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12 Prime ideals Again, the presentation here is somewhat different than the text. In particular, the sections do not match up. (In fact we have done some of the Chapter 12 material in the text in our Chapter 11 notes.) Our goal is to prove that OK has unique factorization into prime ideals for any number field K. We will use this to determine which primes are of the form x2 + 5y2. 12.1 The ideal factorization theorems We are still missing one main ingredient to prove the existence of factorization into prime ideals. The ingredient we need is the following result. Proposition 12.1. Let K be a number field and I; J be nonzero proper ideals of OK . If J jI and 0 0 I 6= J , then I = JJ for some nonzero proper ideal J of OK . In other words, this notion of J jI for ideals, means I really factors as a product I = JJ 0. This is an immediate consequence of Corollary 12.6 below. Stillwell gives what is perhaps a simpler argument for this in Section 12.5 when K is an imaginary quadratic field. For the moment, let’s take it for granted and see how it implies existence of prime factorization for ideals. Theorem 12.2. (Existence of factorization into prime ideals) Let K be a number field and I be a nonzero proper ideal of OK . Then I = p1 ··· pk where pi’s are prime ideals of OK . Proof. Either I is maximal (and therefore prime) or not. If so we’re done, so suppose not. Then there is some prime ideal p1 which contains I by Prop 11.15 and Cor 11.19. By the above proposition, we can write I = p1J1 for some ideal nonzero proper ideal J1. Further J1 ⊇ I. We repeat this argument so that at each stage we get I = p1p2 ··· pnJn and I ⊆ J1 ⊆ J2 ⊆ · · · ⊆ Jn ⊆ OK . This has to terminate at some point because we can only nest in a finite number of ideals in between I and OK (since OK is finitely generated—we used this also in Prop. 11.15), i.e., at some point we have Jk = pk is prime. Theorem 12.3. (Uniqueness of prime ideal factorization) Let I be a nonzero proper ideal of OK and suppose I = p1p2 ··· pr = q1q2 ··· qs where each pi and qj is a prime ideal of OK . Then r = s and, up to reordering the qj’s, pi = qi for 1 ≤ i ≤ r. Stillwell gives a simple argument for this, but just like his argument for unique prime factorization in Z, it is lacking when some of the prime ideals are repeated in the factorization. For a complete proof, we would like to make the following argument. Suppose p1p2 ··· pr = q1q2 ··· qs for some prime ideals pi, qj. By the prime divisor property, i.e., the definition of prime ideals, p1jq1q2 ··· qs implies p1 divides one of the qj’s, say q1. Since p1 and q1 are maximal (prime), p1 divides (contains) q1 implies p1 = q1. Now we would like to be able to multiply both sides by the −1 “inverse” of p1, some p1 , to conclude p2 ··· pr = q2 ··· qs: 88 Repeating this argument would show r = s and pi = qi for each i, giving uniqueness of prime ideal factorizations. −1 Now if we think about ideals in Z, prime ideals are of the form p = (p). Such a p would have −1 to be something like the ideal generated by p , which is not an element of Z. Nevertheless, we can formally construct such ideals, which will be called fractional ideals, to apply the above argument. Precisely what we need is given below in Corollary 12.7. It turns out that the theory of fractional ideals will also provide the proof of Proposition 12.1. 12.2 Fractional ideals and the class group Definition 12.4. Let K be a number field, and I ⊆ K. If aI = fai : i 2 Ig is an ideal of OK for some nonzero a 2 OK , we say I is a fractional ideal of OK . Further if aI is principal, we say I is principal. Denote the set of nonzero fractional ideals of OK by Frac(OK ), and the set of nonzero principal ideals by Prin(OK ). −1 Put another way, the fractional ideals of OK are just subsets of the form a I where a 2 OK , a 6= 0 (possibly a = 1, so this includes the case of usual ideals). It is clear that fractional ideals still satisfy the two defining properties of ideals: (i) closed under addition, and (ii) closed under multiplication by elements of OK . The only thing they don’t satisfy in the definition of ideals is that they may not be contained in OK . Note on terminology: whenever I say an ideal of OK , I will mean an honest ideal in OK . If I mean fractional ideal, I will always say fractional. However, sometimes I will say ordinary ideal to stress that we are talking about honest ideals contained in OK . 1 m Example. The fractional ideals of Z = OQ are n (m) = k n : k 2 Z where m 2 N [ f0g, n 2 N and gcd(m; n) = 1. To see this, it is clear from the definition, together with the classification of a a ideals of Z, that they are subsets of Q of the form n (m) for some n 2 Q and m 2 N [ f0g, i.e., all an m integer multiples of the rational number m . Renaming am to m, we see get all multiples of n . It is clear we may take gcd(m; n) = 1 and n > 0 to give the claim. Hence the fractional ideals of Z are in 1-to-1 correspondence with Q≥0, the set of nonnegative rational numbers. Precisely, the correspondence is a nonnegative rational number corresponds to the fractional ideal which is all integral multiples of that number. Example. Let K be a number field. Any ideal of OK is a fractional ideal. If a 2 K, then aOK is also a fractional ideal. Example. Let K be a number field. Then K is not a fractional ideal of OK , even though it satisfies the properties of being closed under addition and multiplication by elements of OK , because there is no element a of OK such that aK ⊆ OK . A similar non-example (for the same reason), but perhaps 1 a less trivial, is the ring Z[ 2 ] = 2k : a; k 2 Z . One formally defines the product of fractional ideals in the same way as for ideals: IJ = fi1j1 + i2j2 + ··· ikjk : im 2 I; jn 2 J g. The point is every (non-zero) fractional ideal is invertible. −1 1 Exercise 12.1. Let I = (n) be a non-zero ideal of Z. Check the fractional ideal I = n Z is −1 indeed the inverse of I, i.e., II = (1) = Z. Similarly, for any number field K and any non-zero −1 −1 principal ideal I = (α) of OK , show α OK is the inverse of I, i.e., II = (1) = OK . 89 Exercise 12.2. Let K be a number field. Show the principal fractional ideals of OK correspond to the elements of K, up to units. In fact, the product of principal (fractional) ideals corresponds to the product of elements of K (up to units). In the case K = Q, Frac(OK ) is essentially Q>0 the positive rational numbers, which forms a group under multiplication. More generally we have the following. Theorem 12.5. Frac(OK ) is an abelian group under multiplication, and OK is the identity element. Exercise 12.3. Check that OK is the identity element of Frac(OK ), i.e., if I 2 Frac(OK ), show OK ·I = I. Proof. First note that the product of two fractional ideals I and J is again a fractional ideal. This is because there exist a; b 2 OK such that aI and bJ are ideals of OK . Then abIJ is just the product of two ordinary ideals aI and bJ of OK , and thus itself an ideal of OK . Hence multiplication is a binary operation on Frac(OK ). One formally checks from the definition that multiplication is associative and commutative. This is straightforward and I will not write it down. The exercise above shows OK is a multiplicative identity, so it suffices to show every element −1 has an inverse. Let I be an (ordinary) ideal of OK . Define I = fa 2 K : aI ⊆ OK g. Then it is −1 −1 easy to see that I I ⊆ OK . It takes a little more work to show I I = OK , and in the interest of 1 time I will omit it, though it is nothing too difficult (the proof is about 1–1 2 pages, see any text on Algebraic Number Theory). Roughly, one can use a descent-type argument to reduce the proof to −1 −1 the case where I maximal. It easy to see I 6⊆ OK (exercise below), and therefore I I is strictly −1 larger than I. But then the maximality of I implies I I must be OK . 0 Corollary 12.6. Let I, J be ideals of OK . Then J jI () I = JJ for some (ordinary) ideal 0 J of OK . Proof. Let J 0 be the fractional ideal J 0 = J −1I. Then I = JJ 0, and the corollary then reads 0 0 −1 J jI () J is an ordinary ideal of OK , i.e., if and only if J = J I ⊆ OK .
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