.6)]. But by definition of Q i, the least common multiple is simply IJ, which contradicts the choice of a i. We break up the sum defining a as follows: a =(a a i 1 )+a i +(a i a n ). (1)

8 8 CHAPTER 3. DEDEKIND DOMAINS If j i, then a j IQ j IP i, so the first and third terms of the right side of (1) belong to IP i. Since a i / IP i, as found above, we have a/ IP i. In (3.3.7), we found that any nonzero ideal is a factor of a principal ideal. We can sharpen this result as follows Proposition Let I be a nonzero ideal of the Dedekind domain R. Then there is a nonzero ideal I such that II is a principal ideal (a). Moreover, if J is an arbitrary nonzero ideal of R, then I can be chosen to be relatively prime to J. Proof. Let P 1,...,P n be the distinct prime divisors of J, and choose a as in (3.4.1). Then a I, so(a) I. Since divides means contains [see (3.3.5)], I divides (a), so (a) =II for some nonzero ideal I. If I is divisible by P i, then I = P i I 0 for some nonzero ideal I 0, and (a) =IP i I 0. Consequently, a IP i, contradicting (3.4.1) Corollary A Dedekind domain with only finitely many prime ideals is a PID. Proof. Let J be the product of all the nonzero prime ideals. If I is any nonzero ideal, then by (3.4.2) there is a nonzero ideal I such that II is a principal ideal (a), with I relatively prime to J. But then the set of prime factors of I is empty, so I = R. Thus (a) =II = IR = I. The next result reinforces the idea that a Dedekind domain is not too far away from a principal ideal domain Corollary Let I be a nonzero ideal of the Dedekind domain R, and let a be any nonzero element of I. Then I can be generated by two elements, one of which is a. Proof. Since a I, we have (a) I, soi divides (a), say (a) =IJ. By (3.4.2), there is a nonzero ideal I such that II is a principal ideal (b) and I is relatively prime to J. If gcd stands for greatest common divisor, then the ideal generated by a and b is because gcd(j, I ) = (1) The Ideal Class Group gcd((a), (b)) = gcd(ij,ii )=I Let I(R) be the group of nonzero fractional ideals of a Dedekind domain R. If P (R) is the subset of I(R) consisting of all nonzero principal fractional ideals Rx, x K, then P (R) is a subgroup of I(R). To see this, note that (Rx)(Ry) 1 =(Rx)(Ry 1 )=Rxy 1, which belongs to P (R). The quotient group C(R) = I(R)/P (R) is called the ideal class group of R. Since R is commutative, C(R) is abelian, and we will show later that in the number field case, C(R) is finite.

9 3.4. SOME ARITHMETIC IN DEDEKIND DOMAINS 9 Let us verify that C(R) is trivial if and only if R is a PID. If C(R) is trivial, then every integral ideal I of R is a principal fractional ideal Rx, x K. But I R, sox =1x must belong to R, proving that R is a PID. Conversely, if R is a PID and I is a nonzero fractional ideal, then ri R for some nonzero r R. By hypothesis, the integral ideal ri must be principal, so ri = Ra for some a R. Thus I = R(a/r) with a/r K, and we conclude that every nonzero fractional ideal of R is a principal fractional ideal. Problems For Section 3.4 We will now go through the factorization of an ideal in a number field. In the next chapter, we will begin to develop the necessary background, but some of the manipulations are accessible to us now. By (2.3.11), the ring B of algebraic integers of the number field Q( 5) is Z[ 5]. (Note that 5 3 mod 4.) If we wish to factor the ideal (2) = 2B of B, the idea is to factor x mod 2, and the result is x 2 +5 (x +1) 2 mod 2. Identifying x with 5, we form the ideal P 2 =(2, 1+ 5), which turns out to be prime. The desired factorization is (2) = P2 2. This technique works if B = Z[α], where the number field L is Q( α). 1. Show that 1 5 P 2, and conclude that 6 P Show that 2 P2 2, hence (2) P Expand P2 2 =(2, 1+ 5)(2, 1+ 5), and conclude that P2 2 (2). 4. Following the technique suggested in the above problems, factor x mod 3, and conjecture that the prime factorization of (3) in the ring of algebraic integers of Q( 5) is (3) = P 3 P 3 for appropriate P 3 and P With P 3 and P 3 as found in Problem 4, verify that (3) = P 3 P 3.

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