Claborn - Dedekind Domains and Rings of Quotients

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    P AC I F I C JOURNAL OF MATHEMATICSVol. 15, No. 1 1965

    DEDEK IND DOMAINS ANDR INGS OF QUOTIENTSLUTHER CLABORN

    We study the relation of the ideal class group of aDedekind domain Atothat ofAs, where S is a multiplicativelyclosed subset ofA. Weconstruct examples of (a) aDedekinddomain with no principal prime ideal and (b) a Dedekinddomain whch is not theintegral closure of a principal idealdomain. Wealso obtain some qualitative information on thenumber of non-principal prime ideals in an arbitrary Dedekinddomain.If A is a Dadekind domain, S the set of all monic poly-nomials andT the set of all primitive polynomials of A[X],then A[X]

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    60 L U TH E R CLABORNEAa =(P1AS) (PkAsy* =Q i .. .QK

    COROLLARY 1-3. Let A be aDedekind domain and Sbeamulti-plicatively closedset of A. Let C (for C a fractionary ideal of Aor As) denotetheclassof theideal class group to which Cbelongs.Then theassignment C* CAS is a homomorphism of theidealclassgroup of A onto that of As.

    Proof I t is only necessary to note that if C=dA, then CAS=dAs.THEOREM 1-4 Thekernel of is generated by all P , whereParanges over all primes such that P#OS .If Pa S , then P As As. Suppose C is a fractionary idealsuch that C=Pay i.e. C=dP for some d in thequotient fieldofA. Then CAS dPaAs =dAs, andthus CA? is theprincipal class.On the other hand, suppose that C is a fractionary ideal of A

    such that CAS xAs. We may choose x in C. Then C~ xA is anintegral ideal of A, and(C~ xA)As As. In other words, C~1xA =P{i... P{i9 wherePi f)S ,i = 1, ..., . ThenC = P f\ , ~P f,completing theproof.EXAMPLE 1-5. There areDedekind domains with noprime idealsin theprincipal class.Let A be any Dedekind domain which is not a principal ideal

    domain. LetS be themultiplicative set generated by all a, where a ranges over all theprime elements of A. Then by Theorem 1-4,As will have the same class group as A but will have noprincipalprime ideals.

    COROLLARY 1-6 If A is a Dedekind domain which is not aprincipal ideal domain, then A has an infinite number of non-principal prime ideals.Proof. Choose S as in Example 1-5. Then As is not a principalideal domain, hence has an infinite number of prime ideals, none of

    which areprincipal. These are of theform PAS, where P is a (non-principal) prime of A.COROLLARY 1-7 Let A be aDedekind domain with torsion classgroup and let {Pa}be a collectionof primes such that thesubgroupof the ideal class group of A generated by {P0}is not theentire

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    DEDEK I ND DOMAINS AND R I NGS OF QUOTIENTS 61classgroup. Then there are always an infinite number of non-principal primes not in the set {Pa}

    Proof. For each a, chose n such that Pl

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    62 L U T H E R CLABORN2 LEMMA 2-1. Let A be a Dedekind domain. Let S be the

    multiplicatively closed set of A[X] consisting of all monic poly-nomials of A[X]. Let T be the multiplicatively closed set of allprimitive polynomials of A[X] {i.e. all polynomials whose coefficientsgenerate the unit ideal of A). Then A[X]S &nd A\X\T are bothDedekind domains.

    Proof. A[X] is integrally closed and noetherian, and so bothA[X]s and A[X]y are integrally closed and noetherian. Let P be aprime ideal of A[X]. If PnA (0), then P )A = Q is a maximalideal of A. If P QA[X], then passing to A[X]/QA[X], it is easyto see that P= QA[X] +f(X) [X] where f(X) is a suitably chosenmonic polynomial of A[X]. In this case P \ S ,so PA[X]S = A[X]S.Thus if P A (O) and PA[X]^ is a proper prime of A[X]S, thenP=QA[X] where Q = P A. Then height P = height Q= 1. IfP i =(0), then PK[X] is a prime ideal of K[X] (where K denotesthe quotient field of A). Certainly height P = height PK[X] =1, soin any case if a prime P of A[X] is such that PflS = , then heightP ^ 1. This proves that A[X]S is a Dedekind domain. Since S T,A[X]T is also a Dedekind domain by Lemma 1-1.

    REMARK 2-2. A[X]T is customarily denoted by A(X) [1, p. 18].For the remainder of this article, A[X] will be denoted by A1.PROPOSITION 2-3. A1 has the same ideal class group as A. In

    fact, the map C>UAL is a one-to-one map of the ideal class group ofA onto that of A1.

    We can prove that C>C 1 is a one-to-one map of the ideal classof A into that of A by showing that if two integral ideals D and Eof A are not in the same class, neither are DA1 and EA1. Supposethen that DA1 EA1. This implies that there are elements f (X),9i(X), i = 12 in A[X] with (X) monic for i = 12 such that

    DA1-g(X)

    Let a{ be the leading coefficient of f{X) for i = 1,2, and let deD.Then we get a relationd. WL = jl(XL . MXL, g{X) mOnic,(X) g(X) {X) yX }

    where e{X) can be chosen as a polynomial in A[X] all of whose coef-ficients are in E. This leads to d g2(X)'f(X)'g(X) - e(X) /2(X) (X).The leading coefficient on the right is in a2 E. This shows that a -D

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    DEDEK IND DOMAINS AND R INGS OFQUOT IENTS 63D S a2 . Likewise a2 E ax , thus a D = a2 andD = E.To prove the map is onto, thefollowing lemma is needed.

    LEMMA 2-4. Let A be a Dedekind domain with quotient fieldK. Toeach polynomial f{X) ~anXn + V a0 of K[X] assignthe fractionary ideal c(f) =( , ,a0). Then c(fg)=c(f)c(g).Proof Let Vv (for each prime P of A) denote theP-adic valua-tion of A. I t is immediate that Vp(c(f)) min F^ J . Because ofthe unique factorization of fractionary ideals in Dedekind domains, itsuffices to show that Vp(c(fg)) =Vp(c(f))+ Vp(c(g)) for each primeP of A. This will be true if the equation is true in each AP[X].But Ap is a principal ideal domain, and the well-known proof forprincipal ideal domains shows thetruth of thelemma.To complete Prop. 2-3, let P be a prime ideal of A1. Theproofof Lemma 2-1shows that if P[\A (0), then P =QA1 where Q is aprime of A. Thus P =QA1and ideal classesgenerated bythese primesare images of classes of A. Suppose now that P is a prime of A1such that P )A=(0). Let P1 =Pf\A[X]. Then P1 A =(0), andP K[X] is a prime ideal of K[X]. LetP K[X] =f(X)K[X]; wemay choose f(X) in A[X]. Let C = c(f). Suppose that g(X)-f(X)eA[X]. Then because c(fg)=(c(/)) +(c( ^0 for all P, g(X)eC~'

    A[X]. Conversely if g(X)e C- A[X]9 then g(X) f(X) e A[X]. ThusP1 - / (I )K [I ] A[I ] =C-1- A[X]-f{X)A[X],and P = P1A1 - C"1-This gives finally that P =C~A\ and the class is animage of a class of A under our map. Since theideal class group ofA1 is generated by all P where P is a prime of A1, this finishes theproof.

    COROLLARY 2-5. A1 has aprime ideal in eachideal class.Proof Letw be any nonunit of A. Then (wX+(= ( r f + 1)A1) is a prime ideal in theprincipal class. Otherwise letC be any integral ideal in a nonprincipal class D~. C can begenerated by 2elements, so suppose C (c0, Cj); then Q(c0 + X)-K[X] A1 is a prime ideal in C717!"1 =D.PROPOSITION 2-6. If A is a Dedekind domain, then A(X) is aprincipal ideal domain.Proof. Since A(X) = Ai , Corollary 1-3and theproof of Corollary2-5 show that each nonprincipal classofA(X) containsa prime QA(X),where Q is a prime ideal of A of thetype (c0 + c1X)K[X] A1.Clearly QnA[X] =(cQ+CX)K[X] A[X] =C"1- A[X] (c0

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    64 LUTHER CLABORN

    PA[X] for any prime P of A. Thus there is in Q A[X] a primitivepolynomial of A[X\. Thus QA(X) =A(X). Theorem 14 now impliesthat every class of A becomes principal in A{X), i.e. A(X) is aprincipal ideal domain.

    REMARK 2-7. Proposition 2-6 is interesting in light of the factthat the primes of A(X) are exactly those of the form PA(X), whereP is a prime of A [1, p. 18].

    REMARK 2-8. If the conjecture given in Remark 1-10 is true fora Dedekind domain A, it is also true for A1. For suppose A = BM,where M is a multiplicatively closed set of B and B is the integralclosure of a principal ideal domain Bo in a suitable finite extensionfield. Let S, S1, and T be the set of monic polynomials in A[X],B[X], and B0[X] respectively. Then A1 = A[X]S = (BM[X])S =(B[X]M)s = CB[X]) = (J5[-X"*i). I t is easy to see that B[X]s is the integral closureof the principal ideal domain BO[X]T in K(X), where K is the quotientfield of B.

    R E F E R E N C E S1. M.N agata, L ocalrings, New York, I nterscience Publishers, Inc. (1962).2. 0. Zariski and P.Samuel, Commutativealgebra, Vol. I,Princeton, D. VanNostrandCompany (1958).CORNELL COLLEGE