IDEAL FACTORIZATION. Theorem 1.2. For elements α and β in a commutative ring, α β as elements if and only if (α) (β) as ideals.

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1 IDEAL FACTORIZATION KEITH CONRAD 1. Introducton We wll prove here the fundamental theorem of deal theory n number felds: every nonzero proper deal n the ntegers of a number feld admts unque factorzaton nto a product of nonzero prme deals. Then we wll explore how far the technques can be generalzed to other domans. Defnton 1.1. For deals a and b n a commutatve rng, wrte a b f b = ac for an deal c. Theorem 1.2. For elements α and β n a commutatve rng, α β as elements f and only f (α) (β) as deals. Proof. If α β then β = αγ for some γ n the rng, so (β) = (αγ) = (α)(γ). Thus (α) (β) as deals. Conversely, f (α) (β), wrte (β) = (α)c for an deal c. Snce (α)c = αc = {αc : c c} and β (β), β = αc for some c c. Thus α β n the rng. Theorem 1.2 says that passng from elements to the prncpal deals they generate does not change dvsblty relatons. However, rreducblty can change. Example 1.3. In Z[ 5], 2 s rreducble as an element but the prncpal deal (2) factors nontrvally: (2) = (2, 1 + 5)(2, 1 5). To see that nether of the deals (2, 1 + 5) and (2, 1 5) s the unt deal, we gve two arguments. Suppose (2, 1 + 5) = (1). Then we can wrte 1 = 2(a + b 5) + (1 + 5)(c + d 5) for some ntegers a, b, c, and d. Collectng real and magnary parts, 1 = 2a + c 5d, 0 = 2b + c + d. Solvng for d n the second equaton and substtutng that nto the frst, 1 = 2a + 10b + 6c, whch s mpossble snce the rght sde s even. The proof that (2, 1 5) (1) s smlar. For another proof, complex conjugaton s an operaton on deals, a a := {α : α a} that respects addton and multplcaton of deals, and (α, β) = (α, β). In partcular, the conjugate of (2, 1 + 5) s (2, 1 5), so f (2, 1 + 5) = (1) then (2, 1 5) = (1), so the product (2, 1 + 5)(2, 1 5) = (2) s (1)(1) = (1). But (2) (1) snce 2 s not a unt n Z[ 5]. If a b then for some deal c we have b = ac a, so a b. Dvsblty mples contanment. The converse may fal n some rngs (see Example 8.2), but n the rng of ntegers of a number feld t wll turn out that contanment mples dvsblty. So t s useful to thnk about contanment of deals n any rng as a prelmnary form of dvsblty: 1

2 2 KEITH CONRAD a b s somethng lke a b. Consder n ths lght the followng result about contanment n prme deals. Theorem 1.4. In any commutatve rng A, an deal p s prme f and only f for all deals a and b n A, p ab p a or p b. Proof. Frst suppose p s a prme deal. If p ab and p a, pck x a wth x p. For every y b, xy ab p, so by prmalty of p we get x p or y p. Snce x p, y p. Ths holds for all y b, so b p,.e., p b. Now suppose p s an deal that, for any two deals a and b, contans a or b f t contans ab. If x, y A and xy p, then (x)(y) = (xy) p, so (x) or (y) s n p. Thus x or y s n p, so p s prme. Corollary 1.5. Let K be a number feld. In O K, f p p 1 p r where all the deals are nonzero and prme, then p = p for some. Proof. By Theorem 1.4, p p for some. Snce nonzero prme deals n O K are maxmal, p = p. To prove the nonzero proper deals of O K have unque prme factorzaton, we wll frst show how to nvert a nonzero prme deal. The nverse wll be an O K -module that s n K but not n O K (e.g., n Q the nverse of 2Z s (1/2)Z), so n Secton 2 we wll ntroduce the knds of O K -modules we need. That nonzero prme deals have nverses wll be proved n Secton 3, and after collectng a few corollares of ths nvertblty we wll obtan unque factorzaton of deals n O K. Consequences of unque factorzaton (or, n some cases, consequences of the results used to prove unque factorzaton) are dscussed n Secton 4. In Secton 5, the methods for O K are extended to the ntegral closure of F [T ] n a fnte extenson of F (T ) (the functon feld case). Another approach to unque factorzaton s dscussed n Secton 6, whch s ndependent of the other sectons. In Secton 7 we generalze the norm operaton from elements of O K to deals. Fnally, n Secton 8 we dscuss how to extend unque factorzaton of deals to non-maxmal orders. 2. Fractonal Ideals To nvert deals, we ntroduce O K -modules that are deals wth a denomnator. Defnton 2.1. A fractonal deal n K s a nonzero O K -module I K such that for some d O K {0}, di O K. Such a d s called a common denomnator for I. Snce di s an O K -module n O K, di s an deal of O K. Lettng a denote di, we have I = 1 d a. Conversely, f a O K s a nonzero deal and d O K {0} then 1 d a s an O K-module n K wth common denomnator d, so 1 da s a fractonal deal. Example 2.2. Snce Z s a PID, the fractonal deals n Q are subgroups (Z-submodules) of Q havng the form rz for r Q. Examples nclude 1 2 Z and 6 5 Z = Z. Theorem 2.3. The followng propertes of an O K -module I K are equvalent: (1) I s a fractonal deal: for some d O K {0}, di O K, (2) di O K for some d Z {0}, (3) I = xa for some x K and some nonzero deal a n O K, (4) I s a nonzero fntely generated O K -module n K.

3 IDEAL FACTORIZATION 3 Proof. (1) (2): Let I be a fractonal deal and di O K. Any nonzero O K -multple of d s also a common denomnator for I. In partcular, snce d N K/Q (d) n O K we can use N K/Q (d) Z {0} as a common denomnator for I. (2) (3): Snce di s an O K -module n O K, di s an deal. Take a = di and x = 1/d to see that I = xa. (3) (4): Any deal a n O K s fntely generated as a Z-module, and thus fntely generated as an O K -module. Its scalar multple xa s also fntely generated over O K. (4) (1): Wrte I = O K x O K x d where the x s are n K and at least one s nonzero. We want to fnd a d O K {0} such that di O K. Snce O K has fracton feld K, for each x there s a d O K {0} such that d x O K. Let d be the product of the d s, so dx O K for all. Thus di O K. Corollary 2.4. Any fractonal deal n K s a free Z-module of rank [K : Q]. Proof. Ths s true for nonzero deals n O K and a fractonal deal s just a nonzero scalar multple of such an deal. Not all O K -submodules of K are fractonal deals. For nstance, n Q the subgroup Z[1/2] s not a fractonal deal: t admts no common denomnator. The fractonal deals of K are the (nonzero) fntely generated O K -submodules of K. By comparson wth fractonal deals, nonzero deals n O K are called ntegral deals. (Thnk of the terms nteger and fracton.) A fractonal deal of the form xo K for x K s called prncpal. Wrtng x = α/m for α O K and m Z {0}, xo K = 1 m αo K, so a prncpal fractonal deal s the same thng as a nonzero prncpal deal n O K dvded by an nteger. When O K s a PID, all fractonal deals n K are prncpal (and conversely). The O K -modules n K can be added and multpled, wth multplcaton beng commutatve, assocatve, dstrbutng over addton, and havng multplcatve dentty (1) = O K. The sum and product of deals n O K are deals, so the sum and product of fractonal deals are fractonal deals. Defnton 2.5. For a fractonal deal I n O K, set Ĩ = {x K : xi O K }. Ths s more than the common denomnators of I, snce we allow x K and not just x O K {0}. Any common denomnator of I s n Ĩ, so Ĩ {0}. The set Ĩ s an O K- module. In fact, Ĩ s a fractonal deal. To see ths, pck a nonzero y I. Then yĩ O K, so Ĩ (1/y)O K. Therefore Ĩ s a submodule of a fnte free Z-module, so Ĩ s a fntely generated Z-module, hence fntely generated as an O K -module too. Thus Ĩ s a fractonal deal by Theorem 2.3(4). Example 2.6. If a = αo K s a prncpal deal n O K wth α 0 then ã = {x K : xαo K O K } = (1/α)O K and aã = O K = (1). In partcular, ÕK = O K. The mportance of Ĩ s that t s the only possble canddate for a multplcatve nverse to I among the fractonal deals n K: Theorem 2.7. Let I be a fractonal deal n the number feld K. If I admts a fractonal deal nverse then the nverse must be Ĩ. Proof. Suppose there s a fractonal deal J such that IJ = O K. For any y J, yi JI = IJ = O K, so J Ĩ. Multplyng ths ncluson by I, O K IĨ. At the same tme, for any x Ĩ we have xi O K, so ĨI O K.

4 4 KEITH CONRAD Hence IĨ = O K and J = JO K = JIĨ = O KĨ = Ĩ. We can use Theorem 2.7 n two ways: to show I s not nvertble we only have to check II (1), and f we fnd IJ = (1) for some J then we mmedately know J = Ĩ (even f J was not orgnally defned n that way). Example 2.8. If I and J are fractonal deals wth IJ = (x) a prncpal fractonal deal, then 1 x J s a multplcatve nverse for I, so Ĩ = 1 x J. For nstance, n Z[ 5] let I = (3, 1 + 5) and J = (3, 1 5). Check that IJ = (3). Therefore Ĩ = 1 3 J = 1 3 (3, 1 5) = Z[ 5] Z[ 5]. 3 Check that f I 1 I 2 are fractonal deals then Ĩ2 Ĩ1. As a specal case of ths reversed ncluson, f a s a nonzero deal n O K then O K ã. Remark 2.9. In Theorem 2.7, all we used about I s that t s an O K -module nsde of K; the fractonal deal hypothess (.e., I s a fntely generated O K -module) was not explctly nvoked n the proof and the defnton of Ĩ makes sense for all O K-submodules I K. However, the scope of valdty of Theorem 2.7 s not made broader by ths observaton, because an nvertble O K -module n K must be fntely generated: f IĨ = O K then 1 = x 1 y x r y r for some r 1, x I, and y Ĩ. Then for any x I we have x = x 1 = (xy 1 )x (xy r )xr O K x O K x r, so I O K x. The reverse ncluson s mmedate snce I s an O K -module, so I = O K x s a fntely generated O K -module n K and thus s a fractonal deal (Theorem 2.3). 3. Inverses of Prme Ideals Prme factorzaton n Z can be proved by contradcton: f some nteger greater than 1 has no prme factorzaton then let n > 1 be mnmal wthout a prme factorzaton. Of course n s not prme, so n = ab wth a, b > 1. Then a, b < n, so a and b are products of prmes. Hence n = ab s a product of prmes, whch s a contradcton. Unqueness of the prme factorzaton requres more work. We wll use the same dea (contradcton from a mnmal counterexample) to prove nonzero proper deals n O K have prme deal factorzaton. The basc method s to nduct on the ndex [O K : a] = #(O K /a). Lemma 3.1. Every nonzero deal n O K contans a product of nonzero prme deals. Returnng to the ntutve dea that contanment s a prelmnary knd of dvsblty, the dea of ths lemma s somethng lke any nonzero deal dvdes a product of prmes. Proof. Every nonzero deal of O K has fnte ndex. Assume the lemma s false and let a be a nonzero deal of least ndex whch does not contan a product of nonzero prme deals. Then a O K snce O K contans nonzero prme deals, so [O K : a] 2. Snce a can t be a prme deal, there must be x and y n O K that are not n a but xy a. Then the deals (x) + a and (y) + a both properly contan a, so they have smaller ndex n O K than a does. Thus (x) + a and (y) + a contan products of nonzero prme deals: p 1 p r (x) + a, for some nonzero prmes p and q j, so q 1 q s (y) + a (3.1) p 1 p r q 1 q s ((x) + a)((y) + a) = (xy) + xa + ya + a 2 a,

5 IDEAL FACTORIZATION 5 where we used xy a for the last step. By (3.1), a contans a product of nonzero prme deals. Ths s a contradcton. Notce ths proof s close to the proof that every nteger > 1 s a product of prmes. Why ddn t we prove n Lemma 3.1 that every nonzero proper deal equals (rather than merely contans) a product of nonzero prme deals? Because we do not know (yet) that every non-prme deal n O K s a product of deals wth smaller ndex. The followng theorem s the key techncal property of nonzero prme deals n O K. Theorem 3.2. For each nonzero prme deal p of O K, the fractonal deal p defned n Defnton 2.5 satsfes the followng propertes: (1) O K p and the contanment s strct, (2) p p = O K. In partcular, every nonzero prme deal n O K s nvertble as a fractonal deal. Proof. Snce p O K, O K p. We need to fnd an element of p that s not n O K. The element wll arse as a rato y/x wth arbtrary nonzero x p and a carefully chosen y O K. (For example, n the rng Z use p = 2Z, so p = (1/2)Z. An element of p not n Z s n/2 for odd n. We can wrte n/2 = (nm)/(2m) for any nonzero m n Z, and 2m s an arbtrary nonzero element of p, but nm s not arbtrary.) Pck any x p wth x 0. Then p (x). From Lemma 3.1, (x) p 1 p 2 p r for some nonzero prmes p. Use such a product where r s mnmal. If r = 1 then p (x) p 1, so p = p 1 snce both deals are maxmal. Thus p = (x), so p = (1/x)O K O K, whch s what we wanted (wth y = 1). Thus we may suppose r 2. Snce p (x) p 1 p r, p = p for some by Corollary 1.5. Wthout loss of generalty, p = p 1. Then (x) p p 2 p r. By the mnmalty of r, (x) does not contan the product p 2 p r. So there s a y p 2 p r wth y (x). Thus y/x O K. Snce yp pp 2 p r (x) = xo K, (y/x)p O K. Ths shows y/x p, and we already saw that y/x O K. Thus we have settled the frst part. Now we wll show why the second part (whch s more nterestng) follows from the frst part. Pckng x p wth x O K, we have xp O K, so p p + xp O K. Snce p s maxmal, p + xp = p or p + xp = O K. The second opton says p(o K + xo K ) = O K, so O K + xo K s a fractonal deal nverse for p and must be p by Theorem 2.7. We wll show ths opton holds by elmnatng the frst opton by contradcton. Assume p + xp = p, or equvalently xp p. That means multplcaton by x preserves the fntely generated Z-module p. Recall the lnear characterzaton for an element of K to be an algebrac nteger: t les n a subrng of K that s a fntely generated Z-module. The subrng aspect s needed only to be sure that multplcaton by the partcular element preserves the subrng. But the condton xp p has exactly the same feature, even though x need not le n p (and p s not a subrng of K but an deal of O K ). Therefore snce p s a fntely generated Z-module, the lnear characterzaton of ntegralty can be appled to see from xp p that x s ntegral over Z. Hence x O K. But x O K by ts very defnton, so we are done. Corollary 3.3. For any nonzero deals a and b n O K and nonzero prme p, pa = pb f and only f a = b.

6 6 KEITH CONRAD Proof. If a = b then of course pa = pb. Conversely, f pa = pb, multply both sdes by p. Snce O K -module multplcaton s assocatve, the p-terms cancel because pp = O K. Corollary 3.4. For any nonzero deal a of O K and nonzero prme p, p a f and only f p a as deals. Proof. If p a, obvously p a. Conversely, suppose p a. If we are gong to be able to wrte pb = a for some deal b then t must be the case that b = pa. Thus, defne b = pa. Ths s a nonzero O K -module n K. Snce p a, multplyng by p shows O K pa = b, so b s an O K -module n O K,.e., b s an deal n O K. Easly pb = a, so p a as deals. Corollary 3.5. For any nonzero deal b and nonzero prme p, pb b wth strct ncluson. Ths means multplcaton by a prme deal shrnks the deal. Proof. Easly pb b. If pb = b then b = p(pb) = p 2 b, and smlarly b = p k b for all k 1. Therefore b p k for all k 1, so O K p p 2 p 3 p k b. The ndex [O K : b] s fnte, so the descendng chan of powers of p must stablze: p k+1 = p k for some k. Then cancellng k factors of p usng Corollary 3.3 mples p = O K, whch s a contradcton. Thus pb b. Now we are (fnally) ready to prove unque prme factorzaton of deals n O K. All the hard work s over! Theorem 3.6. Every nonzero proper deal of O K s unquely a product of nonzero prme deals n O K. Proof. Exstence: We wll prove by nducton on r 1 that f a nonzero proper deal a O K contans a product of r nonzero prme deals then t equals a product of nonzero prme deals. (Every nonzero proper deal fts ths condton for some r snce, by Lemma 3.1, every nonzero deal n O K contans a product of nonzero prme deals.) When r = 1, a p for a nonzero prme p, so p s maxmal and a = p. Assumng the result for r, suppose a p 1 p r+1. Snce a s a proper deal, a p for some maxmal deal p. Then p p 1 p r+1, so p = p for some. All nonzero prme deals n O K are nvertble, so multplyng through the ncluson p a p 1 p r+1 by p 1 gves O K p 1 a p 1 p p r+1. (The hat means omt ths term. ) Therefore by nducton, p 1 a s a product of nonzero prme deals, hence a s a product of nonzero prme deals by multplyng back by p. Unqueness: Suppose p 1 p r = q 1 q s wth nonzero prmes p and q j. To show r = s and p = q after rendexng, we can cancel any common prme deals on both sdes (Corollary 3.3) and thus may suppose p q j for all and j. Snce p 1 p 1 p r = q 1 q s, p 1 s some q j by Corollary 1.5. Ths s a contradcton. Now we have a smpler proof of Corollary 3.5: f pb = b then p = (1) by wrtng b as a product of prmes and cancellng common prme factors on both sdes. 4. Consequences Usng the exstence of prme deal factorzatons and the nvertblty of nonzero prme deals, we can extend some propertes of nonzero prme deals to other (ntegral or fractonal) deals.

7 IDEAL FACTORIZATION 7 Theorem 4.1. For any fractonal deal I n K, Ĩ as n Defnton 2.5 s an nverse for I: IĨ = O K. Proof. If I = O K then Ĩ = O K. Now suppose I = a s a nonzero proper deal of O K. Then we can wrte a = p 1 p r for some nonzero prme deals p and we already know p p = O K. For x K, Thus x ã xa O K (x)p 1 p 2 p r O K (x)p 2 p r p 1 (multply both sdes by p 1 ). (x) p 1 p 2 p r x p 1 p 2 p r. ã = p 1 p 2 p r. Snce p p = O K for any nonzero prme deal p, we obtan aã = O K. If I s a fractonal deal whch s not an deal of O K, let d be a common denomnator of I: di O K. Denote di as a, so I = (1/d)a. Then Ĩ = {x K : xi O K} = {x K : (x/d)a O K } = dã, so IĨ = Idã = (1/d)adã = aã = O K. From now on we wrte Ĩ as I 1. That s, I 1 = {x K : xi O K }. We defne negatve powers of I as postve powers of I 1 n the obvous way and the usual rules of exponents (such as I n 1 I n 2 = I n 1+n 2 ) apply wth arbtrary ntegral exponents. Corollary 4.2. The fractonal deals n K form a commutatve group under multplcaton that s freely generated by the nonzero prme deals n O K. Proof. Multplcaton of O K -modules n K s commutatve and assocatve, O K = (1) s the dentty, and the product of two fractonal deals s a fractonal deal (a product of fntely generated O K -modules s fntely generated). We have just seen that fractonal deals have fractonal deal nverses, so the fractonal deals are a multplcatve group. To show the nonzero prme deals generate the group, let I be a fractonal deal. For some d O K {0}, di O K, so di s a nonzero deal n O K. Factor ths nto prme deals: di = p 1 p r. Snce di = (d)i, we factor (d) nto prme deals, say as (d) = q 1 q s, and then obtan I = (d) 1 (d)i = q 1 1 q 1 s p 1 p r. Unque factorzaton says the nonzero prme deals admt no nontrval multplcatve relatons, so they generate the group of fractonal deals freely. Any fractonal deal can be wrtten as a product p a 1 1 par r wth a Z, and by unque factorzaton ths s an ntegral deal (that s, an deal n O K ) f and only f every exponent a s nonnegatve. Now we can extend Corollares 3.3, 3.4, and 3.5 to non-prme deals n O K. The results are collected nto one theorem.

8 8 KEITH CONRAD Theorem 4.3. Let a be a nonzero deal n O K. (1) For any nonzero deals b and c n O K, ab = ac f and only f b = c. (2) For any nonzero deal b n O K, a b f and only f a b as deals. (3) If a (1) then for any nonzero deal b n O K, ab b wth strct ncluson. Proof. For (1), the drecton ( ) s trval, and multplcaton by a 1 settles ( ). For (2), ( ) s trval. For ( ), let c = a 1 b O K, so c s an deal of O K and ac = b. For (3), easly ab b. If ab = b then a = (1) by cancellng b. Corollary 4.4. Every nonzero deal n O K has a prncpal deal multple. Proof. Let a be a nonzero deal n O K. Pck α 0 n a. Then (α) a, so a (α) by Theorem 4.3. Ths says (α) s a multple of a. We can speak of the greatest common dvsor and least common multple of two nonzero deals n O K : gcd(a, b) s a common deal dvsor that all other common deal dvsors dvde, and lcm(a, b) s a common multple that dvdes all other common multples. Corollary 4.5. For nonzero deals a and b n O K, wrte a = pm and b = pn wth the p s beng dstnct nonzero prme deals. Then gcd(a, b) = a + b = pmn(m,n ) and lcm(a, b) = a b = pmax(m,n ). In partcular, (a + b)(a b) = ab. Proof. Snce Theorem 4.3 says dvsblty among ntegral deals s the same as contanment, the greatest common dvsor of two ntegral deals s the smallest deal contanng both of them, whch s ther sum. The least common multple of two ntegral deals s the largest deal contaned nsde both deals, whch s ther ntersecton. The prme factorzaton formulas wth exponents mn(m, n ) and max(m, n ) also ft the condtons to be the greatest common dvsor and least common multple usng unque prme deal factorzaton. Snce max(m, n ) + mn(m, n ) = m + n, (a + b)(a b) = ab because the exponent of any prme deal on each sde s the same. The formulas gcd(a, b) = a + b and lcm(a, b) = a b permt the noton of gcd and lcm to extend to the zero deal: gcd(a, (0)) := a and lcm(a, (0)) := (0). We wll have no use for ths. Defnton 4.6. When a and b are deals n O K wth a + b = (1), we say a and b are relatvely prme. In O K, the condtons a + b = (1) and a b = ab are equvalent, both expressng the fact that a and b have no common deal factors except the unt deal. In a general commutatve rng, these condtons are not equvalent (e.g., n Z[T ] we have (2) + (T ) = (2, T ) (1) but (2) (T ) = (2T ) = (2)(T )). However, the condton a + b = (1) does mply a b = ab n any commutatve rng. Indeed, f a + b = (1) then a b = (a b)(1) = (a b)(a + b) = (a b)a + (a b)b ba + ab = ab, and the reverse ncluson ab a b s trval, so a b = ab. Theorem 4.7. Let A be a commutatve rng. If deals a and b n A satsfy a + b = (1) then A/ab = A/a A/b as rngs. Proof. Snce a + b = (1), a b = ab by the computaton precedng the theorem. Let f : A A/a A/b by f(x) = (x mod a, x mod b). Ths s a rng homomorphsm. The kernel s a b = ab, so f nduces an njectve map f : A/ab A/a A/b. To show f

9 IDEAL FACTORIZATION 9 s onto, wrte 1 = α + β for α a and β b. Then, for any par (c 1, c 2 ) A 2, the element x = c 2 α + c 1 β n A satsfes x c 1 mod a and x c 2 mod b. Remark 4.8. Theorem 4.7 s called the Chnese remander theorem. It extends by nducton from two deals whose sum s (1) to any fnte set of deals that parwse add to the unt deal. In terms of solvng smultaneous congruences, f a 1,..., a r are deals wth a + a j = (1) for j, and c 1,..., c r are n the rng, the congruences x c 1 mod a 1,..., x c r mod a r admt a common soluton (unquely modulo a 1 a r ). The followng corollary apples ths dea to strengthen Corollary 4.4. Corollary 4.9. For any nonzero deals a and b n O K, there s an deal c such that ac s prncpal and c s relatvely prme to b. Ths becomes Corollary 4.4 when b = (1): all deals n O K are relatvely prme to (1). Proof. Wrte a = p e 1 1 per r and b = p f 1 1 pfr r, wth e, f 0. (The exponents f wll play no role n what follows. What really matters s that the p s denote the prmes showng up n ether a or b; only ther multplctes n a wll matter.) By Corollary 3.5, p e +1 s a proper subset of p e. Pck y p e p e+1, so y 0 mod p e and y 0 mod p e +1. (If e = 0 ths just means y O K p.) Use Theorem 4.7 to choose x y mod p e +1 for all. Then for all, x 0 mod p e and x 0 mod p e +1, so p e dvdes (x) and p e +1 does not dvde (x). Thus (x) has the same prme-deal dvsblty as a at each p, so (x) and ab have greatest common dvsor a. Wrtng (x) = ac, the deal c s not dvsble by any p snce (x) and a are dvsble by the same power of each p. So c s relatvely prme to b. Remark Let s solate a result we obtaned n the proof: gven any set of prme deal powers p e n O K (the p s are dstnct) there s an x O K such that (x) s dvsble by p e but not p e +1 for all. Ths says we can construct a prncpal deal (x) dvsble by a preselected (fnte) set of prmes to preselected multplctes, ncludng multplcty 0. However, n ths constructon we have no control over whch other prme deals may occur n (x). Usually (x) wll be dvsble by prmes besdes the p s. Theorem The rng O K s a unque factorzaton doman f and only f t s a prncpal deal doman. Proof. It s a general theorem of algebra that any PID s a UFD. We now show that f O K s a UFD then t s a PID. It suffces to show when O K s a UFD that every prme deal s prncpal, snce any nonzero deal s a product of prme deals. For any rreducble π n a UFD, the deal (π) s prme (check!). Now we wll show f O K s a UFD that every nonzero prme deal p n O K s (π) for some rreducble π. Pck α p wth α 0. Snce (α) p, p dvdes (α). By hypothess, α has an rreducble factorzaton n O K. Wrte α = π 1 π r, where the π s are rreducble n O K. Then (α) = (π 1 ) (π r ). Each of the deals (π ) s prme, and p (α), so by unque prme deal factorzaton n O K we must have p = (π ) for some. Thus all prme deals are prncpal, so O K s a PID. For general domans, a UFD need not be a PID: Z[T ] s an example.

10 10 KEITH CONRAD 5. Analogues n F [X] Many propertes of Z can be carred over to F [X], where F s a feld. Both Z and F [X] have dvson wth remander, and thus are PIDs. The table below ndcates some further smlartes. In partcular, the unts of F [X] are the nonzero constants F, so whle Z has only fntely many unts, F [X] has fntely many unts only when F s a fnte feld. Ths s a small ndcaton that analoges between Z and F [X] are strongest when F s fnte, but here we allow any F (such as Q or R, not just fnte felds). Z F [X] Prme Irreducble ±1 F Postve Monc Q F (X) We want to adapt the methods from number felds to the functon feld case: f K s a fnte extenson of F (X), does the ntegral closure of F [X] n K have unque factorzaton of deals? Example 5.1. In the feld C(X, X 3 1), the ntegral closure of C[X] s C[X, X 3 1]. A key dea runnng through the proofs n the prevous two sectons was nducton on the ndex of nonzero deals n a rng of ntegers. We can t drectly use ths dea for the ntegral closure of F [X], snce deals n F [X] don t have fnte ndex f F s an nfnte feld. For example, representatves n Q[X]/(X 3 2) are a + bx + cx 2 wth ratonal a, b, c, and there are nfntely many of these. However, there s somethng fnte about ths example: t s fnte-dmensonal over Q wth dmenson 3. More generally, f f(x) has degree d 0 n F [X] then F [X]/(f) has dmenson d as an F -vector space (wth bass 1, X, X 2,..., X d 1 ). So f we count dmenson over F rather than count ndex n F [X], then F [X]/(f) has a fnteness property we can take advantage of. 1 Let K/F (X) be a fnte separable extenson of degree n and let A be the ntegral closure of F [X] n K. Ths s an analogue of the rng of ntegers of a number feld. For any α A, Tr K/F (X) (α) and N K/F (X) (α) are n F [X]. More generally, the characterstc polynomal of α s n F [X][T ]. Theorem 5.2. Wth notaton as above, A s a fnte free F [X]-module of rank n, any nonzero deal a n A s a fnte free F [X]-module of rank n, and A/a s fnte-dmensonal over F. Proof. The proof that a rng of ntegers s a fnte free Z-module uses the nonvanshng of dscrmnants and the fact that Z s a PID. Because K/F (X) s separable, the dscrmnant of any F (X)-bass of K s nonzero. Snce F [X], lke Z, s a PID, the proof that a rng of ntegers s a fnte free Z-module carres over to show A s a fnte free F [X]-module. Specfcally, there s an F (X)-bass of K that s nsde A, say e 1,..., e n, and then n =1 F [X]e A n =1 F [X]e /d where d = dsc K/F (X) (e 1,..., e n ) F [X]. Snce A s n between two free F [X]-modules of rank n, t too s free of rank n. Now let a be a nonzero deal n A. Pck α a wth α 0, so (α) a. The deal (α) s fnte free as an F [X]-module wth rank n: lettng γ 1,..., γ n be an F [X]-bass of A, 1 Don t confuse F [X] wth Z[X]. Quotent rngs Z[X]/(f) wth f 0 need not be fnte free Z-modules: Z[X]/(2X 1) = Z[1/2] has no Z-bass.

11 IDEAL FACTORIZATION 11 αγ 1,..., αγ n s an F [X]-bass of αa. Therefore a les between two fnte free F [X]-modules of rank n, so t also s fnte free of rank n as an F [X]-module. From the structure of fntely generated modules over a PID, there s an F [X]-bass y 1,..., y n of A and nonzero f 1,..., f n n F [X] such that f 1 y 1,..., f n y n s an F [X]-bass of a, so n n A/a = ( F [X]y )/( F [X]f y ) n = (F [X]/(f ))y. =1 =1 Each F [X]/(f ) has fnte dmenson over F and there are fntely many of these, so A/a s fnte-dmensonal over F. Corollary 5.3. Every nonzero prme deal n A s a maxmal deal. Proof. For any nonzero prme deal p of A, A/p s a doman that s fnte-dmensonal over F. A doman that s fnte-dmensonal over a feld s tself a feld, so p s maxmal. Defne a fractonal A-deal I to be a nonzero A-module n K wth a common denomnator: ai A for some a A {0}. All the theorems n Secton 2 carry over to fractonal A-deals n K. For nstance, fractonal A-deals are precsely the nonzero fntely generated A- modules n K and each s a free F [X]-module of rank n = [K : F (X)]. The proof of Lemma 3.1 carres over to show every nonzero deal n A contans a product of prme deals: n the nductve step, nstead of sayng one deal contanng another has smaller ndex as n the number feld case we now say that f a a A wth a a then dm F (A/a ) < dm F (A/a), so the F -dmenson (rather than the ndex) s smaller. The proofs of Corollares 3.3 and 3.4 go through for A wth no change at all. In the proof of Corollary 3.5, replace the fnteness of the ndex of the deal b n the rng of ntegers wth the fnteness of dm F (A/b). Fnally, Theorem 3.6 about unque factorzaton of deals carres over to A wth no new deas requred. Every result n Secton 4 apples to A by the same proofs (e.g., n A contanment of deals s the same as dvsblty). The only change needed s n the analogue of Theorem 4.11, where the proof that every nonzero nonunt α A s a product of rreducble elements n A should proceed by nducton on deg(n K/F (X) (α)). Example 5.4. The rng C[X, X 3 1] has unque factorzaton of deals. There s one mnor blemsh n our treatment here: t was assumed from the outset that the fnte extenson K/F (X) s separable. The reason for ths assumpton s that wthout t dscrmnants vansh (snce the trace map Tr K/F (X) s dentcally 0), so our proof that the ntegral closure of A n F [X] s a free F [X]-module breaks down. Ths doesn t mean A can t have unque factorzaton of deals, but our method of proof certanly doesn t work anymore. Is there a counterexample to unque factorzaton of deals f K/F (X) s nseparable? We wll fnd out n the next secton. =1 6. Unque factorzaton of deals by commutatve algebra Our proof of unque factorzaton of deals reled on the fnteness of O K /a n the number feld case and the fnteness of dm F (A/a) n the functon feld case. Usng some concepts from commutatve algebra allows for another approach to unque factorzaton of deals that s applcable more broadly. Proofs n ths secton are only sketched.

12 12 KEITH CONRAD Defnton 6.1. Let A be a commutatve rng. An A-module s called Noetheran f all of ts submodules are fntely generated. We call A a Noetheran rng f t s Noetheran as an A-module,.e., all the deals of A are fntely generated. The Noetheran property for modules s preserved under many constructons. For nstance, submodules and quotent modules of a Noetheran module are Noetheran, and any fnte drect sum of Noetheran modules s Noetheran. (These are derved from the followng general fact: f M s an A-module and N M s a submodule, then M s a Noetheran A-module f and only f N and M/N are Noetheran A-modules.) Any PID s a Noetheran rng, snce all deals n a PID are sngly generated, but Noetheran rngs are stable under many more rng-theoretc constructons than PIDs. For nstance, f A s Noetheran then so s A[T ] and A/a for any deal a. Thus Z[T 1,..., T n ] and F [T 1,..., T n ] for any feld F are Noetheran rngs for any n 1: all deals n these rngs are fntely generated. By comparson, f A s a PID then A[T ] s never a PID (unless A s a feld), although A[T ] s Noetheran. In any number feld K, the deals n O K are fntely generated Z-modules, and thus are fntely generated O K -modules. Therefore O K s a Noetheran rng. Smlarly, the ntegral closure of F [X] n a fnte separable extenson of F (X) s a Noetheran rng. Here s the analogue of Lemma 3.1 for Noetheran domans. Lemma 6.2. In a Noetheran doman that s not a feld, any nonzero proper deal contans a product of nonzero prme deals. Proof. See [6, p. 626]. The proof of Lemma 6.2 s lke that of Lemma 3.1, except studyng a counterexample wth least ndex s replaced by Noetheran nducton, whch s a standard technque that wll be found n any textbook whch dscusses Noetheran modules. Noetheran nducton can also be used to prove that n a Noetheran doman every nonzero nonunt can be wrtten as a product of rreducble elements (usually not unquely). Defnton 6.3. An ntegral doman s called a Dedeknd doman f t has the followng propertes: (1) Noetheran, (2) ntegrally closed, (3) all of ts nonzero prme deals are maxmal. The rng of ntegers n any number feld s an example of a Dedeknd doman. So s C[x, y]/(f) for any rreducble polynomal f such that the curve f(x, y) = 0 n C 2 s smooth [7, p. 56]. (Smooth means there s no soluton n C 2 to f(x, y) = 0 where both partal dervatves of f vansh.) The rng Z[X] s not Dedeknd, snce (X) s a nonzero prme deal n Z[X] that s not maxmal. Hstorcally, the rng of ntegers of a number feld and the rng of polynomal functons on a smooth plane curve were the frst mportant examples of Dedeknd domans. Every PID s both a UFD and a Dedeknd doman, and the converse turns out to be true too: a rng that s a UFD and a Dedeknd doman s a PID. Remark 6.4. A feld, strctly speakng, fts the condtons to be a Dedeknd doman: the thrd property holds snce t s vacuous, as there aren t any nonzero prme deals at all. To say UFDs Dedeknd domans = PIDs, we need felds to be Dedeknd domans snce a feld s a PID and a UFD (borng ones, of course). On the other hand, for techncal reasons

13 IDEAL FACTORIZATION 13 related to algebrac geometry t t better to regard felds as not beng Dedeknd domans. That s, a Dedeknd doman should have at least one nonzero prme deal. At the level of our treatment here, whether you want to regard a feld as beng a Dedeknd doman or not has at most trval effects on the valdty of theorems about Dedeknd domans, so we won t stress ths pont. Defnton 6.5. In a doman A, a fractonal A-deal s a nonzero A-module I n the fracton feld of A that admts a common denomnator: di A for some d A {0}. Theorem 6.6. If A s a Dedeknd doman then any nonzero prme deal p A s nvertble as an A-module. The nverse s the A-module p consstng of x n the fracton feld of A such that xp A, and p s a fractonal A-deal. Proof. The proof s essentally the same as that of Theorem 3.2 (usng Lemma 6.2 n place of Lemma 3.1). Ths s left to the reader to check; each property n the defnton of a Dedeknd doman s needed n the proof. (The Noetheran property was already used n Lemma 6.2.) Theorem 6.7. In a Dedeknd doman A, any nonzero proper deal has unque factorzaton as a product of nonzero prme deals. The fractonal A-deals are a commutatve group under A-module multplcaton that s freely generated by the nonzero prme deals of A. Proof. The exstence of a prme deal factorzaton follows from Lemma 6.2 and Theorem 6.6 n the same way Theorem 3.6 follows from Lemma 3.1 and Theorem 3.2. The unqueness of prme deal factorzaton follows from Theorem 6.6 by exactly the same argument we gave earler n the number feld settng. In a Dedeknd doman, f p p 1 p r then p = p for some by the same proof as Corollary 1.5. Theorem 6.8. The ntegral closure of a Dedeknd doman n a fnte extenson of ts fracton feld s also Dedeknd. Proof. See [6, pp ]. Notce we do not need to assume the feld extenson s separable! The proof treats frst the case of a separable extenson (very much lke the number feld settng), then a purely nseparable extenson (usng specal features of characterstc p), and then the general case by wrtng a fnte extenson as a tower of a purely nseparable extenson on top of a separable extenson. Theorem 6.8 wth base rng Z shows O K s Dedeknd for a number feld K and therefore O K has unque factorzaton of deals by Theorem 6.7. Theorem 6.8 s often proved n textbooks only n the case when the feld extenson s separable (or, even more smply, has characterstc 0). What s specal about the separable case s that the trace-parng and dscrmnants are avalable. Moreover, n the case of a separable extenson one can prove a lttle more than Theorem 6.8: the ntegral closure of the ntal Dedeknd doman A s a fntely generated A-module. Ths smplfes the proof of Theorem 6.8. When the extenson s not separable, the ntegral closure of A may not be fntely generated as an A-module [2, exer. 3, p. 461]. We ntroduced Dedeknd domans n Defnton 6.3 by three techncal propertes. These rngs can be characterzed n many other ways, such as the followng. Theorem 6.9. The followng condtons on a doman A are equvalent: (1) A s Dedeknd, (2) every nonzero proper deal n A s a product of prme deals,

14 14 KEITH CONRAD (3) every nonzero proper deal n A s a unque product of prme deals, (4) every nonzero deal n A s nvertble as a fractonal A-deal, (5) every nonzero deal n A s a projectve A-module, (6) A s Noetheran and the localzaton A p s a PID for all nonzero prme deals p. Proof. See [5, pp ]. Remark Another lst of propertes equvalent to beng Dedeknd s: Noetheran, ntegrally closed, and (a + b)(a b) = ab for all deals a and b. I have never found ths characterzaton to be useful. Dedeknd domans are not preserved under some mportant operatons, e.g., f A s Dedeknd then A[T ] s not Dedeknd (unless A s a feld and you consder felds to be Dedeknd domans ths s a case where felds not beng Dedeknd domans makes for a cleaner statement: A Dedeknd mples A[T ] s not Dedeknd, always). There s a class of domans wth unque factorzaton features that encompasses both Dedeknd domans and UFDs as specal cases: Krull rngs. There s a dscusson of Krull rngs n [2, Chap. VII, Sect. 1] and [8, Sect. 12]. (They are also the man object of study n [1, Chap. 3], where they are called rngs wth a theory of dvsors. ) When A s a Krull rng, A[T ] s also a Krull rng. Usng termnology from commutatve algebra, the Dedeknd domans are the 1-dmensonal Krull rngs (ths excludes felds) and the UFDs are the Krull rngs wth trval class group. 7. Ideal Norms We return to the settng of a number feld K and use the fnte ndex of deals to develop analogues for O K of concepts from elementary number theory. Defnton 7.1. For a nonzero deal a n O K, ts (deal) norm s Na = #(O K /a) = [O K : a]. On prncpal deals ths noton of norm s compatble wth the rng-theoretc norm on a generator: Theorem 7.2. For α O K {0}, N((α)) = N K/Q (α). Proof. From the structure of modules over a PID, there s a Z-bass e 1,..., e n of O K and nonzero ntegers a 1,..., a n such that a 1 e 1,..., a n e n s a Z-bass of the deal (α). Then O K /(α) = n =1 (Z/a Z)e, so N((α)) = n =1 a = a 1 a n. Next we compute N K/Q (α) n a clever way and wll get the same value as N((α)), up to sgn. The feld K has three Q-bases: {e }, {a e }, and {αe }. Consder the commutatve dagram of Q-lnear maps K m α K (7.1) e e a e αe K e a e K where the map along the top s multplcaton by α and the maps along the sdes and bottom nterchange Q-bases of K as ndcated. (The map on the left s the dentty.) That the

15 IDEAL FACTORIZATION 15 dagram commutes follows by examnng the effect on each e both ways: on the top t goes to αe and the other way the effect on t s e e a e αe, whch s exactly what m α does to e. So by lnearty the two ways around the dagram are the same. Snce all maps are from K to K we can speak about ther determnants, and the determnant along the top s the product of the other three determnants. The determnant along the top s N K/Q (α), by defnton. The determnant on the left s 1. The matrx representng the lnear map on the bottom s dagonal wth a s on the man dagonal, so ts determnant s a 1 a n. What about the determnant on the rght? It s not clear what a matrx for t s, so how can we compute the determnant? The key pont s that {a 1 e 1,..., a n e n } and {αe 1,..., αe n } are not just Q-bases of K but Z-bases of a common free Z-module (the deal (α)). As bases of a fnte free Z-module, changng from one bass to the other s a lnear map whose determnant s nvertble over Z, hence the determnant s ±1. Puttng t all together, det(m α ) = ±a 1 a n, so N K/Q (α) = a 1 a n = N((α)). Example 7.3. The sze of Z[ 2]/( ) s = 82. See [4, pp ] for a modfed defnton of the deal norm that can take negatve values so a nonzero prncpal deal (α) has deal norm N K/Q (α), even wth the correct sgn. Snce the rng O K /a has sze Na, Na 0 mod a. Contanment mples dvsblty, so a (Na) as deals. Ths s analogous to the elementwse dvsblty relaton α N K/Q (α) for α O K. The followng two theorems are the man propertes of norms of deals n O K. Theorem 7.4. The norm of a nonzero prme deal p s a prme power n Z +. If Np = p f then p (p). Proof. Let p be a nonzero prme deal n O K. The rng O K /p s a fnte feld, so from the theory of fnte felds t has prme power sze. Hence Np s a prme power n Z +. Let p be the characterstc of O K /p, so p = 0 n O K /p. That means p p, so (p) p, so p (p). We call O K /p the resdue feld at p. The number f n Theorem 7.4 can be nterpreted as dm Z/pZ (O K /p) and therefore s called the resdue feld degree. Theorem 7.5. For nonzero deals a and b n O K, N(ab) = Na Nb. Proof. We gve two proofs. Frst proof: Snce every nonzero deal s a product of prme deals t suffces to show N(ap) = Na Np for any nonzero a and nonzero prme p. Snce O K a ap, [a : ap] = N(ap)/ Na and what we want s equvalent to #(a/ap) = #(O K /p). The O K -module a/ap s klled by multplcaton by elements of p, so a/ap s naturally an (O K /p)-vector space. That s the key pont. We wll show a/ap s 1-dmensonal over O K /p, so ts sze s #(O K /p). Snce ap s a proper subset of a, there s an x a wth x ap, so x 0 n a/ap. We expect, f a/ap s to be 1-dmensonal over O K /p, that any nonzero element s a spannng set over O K /p. Let s verfy {x} s such a set. Snce (x) a and (x) ap, a dvdes (x) but ap does not dvde (x) (contanment s equvalent to dvsblty). Ths mples gcd((x), ap) = a, so a = (x) + ap by Corollary 4.5. So for any α a we have α = βx + γ where β O K and γ ap. Thus n a/ap, α = β x, whch shows a/ap s spanned as an (O K /p)-vector space by the sngle nonzero element x: the dmenson of a/ap over O K /p s 1. Hence N(ap) = Na Np.

16 16 KEITH CONRAD Second proof: We wll show a/ab = O K /b as O K -modules, so [a : ab] = Nb, whch mples N(ab) = [O K : ab] = [O K : a][a : ab] = Na Nb. By Corollary 4.9, there s a c prme to b such that ac s prncpal. Let ac = (x). Then (x) + ab = a(c + b) = a(1) = a. Reducng the equaton (x) + ab = a modulo ab shows a/ab s spanned by x as an O K - module. Therefore the map f : O K a/ab gven by α αx = αx s O K -lnear, onto and ts kernel s {α O K : αx ab}. Snce for α n the kernel of f we have αx ab (α)(x) ab (α)ac ab (α)c b, (α) = (α)(1) = (α)(c + b) = (α)c + (α)b b + b = b, so α b. Hence ker f b. The reverse ncluson s easy snce for y b, yx ab because x a. Remark 7.6. In a doman where all nonzero deals have fnte ndex, the multplcatvty of the deal norm (f true) mples the Dedeknd property. See [3]. Corollary 7.7. Let a be an deal n O K and p be a prme number. The deal a s dvsble by a prme factor of (p) f and only f Na s dvsble by p. Proof. Wrte a = p a 1 1 par r wth dstnct p s. Some p s a factor of (p) f and only f Np s a power of p. Snce Na s the product of Np a, p Na f and only f some p has norm a power of p, whch s equvalent to p (p). The next theorem gves a lnk between the norm of an deal and the rng-theoretc norm of the elements n that deal. In partcular, t gves another nterpretaton of Na. Theorem 7.8. Let a be a nonzero deal n O K. The postve nteger Na s the greatest common dvsor of the ntegers N K/Q (α) for all α a. Proof. If a = (1) then the result s clear, so we take a (1). Let d be the gcd of N K/Q (α) as α runs over a. For any nonzero α a, a (α) so Na N((α)), so Na N K/Q (α) by Theorem 7.2. Therefore Na d, so any prme power dvdng Na also dvdes d. To show Na = d we wll show for every prme number p that the hghest power of p dvdng Na s also the hghest power of p dvdng N K/Q (α) for some α a (dependng on p). Snce d N K/Q (α), t follows that the hghest power of p dvdng Na and d agree (why?). Varyng ths over all prmes mples Na = d. In O K, let (p) = p e 1 1 per r wth dstnct prme deals p. Wrte a = p c 1 1 pcr r b where c 0 and b s not dvsble by any p. Snce the p s are all the prme deals dvdng (p), Nb s not dvsble by p. Wrtng Np = p f, Na = p c 1f 1 + +c rf r Nb, so the multplcty (.e., exponent) of p n Na s c 1 f c r f r. We need to fnd α a such that N K/Q (α) has the same p-power dvsblty. By Remark 4.10, there s an α O K {0} such that (α) s dvsble by p c but not p c +1 for all and s also dvsble by b. It follows that (α) s dvsble by a, so α a, and also N((α)) = N K/Q (α) s dvsble by p wth multplcty c f. Snce the deal norm s multplcatve, t extends to a multplcatve functon on all fractonal deals of K n a unque way. If I s a fractonal deal, N(I) Q, so there s not a combnatoral meanng for the norm of I (but see Theorem 7.9). When I s a

17 IDEAL FACTORIZATION 17 prncpal fractonal deal, ts deal norm concdes wth the absolute value of the norm of any generator: N(xO K ) = N K/Q (x) for all x K. Ths s proved by wrtng x = y/z for y and z n O K {0}, so (x) = (y)(z) 1 and now apply the deal norm to both sdes and use Theorem 7.2. Theorem 7.9. For any fractonal deals J I, [I : J] = N(J)/ N(I). Proof. Pck d O K {0} such that di and dj are both n O K. Then dj di O K, and di/dj = I/J as O K -modules, so [I : J] = [di : dj] = [O K : dj] [O K : di] = N(dJ) N((d)) N(J) = N(dI) N((d)) N(I) = N(J) N(I). Example For any nonzero deal a O K, we have O K a 1 so [a 1 : O K ] = N(O K )/ N(a 1 ) = 1/(1/ Na) = Na = [O K : a]. More generally, f J I are fractonal deals then Theorem 7.9 shows [I : J] = [IM : JM] for any fractonal deal M. Usng the deal norm, especally ts multplcatvty, many concepts from elementary number theory for Z can be carred over to O K. For example, snce O K /p s a feld of sze Np, we have an analogue of Fermat s lttle theorem: α 0 mod p = α Np 1 1 mod p. Defne ϕ K (a) = #(O K /a), whch generalzes the Euler functon ϕ(m) = #(Z/mZ). The standard formula ( 1 1 ). p ϕ(m) = m p m for the classcal ϕ-functon has an O K -analogue: ϕ K (a) = Na p a ( 1 1 Np where the product runs over the prme deals dvdng a. The proof s left to the reader. There s an mportant dfference between the groups (Z/mZ) and (O K /a). When m = p k s an odd prme power the group (Z/p k Z) s cyclc for all k 1. However, when a = p k s a prme power deal and k > 1, the group (O K /p k ) s often not cyclc. For example, 3 s prme n Z[] and (Z[]/3 k ) s not cyclc when k > 1. (When k = 1, (O K /p k ) = (O K /p) s cyclc snce ths s the multplcatve group of a fnte feld.) In a dfferent drecton, the Remann zeta-functon ζ(s) = 1 n s = 1 1 1/p s n 1 p generalzes to the zeta-functon of K: ζ K (s) = a 1 Na s = p ), 1 1 1/ Np s, where the seres runs over nonzero deals of O K and the product runs over the nonzero prme deals. The seres and product are absolutely convergent for Re(s) > 1. (Unlke n Z +, there may be no deals of some norm and several deals of some norm 2, so the proof that ζ K (s) 2 In Z[], no deal has norm 3 and the deals (1 + 8), (1 8), (4 + 7), and (4 7) all have norm 65.

18 18 KEITH CONRAD converges on Re(s) > 1 s not smply a recopyng of the proof for the Remann zeta-functon.) The equalty of the seres and product for ζ K (s) comes from unque factorzaton of deals and the multplcatvty of the deal norm. The study of dstrbuton of prme numbers nvolves analytc propertes of ζ(s), and smlarly studyng the dstrbuton of prme deals of O K uses analytc propertes of ζ K (s). There s an analytc contnuaton of ζ K (s) to the whole complex plane except for a smple pole at s = 1, and t s beleved (generalzed Remann hypothess) that the only zeros of ζ K (s) n the vertcal strp 0 < Re(s) < 1 le on the lne Re(s) = 1/2. 8. Analogues for Orders Let K be a number feld and O be an order n K. That s, O s a subrng of K whch s fntely generated as a Z-module and contans a Q-bass of K. Concretely, orders n K are the subrngs of O K wth fnte ndex. A typcal example s Z[α] where α s an algebrac nteger of K such that K = Q(α) and Z + a where a s a nonzero deal n O K. Are the arguments we developed to prove unque factorzaton of deals n O K vald f we apply them to an order O when O O K? To start off, note that orders n K share several features wth O K : (1) any order n K has fracton feld K, (2) there s a Z-bass and t can be chosen to nclude 1, (3) all nonzero deals n an order are fntely generated as Z-modules (even free wth rank n = [K : Q]), (4) all nonzero deals n an order have fnte ndex n the order, (5) all nonzero prme deals n an order are maxmal, (6) for x 0 n an order O, the ndex [O : xo] equals N K/Q (x), (7) for any x n an order O, x N K/Q (x) n Z[x] O. (8) the unts of an order O are {u O : N K/Q (u) = ±1}, Propertes 1, 3, 4, and 5 are what we used about O K n the results up through and ncludng Theorem 3.2(1), so those results all apply to orders and not just to O K. For nstance, defne a fractonal O-deal as a nonzero O-module n K wth a common denomnator from O {0}. Then fractonal O-deals are the same thng as nonzero fntely generated O-modules n K (analogue of Theorem 2.3(4)) and the only possble nverse of a fractonal O-deal I s Ĩ = {x K : xi O}, whch s always a fractonal O-deal (analogue of Theorem 2.7). Any nonzero deal n O contans a product of nonzero prme deals (analogue of Lemma 3.1). When p s a nonzero prme deal n O, O p and ths contanment s strct (analogue of Theorem 3.2(1)). When we try to prove Theorem 3.2(2) wth O n place of O K, we run nto a problem! For x p O, xp + p s ether p or O. If we try to show the frst opton s not true by contradcton, then we want to show from xp p that x O. Snce p s a fntely generated Z-module, when xp p we know x s ntegral over Z, so x O K, but there s no reason to expect x O (the ntegral closure of Z n K s O K, not O), so we can t get a contradcton. The three corollares of Theorem 3.2 for O K are formal consequences of p p = (1): they would apply to any nonzero prme deal p O such that p p = O (usng the same proofs). The proof of unque prme deal factorzaton n O K (Theorem 3.6) would also apply to an order whose nonzero prme deals all satsfy Theorem 3.2(2). Therefore Theorem 3.2(2), the nvertblty of all nonzero prme deals, s really the key step n our proof of unque prme deal factorzaton n O K.

19 IDEAL FACTORIZATION 19 The followng theorem mples that any order n K smaller than the maxmal order O K doesn t satsfy Theorem 3.2(2) for at least one nonzero prme deal. Theorem 8.1. If a doman has cancellaton of deals,.e., always ac = bc mples a = b when c (0), then the doman s ntegrally closed. Proof. Let A be a doman wth cancellaton of deals. Suppose an x n the fracton feld of A s ntegral over A. We want to show x s n A. Wrte x = a/b where a and b are n A wth b 0. Snce x s ntegral over A, x n + c n 1 x n c 1 x + c 0 = 0 wth n 1 and c A. Let R = A[x] = A + Ax + + Ax n 1. Ths s a rng and a nonzero A-module n the fracton feld of A. Snce x has denomnator b, by the defnton of R we have b n 1 R A, so R has common denomnator b n 1. Therefore a := b n 1 R = Ab n 1 + Ab n 2 a + + Aa n 1 s a nonzero A-module n A,.e., a s a nonzero deal n A. Snce R s a rng, R 2 = R, so a 2 = b 2(n 1) R 2 = b n 1 b n 1 R = (b) n 1 a. Therefore by cancellaton of nonzero deals n A, a = (b) n 1 = b n 1 A, so b n 1 R = b n 1 A. Ths mples R = A, so x R = A. A doman wth unque factorzaton of deals wll have cancellaton of deals and thus wll be ntegrally closed. Therefore, snce no order n a number feld s ntegrally closed except for the maxmal order (the full rng of ntegers), non-maxmal orders do not have unque factorzaton of deals: they must contan a non-nvertble prme deal p (0). We gve next a basc example of ths. Example 8.2. Let K be any number feld feld wth degree n > 1 and p be any prme number. Set O = Z + po K. We wll show ths order has an deal wth no prme deal factorzaton and t has an explct nonzero prme deal that s not nvertble. Set b = po = pz + p 2 O K, p = po K = bo K, so b p O. Note p s an deal n both O and O K. We wll show (1) the chan of deals has ndces as ndcated, p 2 p b pn 1 p p O pn 1 O K, (2) p s a prme deal n O that s not nvertble as a fractonal O-deal, (3) bp = p 2, (4) b s not a product of prme deals n O, (5) p does not dvde b. (1): Let {1, e 2,..., e n } be a Z-bass of O K, so O K = Z Ze 2 Ze n, O = Z Zpe 2 Zpe n, po K = Zp Zpe 2 Zpe n, po = Zp Zp 2 e 2 Zp 2 e n, p 2 O K = Zp 2 Zp 2 e 2 Zp 2 e n.