Generalizations of Samuel's criteria for a ring to be a unique factorization domain
aa r X i v : . [ m a t h . A C ] F e b GENERALIZATIONS OF SAMUEL’S CRITERIAFOR A RING TO BE A UNIQUE FACTORIZATION DOMAIN
DANIEL DAIGLE, GENE FREUDENBURG, AND TAKANORI NAGAMINE
Abstract.
We give several criteria for a ring to be a UFD, including generalizations of somecriteria due to P. Samuel. These criteria are applied to construct, for any field k , (1) a Z -gradednon-noetherian rational UFD of dimension 3 over k , and (2) k -affine rational UFDs defined bytrinomial relations. Introduction
Let A be a unique factorization domain (UFD). This paper considers ring extensions of thefollowing two types.(i) A r x s where ax “ b for relatively prime a, b P A zt u such that aA and aA ` bA are primeideals.(ii) A r x s where A has a Z -grading and x n “ F for some positive integer n and homogeneousprime F P A with degree relatively prime to n .In his 1964 treatise on UFDs, Samuel [21] studied each of these kinds of extension.In case (i) Samuel demonstrated that, if A is noetherian, then A r x s is a UFD ([21], Proposi-tion 7.6). We show that the noetherian condition can be weakened to a local condition, namely, ifthe hypotheses of (i) are satisfied and Ş i ě p aA ` bA q i “ p q , then A r x s is a UFD ( Theorem A “ R r X , . . . , X n s , a polynomial ring over aUFD R , where A is Z -graded by positive weights over R . He showed that, if either n is congruent to1 modulo deg F , or every finitely generated projective R -module is free, then A r x s is a UFD ([21],Theorem 8.1). We show, more generally, that A r x s is a UFD whenever the conditions of (ii) hold( Theorem
Section k which motivated our work togeneralize Samuel’s criteria in the first place. In Section
4, we construct a Z -graded non-noetherianUFD over k with Krull dimension 3 and quotient field k p q , the field of rational functions in threevariables over k . This example is similar to one given in [5], but the existence of a Z -grading allowsfor a much simpler proof. Section k -affine rings defined by trinomial relations. Theorem k -varieties which admit a torus action of complexity one. For an algebraically closedfield k of characteristic zero, these rings were studied in [11].The Bourbaki volume [1] claims that A r x s is a UFD whenever the conditions of (i) above hold.However, this assertion is wrong. We construct a counterexample in Section 6 . Date : February 15, 2021.2010
Mathematics Subject Classification.
Key words and phrases.
Krull domain, divisor class group, unique factorization domain, non-noetherian ring.The work of the first author was supported by grant 04539/RGPIN/2015 from NSERC Canada.The work of the third author was supported by JSPS Overseas Challenge Program for Young Researchers (No.201880243) and JSPS KAKENHI Grant Numbers JP18J10420 and JP20K22317. . Preliminaries
All rings are commutative with unity and any domain is understood to be an integral domain.For the ring A , A ˚ is the group of units of A , dim A is the Krull dimension of A and, for an ideal I Ă A , ht p I q is the height of I . If f P A then A f “ S ´ A where S “ t f n | n P N u . For the integer n ě A r n s is the polynomial ring in n variables over A , and A r˘ n s is the ring of Laurent polynomialsin n variables over A . When A is an integral domain, frac p A q is its field of fractions, and if A Ă B are integral domains, then tr . deg A B is the transcendence degree of frac p B q over frac p A q . For thefield K , K p n q denotes the field of fractions of the polynomial ring K r n s .2.1. Krull Domains and Divisor Class Groups.
Given an integral domain A , we write P p A q forthe set of height 1 prime ideals of A . Krull domains can be characterized as follows ([1], Chap. VII, §
1, n ˝
7, Thm 4):
Theorem 2.1.
A ring A is a Krull domain if and only if it is an integral domain satisfying each ofthe following three conditions. (a) For each p P P p A q , A p is a DVR. (b) A “ Ş p P P p A q A p (c) For each x P A zt u , p P P p A q | x P p ( is a finite set. Let A be a Krull domain, K “ frac p A q , and Div p A q the free abelian group on the set P p A q .Elements of Div p A q are formal sums ř p P P p A q n p p with n p P Z for all p P P p A q and n p “ p . For each p P P p A q , A p is a discrete valuation ring and we denote by v p : K ˚ Ñ Z thecorresponding normalized valuation (i.e., the valuation with v p p K ˚ q “ Z ). Given x P K ˚ , the set p P P p A q | v p p x q ‰ ( is finite by Theorem (c) ; sodiv A : K ˚ Ñ Div p A q , div A p x q “ ř p P P p A q v p p x q p p x P K ˚ q defines a group homomorphism. The elements of Div p A q are called the divisors of A and thoseof Prin p A q “ div A p x q | x P K ˚ ( are called the principal divisors of A . The quotient groupCl p A q “ Div p A q{ Prin p A q is the divisor class group of A . See [17], § Proposition 2.2. ( [21], Cor. (a) to Prop. 4.1)
The intersection of a finite number of Krull domains(within a common field) is a Krull domain.
Proposition 2.3. ([21], Prop. 4.2 and Thm. 6.3) If A is a Krull domain and S Ă A zt u is amultiplicatively closed set, then S ´ A is a Krull domain. If S is generated by prime elements of A ,then Cl p S ´ A q – Cl p A q . Proposition 2.4.
Let A be an integral domain, let p , . . . , p n P A zt u be primes, and let S Ă A zt u be the multiplicatively closed set generated by p , . . . , p n . (a) A “ S ´ A X A p p q X ¨ ¨ ¨ X A p p n q (b) Assume that A p p i q is a DVR, ď i ď n . If S ´ A is a Krull domain, then A is a Krulldomain.Proof. Let x P S ´ A X A p p q X ¨ ¨ ¨ X A p p n q . Since x P S ´ A , we may write x “ a { s where a P A and s P S are chosen in such a way that for each i satisfying p i | s , we have p i ∤ a . If s R A ˚ then thereexists i such that p i | s (so p i ∤ a ); since x P A p p i q , we have x “ a i { s i for some a i P A and s i P A zp p i q ;thus as i “ a i s , so p i | p as i q where p i ∤ a and p i ∤ s i , a contradiction. So s P A ˚ and hence x P A .This proves (a). Part (b) follows from part (a) and Proposition (cid:3)
Proposition 2.5. ([8], Prop. 6.1)
A ring A is a UFD if and only if it is a Krull domain satisfying Cl p A q “ . .2. Nagata’s Criterion.
Let A be an integral domain. It is well known that, if A is a UFD, thenevery localization of A is a UFD. A partial converse is given by Nagata. Recall that an integraldomain is atomic if every nonzero element factors as a finite product of irreducible elements. Recallalso that every Krull domain is atomic. Theorem 2.6.
Let A be an integral domain and S Ă A a multiplicatively closed set generated by aset of prime elements, R S . Assume that S ´ A is a UFD. If A is an atomic domain, then A is aUFD. Nagata’s original formulation assumed A to be noetherian ([20], Lemma 2). Samuel extended itto the case A is a Krull domain ([21], Corollary to Theorem 6.3). Then Kaplansky generalized it tothe case where A satisfies the ACCP ([13], Theorem 177). The version of the criterion stated above,together with an elementary proof, is due to Cohn ([2], § Samuel’s Criterion.
Recall that, in an integral domain A , elements a, b P A zt u are rela-tively prime if aA X bA “ abA . Theorem 2.7. ([21], Proposition 7.6.)
Assume that A is an integral domain and a, b P A zt u arerelatively prime. Let A r X s – A r s and A “ A r X s{p aX ´ b q , and consider the subring A r b { a s of frac p A q . (a) The kernel of the A -surjection A r X s Ñ A r b { a s , X ÞÑ b { a equals p aX ´ b q . Consequently, p aX ´ b q is a prime ideal of A r X s and A – A r b { a s . (b) If A is a noetherian normal domain and aA and aA ` bA are prime ideals, then A is anoetherian normal domain and Cl p A q – Cl p A q . (c) If A is a noetherian UFD and aA and aA ` bA are prime ideals, then A is a noetherianUFD. The statement of this result in Samuel’s paper does not assume a, b ‰ a “ b “
0. Part (a) of this theorem issomewhat stronger than Samuel’s original formulation; it states the full consequences of Samuel’sproof. We also need:
Lemma 2.8.
Assume that A is an integral domain and f, g P A zt u are relatively prime. (a) f m and g n are relatively prime for each pair of integers m, n ě . (b) f and g ` f h are relatively prime for each h P A such that g ` f h ‰ . (c) Given h P A , if gh P f A , then h P f A . (d) Let B be an integral domain containing A . If B is a free A -module, then f and g arerelatively prime in B . The reader can easily verify parts (a), (b) and (c) of this lemma; part (d) is [9], Lemma 2.46.Note that part (d) includes the case B “ A r n s for some n ě Prime Avoidance.
The following lemma represents an instance of the Prime AvoidanceLemma. Lemma 2.9.
Given n P N and a , . . . , a n , b, c P Z with gcd p a , . . . , a n , b, c q “ , there exist m , . . . , m n P Z with gcd p c, b ` m a ` ¨ ¨ ¨ ` m n a n q “ .Proof. The lemma is clearly true if | c | “
1, so assume that | c | ě
2. Let p a , . . . , a n q “ d Z , andlet p , . . . , p s be the distinct prime factors of c ( s ě b ` d Z Ă p Z Y ¨ ¨ ¨ Y p s Z .Let t p , . . . , p s u “ S Y S , where elements of S divide b , and elements of S do not divide b . Byhypothesis, S ‰ ∅ . Set Σ i “ Ť p P S i p Z ( i “ ,
2) and let λ “ b ` dσ , where σ is the product of theelements of S . Then b P Σ , and by hypothesis, either λ P Σ or λ P Σ . If λ P Σ , then dσ P Σ implies d P Σ , a contradiction, since gcd p d, b, c q “
1. If λ P Σ , then b P Σ , also a contradiction. „ hochster/615W17/supDim.pdf herefore, b ` d Z is not contained in p Z Y ¨ ¨ ¨ Y p s Z . Pick an element b ` ř ni “ m i a i not in Ť sj “ p j Z ,where m i P Z . Then gcd ` c, b ` ř ni “ m i a i ˘ “ (cid:3) Z -Gradings.Lemma 2.10. Let A be an integral domain with a non-trivial Z -grading and let α be a nonzerohomogeneous element of A . There exist homogeneous elements f P αA zt u and w P A ˚ f such that A f “ p A f q r w, w ´ s – p A f q r˘ s where p A f q denotes the subring of A f of degree zero elements.Proof. Let d “ gcd i P Z | A i ‰ ( and note that d ą a, b P A zt u such that deg p a q ´ deg p b q “ d . Define f “ αab P αA zt u and w “ a { b , then w P A ˚ f and w is homogeneous of degree d . Given any homogeneous x P A f , we have deg p x q “ id for some i P Z , so x { w i P p A f q and hence x P p A f q r w, w ´ s ,showing that A f “ p A f q r w, w ´ s . Since deg p w q ą w is algebraically independent over p A f q , so p A f q r w, w ´ s – p A f q r˘ s . (cid:3) Criteria for a Ring to be a UFD
First Criterion.
Let A be an integral domain, and let p a, b q P A . Define the set P p A, p a, b qq “ set of prime elements p of A satisfying a P pA and pA ` bA ‰ A .The pair p A, p a, b qq is said to satisfy condition P if it satisfies each of the following four conditions. P (i): a, b are nonzero and relatively prime, and a is either a unit or a product of primes of A . P (ii): pA ` bA is a prime ideal for every p P P p A, p a, b qq . P (iii): q R pA ` bA for every non-associate pair p, q P P p A, p a, b qq . P (iv): Ş i ě p pA ` bA q i “ p q for every p P P p A, p a, b qq .Note that if a is a power of a prime then condition P (iii) is satisfied, since there are no non-associatepairs p, q P P p A, p a, b qq in this case. The following generalizes Theorem
Theorem 3.1.
Let A be a Krull domain, let p a, b q P A , and define the ring A “ A r X s{p aX ´ b q where A r X s – A r s . Assume that p A, p a, b qq satisfies condition P . (a) A is a Krull domain and Cl p A q – Cl p A q . (b) A is a UFD if and only if A is a UFD. Several preliminaries are needed for the proof of this theorem.Recall that if W is an ideal of a ring A and t P A , one defines the ideal p W : t q by: p W : t q “ x P A | tx P W ( Notation 3.2.
Any triple p b, s, t q of elements of a ring A determines an ideal W p b, s, t q of A ,defined as follows. First, define sequences J n and W n of ideals of A ( n ě
0) by setting W “ A , J “ p W : t q “ A , and: W i “ bJ i ´ ` s i A, J i “ p W i : t q for all i ě W i ` Ă W i and J i ` Ă J i for each i ě
0. Set W p b, s, t q “ Ş i ě W i . Lemma 3.3.
Let A be a domain, a, b P A zt u relatively prime, and let s, t P A be such that a “ st and s, t are relatively prime. Let A r X s “ A r s and I “ p aX ´ b q A r X s . Then the following areequivalent: (a) Ş i ě p s i A r X s ` I q “ I , (b) A X Ş i ě p s i A r X s ` I q “ . oreover, A X Ş i ě p s i A r X s ` I q “ W p b, s, t q .Proof. Since I X A “
0, it is clear that (a) implies (b). To prove the converse, suppose that (b) istrue and consider f P Ş i ě p s i A r X s ` I q . By the division algorithm, there exist N ě Q P A r X s and r P A such that a N f “ p aX ´ b q Q ` r . Note that r “ ´p aX ´ b q Q ` a N f where both aX ´ b and f belong to Ş i ě p s i A r X s ` I q ; so r P A X Ş i ě p s i A r X s ` I q . Since (b) is true, we have r “ a N f “ p aX ´ b q Q ; since a, b are relatively prime, Lemma a N | Q in A r X s , so f P I ,proving that (a) is true. This shows that (a) and (b) are equivalent.To prove that A X Ş i ě p s i A r X s ` I q Ă W p b, s, t q , consider an element r P A X Ş i ě p s i A r X s ` I q .Fix n ą r P W n .We have r “ p aX ´ b q U ` s n V for some U, V P A r X s . Write U “ ř i “ u i X i and V “ ř i “ v i X i where u i , v i P A for all i ě u i “ “ v i for i "
0. Then(1) ´ bu ` s n v “ r and au i ´ ´ bu i ` s n v i “ i ą s i | u i for all i P t , . . . , n u .We proceed by induction on i , the case i “ i is such that 0 ă i ď n and s i ´ | u i ´ ; then (1) implies that s i | bu i , so s i | u i . This proves (2). Define p u , . . . , u n q P A n ` by u i “ u i { s i (0 ď i ď n ). Dividing the second part of (1) by s i gives(3) tu i ´ ´ bu i ` s n ´ i v i “ i P t , . . . , n u .By descending induction on i , we shall now prove that(4) u i P J n ´ i ´ for all i P t , . . . , n ´ u .Since u n ´ P A “ J , the case i “ n ´ i is such that 0 ă i ď n ´ u i P J n ´ i ´ ,then (3) gives tu i ´ “ bu i ´ s n ´ i v i P bJ n ´ i ´ ` s n ´ i A “ W n ´ i , so u i ´ P p W n ´ i : t q “ J n ´ i . Thisproves (4). It follows that u “ u P J n ´ and consequently that r “ ´ bu ` s n v P bJ n ´ ` s n A “ W n . Since n is arbitrary, r P W p b, s, t q . This shows that A X Ş i ě p s i A r X s ` I q Ă W p b, s, t q .For the reverse inclusion, consider an element r P W p b, s, t q . Fix n ą
0, and let us prove that r P s n A r X s ` I . It is convenient to define J ´ “ A , and to note that the relations W i “ bJ i ´ ` s i A and J i “ p W i : t q are also valid for i “
0. We first prove that assertions P p q , . . . , P p n q are true,where for each j P t , . . . , n u we define P p j q : There exist p u , . . . , u j q , p v , . . . , v j q P A j ` satisfying: p a q r “ ´ bu ` s n v , p b q if j ą then tu i ´ ´ bu i ` s n ´ i v i “ for all i P t , . . . , j u , p c q u j P J n ´ j ´ . Since r P W p b, s, t q , we have r P W n “ bJ n ´ ` s n A and so we can choose u P J n ´ and v P A such that r “ ´ bu ` s n v . So P p q is true. Consider j P t , . . . , n u such that P p j ´ q is true. Then we have p u , . . . , u j ´ q , p v , . . . , v j ´ q P A j satisfying the above conditions. Since u j ´ P J n ´ j “ p W n ´ j : t q , we have tu j ´ P W n ´ j “ bJ n ´ j ´ ` s n ´ j A , so we can choose u j P J n ´ j ´ and v j P A such that tu j ´ ´ bu j ` s n ´ j v j “
0. Then p u , . . . , u j q , p v , . . . , v j q P A j ` satisfy therequirements of P p j q . By induction, it follows that P p q , . . . , P p n q are true. Since P p n q is true,there exist p u , . . . , u n q , p v , . . . , v n q P A n ` satisfying r “ ´ bu ` s n v and (3). Then p aX ´ b q ` ř n ´ i “ u i s i X i ˘ ` s n ` ´ bu n X n ` ř ni “ v i X i ˘ “ r, so r P s n A r X s ` I and we are done. (cid:3) Lemma 3.4.
Let A be a ring and b, s, t P A . (a) If t P Ş i ě p bA ` s i A q then W p b, s, t q “ Ş i ě p bA ` s i A q . (b) If bA ` sA ` tA “ A , or if b, s, t is an A -regular sequence, then W p b, s, t q “ Ş i ě p bA ` sA q i . roof. (a) If t P Ş i ě p bA ` s i A q then one can prove by induction that W i “ bA ` s i A and J i “ A for all i ě
0, so the conclusion follows.(b) Let J “ bA ` sA . Observe that if the condition(5) p J i : t q “ J i for all i ě W i “ J i “ J i for all i ě
0, from which we get thedesired conclusion. So it’s enough to prove (5). If bA ` sA ` tA “ A then J i ` tA “ A for all i ě b, s, t is an A -regular sequence. Also, thecase i “ i “ b, s, t is an A -regular sequence. By part (i) of Thm 27 on page 98 of [16], it suffices to prove that b, s is an A -quasiregular sequence. The fact that b, s is A -quasiregular follows from part (ii) of thesame Theorem together with the fact that b, s is an A -regular sequence. (cid:3) Proof of Theorem . Assume that p A, p a, b qq satisfies P . Then A is an integral domain, by Theo-rem A is a subring of A . We may assume that a R A ˚ , otherwise the claim is trivial. So,by P (i), a is a nonempty product of primes. Let P “ P p A, p a, b qq , and let Q be the set of primes p P A such that a P pA (note that P Ă Q ). If p P Q then: A { pA – A r X s{p aX ´ b, p q – A {p pA ` bA qr X s – p A {p pA ` bA qq r s Therefore, if p P P , then p is a prime element of A ; and if p P Q z P , then p is a unit of A .Let S Ă T be the multiplicative sets of A generated by P and Q respectively. Since a P T , wehave T ´ A “ T ´ A ; since Q z P Ă A , we have S ´ A “ T ´ A ; thus S ´ A “ T ´ A . Therefore,by Proposition S ´ A is a Krull domain.Since A is a Krull domain, the set t pA | p P P u is finite; consequently, t pA | p P P u is finite. By Proposition A p q is a DVR for each p P P , then A is a Krull domain. We thus need to show:(6) č i ě p i A “ p q for every p P P Let I “ p aX ´ b q A r X s . For each p P P , define elements s p p q and t p p q of A by the conditions: a “ s p p q ¨ t p p q , s p p q is a power of p , and t p p q R pA Given p P P , the condition Ş i ě p i A “ p q is equivalent to Ş i ě s p p q i A “ p q , which is equivalentto Ş i ě p s p p q i A r X s ` I q “ I , which (by Lemma W p b, s p p q , t p p qq “ p q . Socondition (6) is equivalent to:(7) W p b, s p p q , t p p qq “ p q for every p P P We show that condition P (iii) implies that W p b, s p p q , t p p qq “ Ş i ě p pA ` bA q i for every p P P . Bycondition P (iv), this suffices to prove (6).Given p P P , let s “ s p p q and t “ t p p q . In view of Lemma (b) and of the fact that č i ě p pA ` bA q i “ č i ě p sA ` bA q i it suffices to show that bA ` sA ` tA “ A or b, s, t is an A -regular sequence. Assume that bA ` sA ` tA ‰ A and let us prove that b, s, t is A -regular. Since s “ p e for some e ě
1, it suffices to show that b, p, t is A -regular (see [16], Thm 26, p. 96). Note that p : “ bA ` pA is a prime ideal, because p P P and P (ii) is true. Since t R A ˚ (because bA ` sA ` tA ‰ A ), and since a is a product of primes, wehave t “ q ¨ ¨ ¨ q n for some q , . . . , q n P Q . If q i P P then q i R p by condition P (iii); if q i P Q z P then q i A ` p Ą q i A ` bA “ A , so again q i R p . So t R p . Clearly, A b ÝÑ A (multiplication by b ) is injective.Since p is prime and p ∤ b , A { bA p ÝÑ A { bA is injective. Since p is a prime ideal and t R p , A { p t ÝÑ A { p is injective. So b, p, t is A -regular, and equation (6) is confirmed.Therefore, A p q is a DVR for each p P P , and A is a Krull domain. In addition, by Proposition p A q – Cl p T ´ A q – Cl p S ´ A q – Cl p A q his proves assertion (a), and assertion (b) follows immediately from (a) and Proposition (cid:3)
Corollary 3.5.
Let A be a noetherian UFD, and let p a, b q P A . Define the ring A “ A r X s{p aX ´ b q where A r X s – A r s . If p A, p a, b qq satisfies P (i) and P (ii) , then A is a noetherian UFD.Proof. As in the proof of
Theorem A is an integral domain, each p P P p A, p a, b qq is prime in A ,and S ´ A “ T ´ A . Since A is noetherian, it follows by Nagata’s Criterion that A is a UFD. (cid:3) Second Criterion.Theorem 3.6.
Consider Krull domains A Ă B such that B is finitely generated as an A -module.There exist group homomorphisms ¯ i : Cl p A q Ñ Cl p B q and ¯ N : Cl p B q Ñ Cl p A q such that ¯ N ˝ ¯ i : Cl p A q Ñ Cl p A q is the map γ ÞÑ nγ with n “ r frac p B q : frac p A qs . Bourbaki gives a proof of the special case of this theorem where A and B are assumed to benoetherian (see [1], Chap. VII, §
4, n ˝ A Ă B such that B is finitely generated as an A -module. Let K “ frac p A q and L “ frac p B q , and let n “ r L : K s . Then for each P P P p B q , we have P X A P P p A q .Moreover, for each p P P p A q the set P P P p B q | P X A “ p ( is nonempty and (by Thm 2.1(c))finite, and B P | P P P p B q and P X A “ p ( is the set of all valuation rings R of L satisfying R X K “ A p . Caution: if P P P p B q and p “ P X A then the map v P : L ˚ Ñ Z is not anextension of the map v p : K ˚ Ñ Z , because both v p and v P are normalized by convention (i.e., v p p K ˚ q “ Z and v P p L ˚ q “ Z ); however, there exists a positive integer e p P { p q such that the valuation v P “ e p P { p q v P : L ˚ Ñ e p P { p q Z is an extension of v p . The number e p P { p q is the ramification indexof v P over v p , and we have v P p x q “ e p P { p q v p p x q for all x P K ˚ . We shall also consider theresidual degree f p P { p q of v P over v p , i.e., f p P { p q “ r κ p P q : κ p p qs where κ p P q “ B P { P B P and κ p p q “ A p { p A p . Also note that v P | P P P p B q and P X A “ p ( is a complete system ofextensions of v p to L .Given p P P p A q , we write P | p as an abbreviation for “ P P P p B q and P X A “ p ”. Consider thenorm N L { K : L ˚ Ñ K ˚ . We claim:for each p P P p A q we have ř P | p e p P { p q f p P { p q “ n, (8) for each x P L ˚ and p P P p A q we have v p ` N L { K p x q ˘ “ ř P | p f p P { p q v P p x q ,(9)where ř P | p means ř P P X with X “ P P P p B q | P X A “ p ( . Indeed, let p P P p A q . Then A p Ă B p are Krull domains and B p is finitely generated as an A p -module. Let q denote the maximalideal of A p (so P p A p q “ t q u ). The valuations v Q “ e p Q { q q v Q with Q P P p B p q constitute a completesystem of extensions of v q “ v p to L . Since B p is the integral closure of A p in L and B p is finiteover A p , [1], Chap. VI, §
8, n ˝
5, Thm 2 implies that ř Q P P p B p q e p Q { q q f p Q { q q “ n, and Cor. 3 of that Theorem implies that, for all x P L ˚ , v q ` N L { K p x q ˘ “ ř Q P P p B p q e p Q { q q f p Q { q q v Q p x q “ ř Q P P p B p q f p Q { q q v Q p x q . The rule Q ÞÑ Q X B defines a bijection P p B p q Ñ P P P p B q | P X A “ p ( , and for each Q P P p B p q if we define P “ Q X B then e p Q { q q “ e p P { p q , f p Q { q q “ f p P { p q , and v Q p x q “ v P p x q . This proves(8) and (9).We now proceed with the proof of Theorem roof. Let A Ă B be Krull domains such that B is a finitely generated A -module. Let K “ frac p A q , L “ frac p B q , and n “ r L : K s . Define the map i : P p A q Ñ Div p B q , i p p q “ ř P | p e p P { p q P (for each p P P p A q ) . Extend i to a group homomorphism i : Div p A q Ñ Div p B q . By [1], Chap. VII, §
1, n ˝
10, Prop. 14,we have i p div A p x qq “ div B p x q , for all x P K ˚ . So i induces a group homomorphism ¯ i : Cl p A q Ñ Cl p B q .Define a map N : P p B q Ñ Div p A q by declaring that if P P P p B q and p “ P X A then N p P q “ f p P { p q p . Extend N to a homomorphism of groups N : Div p B q Ñ Div p A q . We claim:(10) N ` div B p x q ˘ “ div A ` N L { K p x q ˘ for each x P L ˚ . Indeed, let x P L ˚ and let φ : P p B q Ñ P p A q be the map P ÞÑ P X A . Then N ` div B p x q ˘ “ N ` ř P P P p B q v P p x q P ˘ “ ř P P P p B q v P p x q N p P q“ ř P P P p B q v P p x q f p P { φ p P qq φ p P q “ ř p P P p A q ř P | p v P p x q f p P { p q p “ ř p P P p A q v p ` N L { K p x q ˘ p “ div A ` N L { K p x q ˘ , where the penultimate equality is (9). So (10) is true and, consequently, N induces a group homo-morphism ¯ N : Cl p B q Ñ Cl p A q .For each p P P p A q , we have N p i p p qq “ N ` ÿ P | p e p P { p q P ˘ “ ÿ P | p e p P { p q N p P q “ ÿ P | p e p P { p q f p P { p q p “ n p by (8). So N ˝ i : Div p A q Ñ Div p A q is multiplication by n , and consequently ¯ N ˝ ¯ i : Cl p A q Ñ Cl p A q is multiplication by n . (cid:3) As a corollary to this theorem, we give the following descent property for integral extensions.
Corollary 3.7.
Consider S Ą R Ă T where R is a Krull domain and S, T are UFDs. Assume that S and T are finitely generated R -modules and that the integers r frac p S q : frac p R qs and r frac p T q : frac p R qs are relatively prime. Then R is a UFD.Proof. By Proposition p S q “ “ Cl p T q and it suffices to show that Cl p R q “
0. Let s “ r frac p S q : frac p R qs and t “ r frac p T q : frac p R qs (so gcd p s, t q “ Theorem f : Cl p R q Ñ Cl p S q , g : Cl p S q Ñ Cl p R q , F : Cl p R q Ñ Cl p T q , G : Cl p T q Ñ Cl p R q such that p g ˝ f qp γ q “ sγ and p G ˝ F qp γ q “ tγ for every γ P Cl p R q . Then sγ “ tγ “ γ “
0) for every γ P Cl p R q , i.e., Cl p R q “ R is a UFD. (cid:3) Third Criterion.
The following criterion generalizes Samuel [21], Thm. 8.1.
Theorem 3.8.
Let A be an integral domain, F P A zt u , and c ą an integer, and define the ring B “ A r Z s{p Z c ´ F q where A r Z s – A r s . Assume that there exists a Z -grading of A such that F is homogeneous and gcd p c, deg F q “ . Then B is an integral domain and frac p B q – frac p A q . Moreover, if F is primein A then the following hold. (a) A is a Krull domain if and only if B is a Krull domain. (b) Assume that
A, B are Krull domains. Then Cl p B q is a direct summand of Cl p A q‘ Cl p A q and Cl p A q is a direct summand of Cl p B q ‘ Cl p B q . Moreover, if one of Cl p A q , Cl p B q is finitelygenerated then Cl p A q – Cl p B q . c) A is a UFD if and only if B is a UFD. Remark 3.9. In Theorem p B q – frac p A q does not mean that the canonicalmap A Ñ B extends to an isomorphism of the fields of fractions. Actually, the canonical mapextends to an embedding frac p A q Ñ frac p B q that satisfies r frac p B q : frac p A qs “ c . Remark 3.10.
Let g be the Z -grading of A in Theorem ω “ deg g F . Extend the Z -grading c g of A to a Z -grading g of A r Z s by letting Z be homogeneous of degree ω . Then Z c ´ F is g -homogeneous and the quotient B “ A r Z s{p Z c ´ F q “ A r z s has the Z -grading induced by g .This Z -grading of B has the property that z P B is homogeneous of degree ω .For the proof, we need the following facts. The proof of each of the first two lemmas is straight-forward, and left to the reader. Lemma 3.11.
Let R be an integral domain, R r x, y s – R r s , λ P R ˚ and a, b P Z positive andrelatively prime. Then R r x, y s{p x a y b ´ λ q “ R r z, z ´ s – R r˘ s where z “ x n { y m for integers m, n with am ` bn “ . Lemma 3.12.
Let p G, `q be an abelian group and W “ À i P G W i a G -graded ring. Given any grouphomomorphism θ : G Ñ W ˚ , the map Φ θ : W Ñ W defined by Φ θ ` ř i P G w i ˘ “ ř i P G θ p i q w i (where w i P W i ) is an automorphism of W , both as a G -graded ring and as a W -algebra. Notation 3.13.
Let
G, H be abelian groups. If n P Z zt u , we write G n ý H as an abbreviation for: There exist group homomorphisms G φ ÝÑ H ψ ÝÑ G such that ψ ˝ φ : G Ñ G is themultiplication by n . We write G ˚ ý H as an abbreviation for: For each r P Z zt u , there exists n P Z zt u such that gcd p n, r q “ and G n ý H . We write rank p G q for the torsion-free rank of G , i.e., the dimension of the Q -vector space Q b Z G . Lemma 3.14.
Let
G, H be abelian groups. (a)
If there exists n P Z zt u such that G n ý H , then rank p G q ď rank p H q . (b) If there exist m, n P Z zt u such that G m ý H , G n ý H and gcd p m, n q “ , then G is a directsummand of H ‘ H . In particular, if G ˚ ý H then G is a direct summand of H ‘ H . (c) Assume that there exist relatively prime integers n, r ‰ satisfying G n ý H and rG “ .Then G is a direct summand of H . (d) Assume that one of
G, H is finitely generated. If G ˚ ý H and H ˚ ý G , then G – H .Proof. (a) Consider homomorphisms G φ ÝÑ H ψ ÝÑ G such that ψ ˝ φ : G Ñ G is the multiplication by n . Then we have Q b Z G φ ÝÑ Q b Z H ψ ÝÑ Q b Z G where ψ ˝ φ : Q b Z G Ñ Q b Z G is the multiplicationby n ; so ψ ˝ φ is an automorphism of Q b Z G , so φ is injective and rank p G q ď rank p H q .(b) For each i P t m, n u , let G φ i ÝÑ H ψ i ÝÑ G be homomorphisms such that ψ i ˝ φ i : G Ñ G isthe multiplication by i . Choose a, b P Z such that am ` bn “ G φ ÝÑ H ‘ H ψ ÝÑ G bydeclaring that φ p g q “ p φ m p g q , φ n p g qq for all g P G and ψ p x, y q “ aψ m p x q ` bψ n p y q for all x, y P H .Then ψ ˝ φ is the identity map of G , so G is a direct summand of H ‘ H .(c) Consider G φ ÝÑ H ψ ÝÑ G such that ψ ˝ φ : G Ñ G is the multiplication by n . Choose u, v P Z such that un ` vr “ ψ : H Ñ G by ψ p y q “ uψ p y q for all y P H . Then x “ unx for all x P G , so ψ ˝ φ is the identity map of G , so G is a direct summand of H .(d) The hypothesis together with (b) implies that G is a direct summand of H ‘ H and H is a directsummand of G ‘ G ; so both G, H are finitely generated. By part (a), we have rank p G q “ rank p H q , o it suffices to show that T p G q – T p H q (where T p G q denotes the torsion subgroup of G ). Since G ÞÑ T p G q is an additive functor, we have T p G q n ý T p H q for each n satisfying G n ý H , so we have T p G q ˚ ý T p H q , and by symmetry T p H q ˚ ý T p G q . Since T p G q is a finite group, there exists an integer r ą rT p G q “
0. By (c), T p G q is a direct summand of T p H q . By symmetry, T p H q is adirect summand of T p G q . Since T p G q , T p H q are finite, they are isomorphic. (cid:3) Proof of Theorem . Let ω “ deg p F q with respect to the given Z -grading of A . With no loss ofgenerality, we may assume that ω ě
0. If ω “ c “ B – A , in which case the result istrivial. So we shall assume throughout that ω ą Z c ´ F is an irreducible element of K r Z s , where K “ frac p A q . In viewof [15], Theorem 9.1, it suffices to show that F cannot be written in the form F “ nξ d with ξ P K , n P Z zt u and d ą c . By contradiction, suppose that F is written in thatform. Write ξ “ U { V where U, V P A . Then F V d “ nU d , which implies that d | ω . Since d | c , d ą p c, ω q “
1, we have a contradiction. So Z c ´ F is an irreducible element of K r Z s . Now let τ be an element of an algebraic closure of K satisfying τ c “ F ; then Z c ´ F isthe minimal polynomial of τ over K . Consider the surjective A -homomorphism φ : A r Z s Ñ A r τ s defined by φ p Z q “ τ . Clearly, p Z c ´ F q A r Z s Ă ker φ . Consider an element G p Z q P ker φ . By thedivision algorithm, there exist Q p Z q , r p Z q P A r Z s such that G p Z q “ Q p Z qp Z c ´ F q ` r p Z q anddeg Z r p Z q ă c . Then r p τ q “
0. Since Z c ´ F is the minimal polynomial of τ over K , we have r p Z q “
0, so G p Z q “ Q p Z qp Z c ´ F q P p Z c ´ F q A r Z s . Thus ker φ “ p Z c ´ F q A r Z s and consequently B “ A r Z s{p Z c ´ F q is isomorphic to A r τ s . So B is a domain. Moreover, the above argument showsthat r frac p B q : frac p A qs “ c .By Lemma f P F A zt u and x P p A f q ˚ (where the Z -grading of A is extended to A f ) such that A f “ p A f q r x, x ´ s – p A f q r˘ s . Interchanging x and x ´ if necessary,we have deg x “ e , where we define e “ gcd i P Z | A i ‰ ( . Since F is homogeneous of degree ω , there exists κ P p A f q such that F “ κx ω { e in A f . Since F | f , we have F P A ˚ f ; so κ P A ˚ f andconsequently κ is a unit of p A f q . Note that the canonical homomorphism π : A r Z s Ñ B is injectiveon A . Let z “ π p Z q , so that B “ A r z s . Let m, n P Z be such that cm ` p ω { e q n “
1. By
Lemma B f “ A f r Z s{p Z c ´ F q“ p A f q r x, x ´ , Z s{p Z c ´ κx ω { e q“ p A f q r x, x ´ , Z s{pp x ´ q ω { e Z c ´ κ q“ p A f q r x, y, y ´ s“ p A f q r y, y ´ s – p A f q r˘ s – A f where y “ z n x m . So B f – A f and in particular frac p B q – frac p A q . This proves the first assertion.Since B { zB – A { F A , it is clear that F is prime in A if and only if z is prime in B . Until the endof the proof, we assume that F is prime in A (and z is prime in B ).(a) Choose integers s, t ą sc ” p mod ω q , tc ” p mod ω q , and gcd p s, t q “ s ą sc ” p mod ω q , and let t “ s ` | ω | ). Let U and V beindeterminates over B and define S “ B r U s{p U s ´ z q “ B r u s and T “ B r V s{p V t ´ z q “ B r v s . Note that gcd p s, ω q “ Remark Z -grading of B such that z ishomogeneous of degree ω ; this allows us to apply the first part of the proof to S and conclude that S is a domain and r frac p S q : frac p B qs “ s . Similarly, T is a domain and r frac p T q : frac p B qs “ t . Alsodefine Ω “ B r Q s{p Q st ´ z q “ B r q s where B r Q s “ B r s ; since gcd p st, ω q “
1, Ω is a domain. Then r X s{p X t ´ u q “ Ω “ T r Y s{p Y s ´ v q and we have the commutative diagram of integral domains T ) ) ❘❘❘❘❘❘ A / / B ❧❧❧❧❧❧ ) ) ❘❘❘❘❘❘ Ω S ❧❧❧❧❧❧ where all homomorphisms are injective. The fact that gcd p s, t q “ S X T “ B (taking the intersection in Ω).Indeed, this follows by considering that Ω “ B r q s is a free B -module with basis t , q, . . . , q st ´ u , andthat S “ B r q t s and T “ B r q s s are free B -modules with bases t q ti u s ´ i “ and t q si u t ´ i “ respectively.Since F “ z c “ u sc , we see that: S “ A r u s – A r U s{p U sc ´ F q Let W “ A r U, U ´ s – A r˘ s . The Z -grading of A extends to a Z -grading of W in which U ishomogeneous of degree 0 and W “ A r U, U ´ s . Since sc ” p mod ω q , there exists d P Z such that sc “ dω `
1. Consider the group homomorphism θ : Z Ñ W ˚ , θ p i q “ U di and the correspondingautomorphism Φ “ Φ θ : W Ñ W defined in Lemma p U sc ´ F q “ U sc ´ U dω F “ U dω p U ´ F q where U dω P W ˚ , so Φ maps the principal ideal p U sc ´ F q W onto the principal ideal p U ´ F q W . SoΦ induces an isomorphism ¯Φ in: S u – / / W {p U sc ´ F q – ¯Φ / / W {p U ´ F q – / / A F . Therefore, S u – A F . Since t satisfies tc ” p mod ω q , the same argument shows that T v – A F .Assume that A is a Krull domain. Since S u – A F , S u is a Krull domain by Proposition S { uS – A { F A , u is prime in S . As an A -module, we have: S “ A ‘ Au ‘ ¨ ¨ ¨ ‘ Au sc ´ Choose h P Ş i P N u i S , and write h “ h ` h u ` ¨ ¨ ¨ ` h sc ´ u sc ´ for h j P A . By hypothesis, h P u sci S “ F i S “ F i A ‘ F i Au ‘ ¨ ¨ ¨ ‘ F i Au sc ´ for each i P N . Therefore, h j P F i A for each j “ , . . . , sc ´ i P N . Since A is a Krulldomain, h j “ j “ , . . . , sc ´
1, so h “
0. Therefore, Ş i u i S “ p q . Since u is prime in S ,it follows that S p u q is a DVR. By Proposition S is a Krull domain. So T is a Krull domain bythe same argument, and B “ S X T is a Krull domain by Proposition A is aKrull domain then so is B .Conversely, assume that B is a Krull domain. Since S “ B r U s{p U s ´ z q “ B r u s and gcd p s, ω q “ S is a Krull domain. As u is prime in S , we get Ş j u j S “
0. Using u sc “ F , we see that Ş i F i A Ă Ş j u j S “
0, so A p F q is a DVR. We also know that A F is a Krulldomain, because S u – A F . By Proposition A is a Krull domain. This proves (a), but let us alsoobserve that, since S u – A F where u is prime in S and F is prime in A , Proposition p S q – Cl p S u q – Cl p A F q – Cl p A q .(b) Assume that A, B are Krull domains. The proof of (a) shows that if s P Z satisfies s ą sc ” p mod ω q then S “ B r U s{p U s ´ z q is a Krull domain and a finite B -module such that r frac p S q : frac p B qs “ s and Cl p S q – Cl p A q . By Theorem p B q Ñ Cl p S q Ñ Cl p B q whose composition Cl p B q Ñ Cl p B q is multiplication by s . This meansthat Cl p B q s ý Cl p S q (see Notation p S q – Cl p A q , we have in fact shown thatCl p B q s ý Cl p A q for every integer s ą sc ” p mod ω q . It follows that(11) Cl p B q ˚ ý Cl p A q . ndeed, let r P Z zt u . Choose any s P Z such that s c ” p mod ω q . Since gcd p ω, s , r q “ Lemma m P Z satisfying gcd p r, s ` mω q “
1; then for N large enough,the number s “ s ` mω ` N ¨| r |¨ ω satisfies s ą
0, gcd p s, r q “ sc ” p mod ω q , so gcd p s, r q “ p B q s ý Cl p A q . This proves (11). Note that (11) says that we have Cl p B q ˚ ý Cl p A q for any pairof rings p A, B q that satisfies the hypotheses of assertion (b). Let s P Z be such that s ą sc ” p mod ω q and consider S “ B r U s{p U s ´ z q ; then, in view of Remark p B, S q satisfies thehypotheses of assertion (b), so (11) implies that Cl p S q ˚ ý Cl p B q . It follows that Cl p A q ˚ ý Cl p B q ,because Cl p S q – Cl p A q was noted at the beginning of the proof of (b). Thus:Cl p A q ˚ ý Cl p B q and Cl p B q ˚ ý Cl p A q . Now assertion (b) follows from
Lemma
Proposition (cid:3)
Corollary 3.15.
Let R be a UFD, a , . . . , a n P N zt u and R r X , . . . , X n s – R r n s , where n ě .Assume that one of the following holds: (1) n ě and gcd p a n , a ¨ ¨ ¨ a n ´ q “ ; (2) n “ and a , a , a are pairwise relatively prime.Then the ring R r X , . . . , X n s{p X a ` ¨ ¨ ¨ ` X a n n q is a UFD.Proof. Let A “ R r X , . . . , X n ´ s and define F P A by F “ ´ X a ´ ¨ ¨ ¨ ´ X a n ´ n ´ . The conditions onthe a i imply that F is irreducible in frac p R qr X , . . . , X n ´ s , and it follows easily that F is irreduciblein A . Let ω “ lcm p a , . . . , a n ´ q and let A have the Z -grading over R for which X i is homogeneousof degree ω { a i . Then F is homogeneous of degree ω and gcd p ω, a n q “ Theorem R r X , . . . , X n s{p X a ` ¨ ¨ ¨ ` X a n n q – A r Z s{p Z a n ´ F q is a UFD. (cid:3) Fourth Criterion.Theorem 3.16.
Let A be a Z -graded integral domain, let a, b P A zt u be relatively prime, and let n ą be such that gcd p n, deg b ´ deg a q “ . Let A r Z s “ A r s . (a) p aZ n ´ b q is a prime ideal of A r Z s . (b) Assume that A is a noetherian UFD. If b is prime in A and p A, p a, b qq satisfies P (ii) , then B : “ A r Z s{p aZ n ´ b q is a UFD and frac p B q – frac p A q .Proof. Extend the Z -grading to A r X s – A r s so that X is homogeneous and deg X “ deg b ´ deg a .Then aX ´ b is homogeneous and the ring A “ A r X s{p aX ´ b q is Z -graded. Write A “ A r x s ,where x is the canonical image of X in A (in particular, x ‰ x is homogeneous anddeg x “ deg b ´ deg a . It follows by Theorem A is an integral domain. It is also clear thatfrac p A q “ frac p A q . We have: B – A r X, Z s{p aX ´ b, Z n ´ X q – A r Z s{p Z n ´ x q Since gcd p n, deg p x qq “ Theorem B is an integral domain. This proves part (a).Assume that the hypotheses of part (b) hold. Then p A, p a, b qq satisfies P p i q and, since A is aUFD, A is a UFD by Corollary A { xA – A { bA , we see that x is a prime element of A .So Theorem B – A r Z s{p Z n ´ x q is a UFD, and that frac p B q – frac p A q “ frac p A q .So part (b) is proved. (cid:3) Fifth Criterion.Theorem 3.17.
Let A “ À i P Z A i be a Z -graded integral domain and let F P A ω zt u , ω P Z . Let A r Z , . . . , Z n s – A r n s for n ě , and let e , . . . e n ě be integers such that gcd p e , . . . , e n , ω q “ . (a) p Z e ¨ ¨ ¨ Z e n n ´ F q is a prime ideal of A r Z , . . . , Z n s b) Define: B “ A r Z , . . . , Z n s{p Z e ¨ ¨ ¨ Z e n n ´ F q If A is a noetherian UFD and F is prime in A , then B is a UFD and frac p B q – frac p A q p n ´ q .Proof. We may assume that n ą
1, otherwise both parts of the claim follow from
Theorem
Lemma m , . . . , m n ´ P Z so that:gcd p e n , ω ´ p m e ` ¨ ¨ ¨ ` m n ´ e n ´ qq “ Z -grading of A to a Z -grading of R : “ A r Z , . . . , Z n ´ s by declaring that Z i is homo-geneous of degree m i , noting that R is an integral domain. Define a, b P R by a “ Z e ¨ ¨ ¨ Z e n ´ n ´ and b “ F . Then a and b are nonzero elements of R with aR X bR “ abR , and deg b “ ω anddeg a “ m e ` ¨ ¨ ¨ ` m n ´ e n ´ . Since gcd p e n , deg b ´ deg a q “ Theorem (a) implies that p aZ e n n ´ b q “ p Z e ¨ ¨ ¨ Z e n n ´ F q is a prime ideal of A r Z , . . . , Z n s . This proves part (a).Assume that the hypotheses of part (b) hold. Note that R is a noetherian UFD. It is easy tocheck that b is prime in R , b ∤ a in R , and for each i P t , . . . , n ´ u , p Z i , b q is a prime ideal of R .So Theorem (b) implies that B is a UFD, and that frac p B q – frac p R q – frac p A q p n ´ q . (cid:3) Application I: Rational UFDs of Dimension Three
A Family of Three-dimensional Affine UFDs.
The following lemma generalizes Lemma10.15 in [8], and Lemma 2 in [4].
Lemma 4.1.
Let D be an integral domain, n ě , u , . . . , u n P D ˚ , R n “ D r Z , . . . , Z n s – D r n ` s ,and a , . . . , a n , b , . . . , b n positive integers such that gcd p a i , b ¨ ¨ ¨ b i q “ for each i P t , . . . , n u .Then I n : “ p u Z a ` Z b , . . . , u n Z a n n ` Z b n n ´ q is a prime ideal of R n and Z n R I n .Proof. We proceed by induction on n , the case n “ I “ p q Ă R “ D r Z s . Let n ě I n ´ is a prime ideal of R n ´ “ D r Z , . . . , Z n ´ s and that Z n ´ R I n ´ . Define a Z -grading of R n ´ over D for which Z i is homogeneous of degree b ¨ ¨ ¨ b i a i ` ¨ ¨ ¨ a n ´ , 0 ď i ď n ´ A : “ R n ´ { I n ´ is a Z -graded integral domain.Let F P A be the image of Z b n n ´ ; since Z n ´ R I n ´ and I n ´ is prime, we have Z b n n ´ R I n ´ and hence F ‰
0; note that deg F “ b ¨ ¨ ¨ b n . By hypothesis, gcd p a n , deg F q “
1. Therefore, by
Theorem (a) , the ring A r Z s{p Z a n ´ u ´ n F q – R n { I n is an integral domain, so I n is prime. If Z n P I n then the image of Z in A r Z s{p Z a n ´ u ´ n F q is zero, which is not the case because F ‰ Z n R I n and the proof is complete. (cid:3) Theorem 4.2.
Let K be a noetherian UFD, n P N , K r Z , . . . , Z n ` s – K r n ` s , u i , v i P K ˚ and a i , b i positive integers such that gcd p a i , b ¨ ¨ ¨ b i q “ , ď i ď n . Define the ring A n “ K r Z , . . . , Z n ` s{p f i Z i ` ` u i Z a i i ` v i Z b i i ´ q ď i ď n where f , . . . , f n P K zt u and the set of prime factors of f i in K is the same for all i “ , . . . , n .Then A n is a UFD and frac p A n q – frac p K r Z , Z sq – p frac K q p q .Proof. Since f i Z i ` ` u i Z a i i ` v i Z b i i ´ “ v i p v ´ i f i Z i ` ` u i v ´ i Z a i i ` Z b i i ´ q , we may assume each v i “
1. Let π n : K r Z , . . . , Z n ` s Ñ A n be the standard surjection. The restriction of π n to K isinjective. To see this, observe that the ideal p f i Z i ` ` u i Z a i i ` Z b i i ´ q ď i ď n of K r Z , . . . , Z n ` s isincluded in p Z , . . . , Z n ` q , and p Z , . . . , Z n ` q X K “ t u . So K Ă A n .Let P K be the set of prime elements of K dividing some (hence all) f i . If P K “ ∅ then f , . . . , f n are units of K and consequently A n – K r Z , Z s “ K r s , in which case the theorem is true. So wemay assume, throughout, that P K ‰ ∅ . Given κ P P K and m P t , . . . , n u , define the ring: Q p m, κ q “ p K { κK qr Z , . . . , Z m s{p ¯ u i Z a i i ` Z b i i ´ q ď i ď m here ¯ u i is the class of u i in K { κK . By Lemma Q p m, κ q is an integral domain. Note that κ ‰ A m because π m is injective on K . Moreover, A m { κA m “ K r Z , . . . , Z m ` s{p κ, f i Z i ` ` u i Z a i i ` Z b i i ´ q ď i ď m – Q p m, κ qr Z m ` s – Q p m, κ q r s . Assuming that m ă n , define h m P A m by h m “ π m p u m ` Z a m ` m ` ` Z b m ` m q ; then A m {p κA m ` h m A m q “ K r Z , . . . , Z m ` s{p κ, u i Z a i i ` Z b i i ´ q ď i ď m ` – Q p m ` , κ q . Also note that h m R κA m . Indeed, the image of h m in A m { κA m – Q p m, κ qr Z m ` s – Q p m, κ q r s isnot zero, because it has the form uZ a m ` m ` ` c with u, c P Q p m, κ q , u ‰ a m ` ą
0. We haveshown that for each κ P P K , the following two statements are true:For each m P t , . . . , n u , κA m is a nonzero prime ideal of A m .(12) For each m P t , . . . , n ´ u , κA m ` h m A m is a prime ideal of A m and h m R κA m .(13)By induction on m , we proceed to show A m is a UFD for all m “ , . . . , n . Since A “ K r Z , Z s – K r s , we have a basis for induction. Consider m P t , . . . , n ´ u such that A m is a UFD. Let usprove the following assertions:(i) f m ` , h m P A m zt u , f m ` A m X h m A m “ f m ` h m A m , and f m ` is a product of primesin A m .(ii) For each prime element p of A m such that f m ` P pA m , we have pA m ` h m A m P Spec A m .Since P K ‰ ∅ , we have f m ` “ κ ¨ ¨ ¨ κ r for some κ , . . . , κ r P P K and r ě
1. By (12), each κ i is a prime element of A m ; so f m ` ‰ A m and f m ` is a product of primes in A m . Since h m R κ A m by (13), we have h m ‰ A m . Actually, (13) gives h m R κ i A m for all i “ , . . . , r , so f m ` and h m are relatively prime in A m and consequently f m ` A m X h m A m “ f m ` h m A m . Thisproves (i). To prove (ii), consider a prime element p of A m satisfying f m ` P pA m . Then, for some i , we have p | κ i in A m . Since κ i is a prime element of A m , we have p “ uκ i for some u P A ˚ m , so pA m ` h m A m “ κ i A m ` h m A m . We have κ i A m ` h m A m P Spec A m by (13), so (ii) is proved.It follows from (i) and (ii) that p A m , p f m ` , h m qq satisfies the conditions P (i) and P (ii) statedjust before Theorem 3.1. Since A m ` “ A m r Z m ` s{p f m ` Z m ` ` h m q , Corollary A m ` is a UFD. By induction, we obtain that A , . . . , A n are UFDs. Since A n is in particular anintegral domain, the assertion frac p A n q – frac p K r Z , Z sq – p frac K q p q is now clear. (cid:3) Let k r x s “ k r s for a field k , u i , v i P k ˚ , and let p p x q , . . . , p n p x q P k r x szt u be such that each p i p x q has the same set of prime factors. Define the affine k -algebra(14) B n “ k r x sr z , . . . , z n ` s{p p i p x q z i ` ` u i z a i i ` v i z b i i ´ q ď i ď n where a , . . . , a n , b , . . . , b n are positive integers such that gcd p a i , b ¨ ¨ ¨ b i q “ i . Using K “ k r x s in Theorem n ě B n is an affine rational UFD of dimension3 over k . Proposition 4.3. If p i p x q R k and a i , b i ě for all i , then the minimum number of generators of B n as a k -algebra is n ` .Proof. It suffices to prove the case where k is algebraically closed. Let d be the minimum numberof generators of B n over k . Then clearly d ď n `
3. Set X “ Spec p B n q Ă A n ` k , affine n space over k . For 1 ď i ď n , let f i “ p i p x q z i ` ` u i z a i i ` v i z b i i ´ . Let J be the Jacobian matrix of p f , . . . , f n q ,namely: J “ ˜ B f i B x , B f i B z j ¸ ď i ď n, ď j ď n ` hen J is a matrix of size n ˆ p n ` q . For 1 ď i ď n and 0 ď j ď n `
1, we have B f i {B x “ p i p x q z i ` and: B f i B z j “ $’’’&’’’% p i p x q p j “ i ` q u i a i z a i ´ i p j “ i q v i b i z b i ´ i ´ p j “ i ` q p otherwise q For a maximal ideal m of B n , we denote by J p m q the Jacobian matrix at m , that is, J p m q “ ˜ B f i B x p m q , B f i B z j p m q ¸ ď i ď n, ď j ď n ` where for g P B n , g p m q means the image of g in B n { m .Take a common prime divisor q p x q P k r x s of p p x q , . . . , p n p x q , which is possible since p i p x q R k and each p i p x q generates the same radical ideal in k r x s . Let m be the maximal ideal of B n generatedby q p x q , z , . . . , z n ` . Since a i , b i ě i , we see that rank p J p m qq “
0, hence we have:dim k p m { m q “ p n ` q ´ rank p J p m qq “ n ` m is n `
3, which implies d ě n ` (cid:3) Example 4.4.
In [4], the authors give the following rings. Let k be a field of characteristic zero,and let p, q be prime integers with p ă q . Given n ě
0, defineΩ n “ k r X, Z , . . . , Z n ` s{ ` XZ i ` ` Z pi ` Z qi ´ ˘ ď i ď n where X, Z , . . . , Z n ` are indeterminates over k . Then for each n ě
0, there exists a locallynilpotent derivation of k r s with kernel isomorphic to Ω n . Moreover, Proposition n is minimally generated by n ` k . Remark 4.5.
One motivation to consider UFDs of the type given in this section comes from thestudy of locally nilpotent derivations of polynomial rings C r n s for the field C of complex numbers.The kernel A of such a derivation is a UFD of transcendence degree n ´ C . It is known that A is quasi-affine [22] and frac p A q is ruled [6]; that A – C r n ´ s if 1 ď n ď A is generallynon-noetherian if n ě n “
4, it is further known that A is rational [6] and that therecan be no a priori bound on the number of generators needed [4]. But the question whether A mustbe affine when n “
4, or even noetherian, is open. We are thus led to study quasi-affine rationalUFDs of transcendence degree 3 over C .4.2. A Graded Rational Non-Noetherian UFD of Dimension 3.
Let k be a field and letΛ “ k r X, Z , Z , Z , . . . s be the polynomial ring in a countably infinite number of variables X, Z i , i ě
0. Define Ω “ k r X, Z , Z , Z , . . . s{p X i Z i ` ` Z i ` Z i ´ q i ě “ k r x, z , z , z , . . . s where x, z i denote the images of X, Z i under the standard surjection of Λ onto Ω. For each n ě f n “ X n Z n ` ` Z n ` Z n ´ P Λ n : “ k r X, Z , . . . , Z n ` s – k r n ` s Since the ring displayed in (14) is a UFD, hence a domain, we see that the ideal I n : “ p f , . . . , f n q is a prime ideal of Λ n . Let I Ă Λ be the ideal I “ p f , f , f , . . . q . Then I “ Ť n ě I n , which impliesthat I is a prime ideal and Ω is an integral domain. In addition:Ω r x ´ s “ k r x, x ´ , z , z s – k r x, x ´ s r s ùñ frac p Ω q – k p q Define ideals in Ω by: p “ p z , z , z , . . . q and m “ p x, z q emma 4.6. m is a maximal ideal of Ω , and x Ω and p are prime ideals properly contained in m .In addition: (a) z i ` z i P x Ω for each i ě , and (b) z i R x Ω for each i ě .Proof. Define the subring R Ă Λ and maximal ideal J Ă R by: R “ k r X, Z i ` Z i ´ s i ě and J “ p X, Z i ` Z i ´ q i ě Then Λ “ R r Z s – R r s and J Λ “ X Λ ` I . It follows thatΩ { x Ω – Λ { J Λ “ p R { J qr ¯ Z s “ k r ¯ Z s – k r s ùñ Ω { x Ω “ k r ¯ z s – k r s where ¯ Z is the image of Z in Λ { J Λ and ¯ z is the image of z in Ω { x Ω. Therefore, x Ω is a primeideal and z R x Ω. For each i ě
1, we have z i ` z i ´ P x Ω. Parts (a) and (b) now follow byinduction. Since f i P p Z , Z , Z , . . . q for each i ě
1, we see hatΩ { p – k r x s – k r s and p is prime. Since Ω { m – k , m is maximal. Part (a) shows p Ă m . Since neither x Ω nor p ismaximal, their containment in m is proper. (cid:3) Define a Z -grading Ω “ À d P Z Ω d by letting x and each z i be homogeneous, deg x “ ´ z i “ i . A monomial µ P Ω is of the form µ “ x r ś i P N z e i i for r, e i P N . Note that eachmonomial is homogeneous, and that the set of all monomials forms a multiplicative monoid. Define: | µ | “ min ! ř i ǫ i ˇˇˇ µ “ x s ś i z ǫ i i , s, ǫ i P N ) Given d P N , let d “ ř i ě d i i be its binary expansion, where d i P t , u for each i and d i “ i . Define the function: σ : N Ñ Ω , d ÞÑ F d “ ź i ě z d i i Since F d P Ω d for d P N , it follows that σ is injective. Note that, since z i R x Ω for all i ě F d R x Ωfor all d P N . Lemma 4.7.
Given d P Z , the set B d “ t x m F n | m, n P N , n ´ m “ d u is a k -basis of Ω d .Proof. Fix d P Z .Suppose that c x m F n ` ¨ ¨ ¨ ` c s x m s F n s “ c i P k ˚ , s ě
2, and distinct p m i , n i q P N with n i ´ m i “ d . Note that m , . . . , m s are distinct; we may assume m “ min i m i . Then F n P x Ω incontradiction to the above stated property. Therefore, elements of B d are k -linearly independent.Let V Ă Ω d be the subspace spanned by B d . We proceed by induction on | µ | to show that everymonomial µ P Ω d is in V . From this it follows that V “ Ω d .Let µ P Ω d be a monomial where µ “ x r ś i z e i i and | µ | “ ř i e i .If | µ | “
0, then µ “ x r F P V .Assume that λ P V whenever λ P Ω d is a monomial with 0 ď | λ | ă | µ | . If max i e i ď
1, then µ “ x r F d ` r P V . Otherwise, let e m “ max i e i ě e m “ a ` b for integers a ě b P t , u . Let P “ µ { z e m m . Then: µ “ P z e m m “ P z bm p z m q a “ p´ q a P z bm p x m ` z m ` ` z m ` q a “ p´ q a P z bm a ÿ j “ ˆ aj ˙ x j m ` z jm ` z a ´ jm ` “ p´ q a a ÿ j “ ˆ aj ˙ x j m ` P z bm z a ´ jm ` z jm ` or each j , we see that M j : “ x j m ` P z bm z a ´ jm ` z jm ` is a monomial in Ω d and that: | M j | ď p ř i ‰ m e i q ` b ` p a ´ j q ` j ă ř i e i “ | µ | By the inductive hypothesis, M j P V . It follows that µ P V . (cid:3) Lemma 4.8. Ş m ě x m Ω “ p q Proof.
Since J “ Ş m ě x m Ω is a graded ideal, it suffices to show that J X Ω d “ d P Z .Arguing by contradiction, assume that 0 ‰ f P J X Ω d for some d P Z . Then, by Lemma f “ ř i ě max p , ´ d q a i x i F d ` i where a i P k for all i . Let m “ min i | a i ‰ ( . Then ř i ą m a i x i F d ` i P x m ` Ω and (since f P J ) f P x m ` Ω, so x m F d ` m P x m ` Ω, so F d ` m P x Ω, a contradiction. (cid:3)
Theorem 4.9. Ω is a UFD.Proof. Let q “ x Ω. By
Lemma q is a discrete valuation ring. Since Ω “ Ω r x ´ s X Ω q , Prop. 2.4 implies that Ω is a Krull domain. Since Ω r x ´ s is a UFD and x is prime, it follows by Nagata’sCriterion that Ω is a UFD. (cid:3) Corollary 4.10. z Ω is a prime ideal properly contained in p .Proof. Since Ω r x ´ s “ k r x, x ´ , z , z s – k r x, x ´ s r s , z is irreducible in Ω r x ´ s . Since z R x Ω, z is irreducible in Ω. By Theorem z is prime in Ω.The inclusion z Ω Ă p is clear. Suppose that z P z Ω. Then z P z Ω r x ´ s where Ω r x ´ s “ k r x, x ´ , z , z s – k r x, x ´ s r s which is not possible. Therefore, z Ω Ĺ p . (cid:3) Theorem 4.11. dim Ω “ and ht p m q “ . Consequently, Ω is not noetherian.Proof. Since frac p Ω q “ k p q we see that tr . deg k p Ω q “
3. Since dim Ω ď tr . deg k p Ω q (see, for example,[14], Theorem 5.5), we have dim Ω ď
3. Since p q Ĺ z Ω Ĺ p Ĺ m is a chain of primes in Ω, it follows that ht p m q “ “
3. Since m is generated bytwo elements, it follows from the Krull dimension theorem ([7], Theorem 10.2) that Ω cannot benoetherian. (cid:3) Application II: Affine UFDs Defined by Trinomial Relations
Let k be a field. Assume that the following data are given.(D.1) An integer n ě n “ n ` n ` ¨ ¨ ¨ ` n r where r ě n i ě ď i ď r .This induces a partition of variables t ij in the polynomial ring k r n s : k r n s “ k r T , T , . . . , T r s , where T i “ t t i , ¨ ¨ ¨ , t in i u , ď i ď r (D.2) A sequence β , . . . , β r where β i “ p β i , . . . , β in i q P Z n i ` , 0 ď i ď r satisfy:If d i “ gcd p β i , . . . , β in i q , 0 ď i ď r , then d , . . . , d r are pairwise relatively prime.This induces a sequence T β , . . . , T β r r P k r T , . . . , T r s of monomials: T β i i “ t β i i ¨ ¨ ¨ t β ini in i , ď i ď r (D.3) A sequence of distinct elements λ , . . . , λ r P k ˚ . Theorem 5.1.
Given the data and notation in (D.1), (D.2) and (D.3) above, the ring B : “ k r T , . . . , T r s{p T β ` λ i T β ` T β i i q ď i ď r is an affine rational UFD of dimension n ´ r ` over k , and the image of T β ` µT β in B is primefor each µ P k ˚ zt λ , . . . , λ r u . roof. In a slight abuse of notation, we use T i and t ij to denote their images in B . We proceed byinduction on r . Define subrings B Ă B Ă ¨ ¨ ¨ Ă B r “ B by: B m “ k r T , . . . , T m s{p T β ` λ i T β ` T β i i q ď i ď m Note that B “ k r T , T s – k r n ` n s , which is a rational UFD. Moreover, it follows easily from Theorem (a) that T β ` µT β is prime in B for each µ P k ˚ . This gives a basis for induction.Assume that, for some m ě B , . . . , B m ´ are rational UFDs over k and that T β ` µT β isprime in B m ´ for every µ P k ˚ zt λ , . . . , λ m ´ u . Let F “ T β ` λ m T β . Given i with 0 ď i ď m ´ d i “ d i β i ` ¨ ¨ ¨ ` d in i β in i for d ij P Z . Put an N -grading on B m ´ by letting t ij behomogeneous with deg p t ij q “ d ij d ¨ ¨ ¨ ˆ d i ¨ ¨ ¨ d m ´ . Then for each monomial T β i i , 0 ď i ď m ´
1, wehave: deg T β i i “ deg p t β i i ¨ ¨ ¨ t β ini in i q “ p d i β i ` ¨ ¨ ¨ ` d in i β in i qp d ¨ ¨ ¨ ˆ d i ¨ ¨ ¨ d m ´ q “ d ¨ ¨ ¨ d m ´ Therefore, F is homogeneous and deg F “ d ¨ ¨ ¨ d m ´ . Since B m “ B m ´ r T m s{p T β m m ` F q andgcd p β m , . . . , β mn m , deg F q “ Theorem (b) implies that B m is a UFD, and that frac p B m q – frac p B m ´ q p n m ´ q . Since B m ´ is rational over k , it follows that B m is rational over k .Given µ P k ˚ zt λ , . . . , λ m u , let G “ T β ` µT β and ¯ B m ´ “ B m ´ { GB m ´ . Since G is ahomogeneous prime of B m ´ , it follows that ¯ B m ´ is a Z -graded integral domain. We have B m { GB m “ B m ´ r T m s{p T β ` µT β , T β m m ` T β ` λ m T β q“ ¯ B m ´ r T m s{p T β m m ` p λ m ´ µ q T β q and Theorem (a) implies that this ring is an integral domain. So G is prime in B m .By induction, it follows that B r “ B is a rational UFD over k and T β ` µT β is prime in B foreach µ P k ˚ zt λ , . . . , λ r u . (cid:3) Example 5.2.
In [19], Mori classified affine UFDs of dimension two over an algebraically closedfield k which admit a nontrivial N -grading. Each such ring is of the form k r x, y, z , . . . , z N s{p x a ` µ i y b ` z c i i q ď i ď N where N ě a, b, c , . . . , c N ě “ µ , . . . , µ N P k ˚ are distinct.These rings conform to the data (D.1) n “ N ` r “ N `
1, where T “ x , T “ y and T i “ z i ´ ,2 ď i ď r ; (D.2) β “ a , β “ b and β i “ c i ´ , 2 ď i ď r ; (D.3) λ i “ µ i ´ , 2 ď i ď r . Remark 5.3.
A UFD B of the type presented in Theorem gradingby Z n ´ r . When k is algebraically closed, this means that the variety X “ Spec p B q admits a torusaction of complexity one. Surfaces and threefolds of this type were classified by Mori [19] and Ishida[12], respectively. More recently, Hausen, Herppich and S¨uss [11] classified all such varieties in termsof Cox rings, under the additional assumption that the characteristic of k is zero. Their descriptionmatches that given in Theorem k is algebraically closed of characteristic zero, every k -affine rational UFD of dimension d that admits a non-degenerate Z d ´ -grading is one of the ringsof Theorem
Theorem k .6. A Counterexample
The Bourbaki volume [1] includes the following exercise (p.549, Exercise 15(b), VII, § A be a Krull domain and a, b two [nonzero] elements of A such that Aa and Aa ` Ab are prime and distinct. Show that A r X s{p aX ` b q is a Krull domain andthat [the divisor class group] C p A r X s{p aX ` b qq is isomorphic to C p A q .” Let p G, `q be an abelian group and R “ À i P G R i a G -graded ring. We say that the grading is non-degenerate if i P G | R i ‰ ( is a generating set of G . The hypothesis that a, b ‰ n this section, we construct a counterexample to this assertion.Let k be a field of characteristic zero and let A “ k r x, y s – k r s . Define a sequence of integers s p n q by s p q “ , s p q “ s p n q “ n ś ď i ď n ´ s p i q for n ě
3. Let A r Z , Z , Z . . . s be the polynomialring in a countably infinite number of variables Z i over A , and define: B “ A r Z , Z , Z , . . . s{p xZ i ` ` y s p i ` q´ Z s p i ` q i ´ Z i ´ q i ě Theorem 6.1.
The following properties hold. (a) B is a non-noetherian UFD of transcendence degree 4 over k , the surjection of A r Z , Z , . . . s onto B is injective on A , x, y P B are non-associated primes of B , and xB ` yB is a maximalideal. (b) Let B r T s – B r s . The ring B “ B r T s{p xT ´ y q is an integral domain which does not satisfythe ascending chain condition on principal ideals. The proof consists of the following series of lemmas.
Lemma 6.2.
Let k be a field, R a k -algebra and R Ă R Ă R Ă ¨ ¨ ¨ finitely generated subalgebrasof R such that R “ Ť n ě R n . Consider J Ă J Ă J Ă ¨ ¨ ¨ where, for each n ě , J n is a primeideal of R n . Then J “ Ť n ě J n is a prime ideal of R . Moreover, consider the following conditions: (i) there exists a positive integer d such that tr . deg k p R n { J n q “ d for all n ě ; (ii) for each n ě , R n { J n is an algebraic extension of the image of R n ´ Ñ R n { J n ; (ii ) for each n ě , R n { J n is a birational extension of the image of R n ´ Ñ R n { J n .Then the following hold. (a) If (i) and (ii) hold then J X R n “ J n and tr . deg k p R { J q “ tr . deg k p R n { J n q for all n ě . (b) If (i) and (ii ) hold then J X R n “ J n and frac p R { J q “ frac p R n { J n q for all n ě .Proof. It is clear that J is a prime ideal of R . Assume that (i) and (ii) hold and let us prove:(15) for all m, n such that 1 ď m ď n , we have J n X R m “ J m .It suffices to prove the case where m “ n ´
1. So fix n ě φ : R n ´ Ñ R n { J n be thecomposition R n ´ ã Ñ R n Ñ R n { J n . Assumptions (i) and (ii) give the first two equalities in:tr . deg k ` R n ´ { J n ´ ˘ “ tr . deg k ` R n { J n ˘ “ tr . deg k ` φ p R n ´ q ˘ “ tr . deg k ` R n ´ {p J n X R n ´ q ˘ . We have J n ´ Ă J n X R n ´ , where both J n ´ and J n X R n ´ are prime ideals of R n ´ ; so theequality tr . deg k ` R n ´ { J n ´ ˘ “ tr . deg k ` R n ´ {p J n X R n ´ q ˘ implies that J n ´ “ J n X R n ´ , asdesired. This proves (15). It follows that(16) J X R n “ J n for all n ě f P J X R n . Then there exists N ě n such that f P J N , so f P J N X R n “ J n by(15), so (16) is proved.Let S “ R { J , let π : R Ñ S be the canonical surjection and let S n “ π p R n q for each n ě S n – R n { J n for all n ě
1. Moreover, assumption (ii) implies that S n is analgebraic extension of S n ´ for every n ě
2; since S “ Ť n ě S n , it follows that S is an algebraicextension of S n (so tr . deg k p R { J q “ tr . deg k p R n { J n q ) for all n ě ) hold then all of the above is true and S n is a birational extension of S n ´ for every n ě
2; so S is a birational extension of S n (so frac p R { J q “ frac p R n { J n q ) for all n ě (cid:3) Lemma 6.3.
The canonical map A Ñ B is injective, B is a domain and frac B “ p frac A q p q “ k p q .Proof. We use
Lemma R “ A r Z , Z , . . . s and, for each n ě R n “ A r Z , . . . , Z n ` s and J n “ p xZ i ` ` y s p i ` q´ Z s p i ` q i ´ Z i ´ q ni “ .Let J “ Ť n ě J n “ p xZ i ` ` y s p i ` q´ Z s p i ` q i ´ Z i ´ q i ě , then R { J “ B . Since J Ă p Z , Z , . . . q and p Z , Z , . . . q X A “ p q , we have J X A “ p q , so A Ñ B is injective. By the same argument, A Ñ R n { J n is injective for each n ě ix n ě
1. For each i P t , . . . , n u , let φ i be the A -automorphism of R n defined by φ i p Z i ´ q “ Z i ´ ` xZ i ` ` y s p i ` q´ Z s p i ` q i and φ i p Z j q “ Z j for all j P t , . . . , n ` uzt i ´ u . Then φ n ˝ ¨ ¨ ¨ ˝ φ maps J n onto p Z , . . . , Z n ´ q , so R n { J n – A r s . So condition (i) of Lemma R n { J n “ A r z , . . . , z n ` s where z i is the image of Z i by the canonical surjection R n Ñ R n { J n .Since A Ñ R n { J n is injective, we have x ‰ R n { J n , so xz n ` ` y s p n ` q´ z s p n ` q n ´ z n ´ “ z n ` P frac p A r z , . . . , z n sq , showing that R n { J n is a birational extension of the image of R n ´ Ñ R n { J n , i.e., condition (ii ) of Lemma B is a domain andfrac B “ frac p R n { J n q “ p frac A q p q “ k p q . (cid:3) Lemma 6.4. B { xB is a domain of transcendence degree over k .Proof. We use
Lemma R “ A r Z , Z , . . . s and, for each n ě R n “ A r Z , . . . , Z n ` s and J n “ p x q ` p xZ i ` ` y s p i ` q´ Z s p i ` q i ´ Z i ´ q ni “ .Let J “ Ť i ě J n , then B { xB “ R { J . Fix n ě
1. Then J n “ p x q ` p y s p i ` q´ Z s p i ` q i ´ Z i ´ q ni “ . Ifwe write V i ´ “ y s p i ` q´ Z s p i ` q i ´ Z i ´ for i “ , . . . , n , then R n “ k r x, y, V , . . . , V n ´ , Z n , Z n ` s and J n “ p x, V , . . . , V n ´ q , so R n { J n “ k r s and J n is a prime ideal of R n of height n `
1. It followsfrom
Lemma J is a prime ideal of R and hence that B { xB is a domain. Let us prove:(17) the composition k r y, Z s ã Ñ R Ñ R { J “ B { xB is injective.To see this, it’s enough to show that k r y, Z sX J “ p q ; so it’s enough to show that k r y, Z sX J n “ p q for all n ě
1. Let n ě
1. If y P J n then we see that Z , . . . , Z n ´ P J n (and x, y P J n ) so ht J n ą n ` y R J n . If Z P J n then (using that J n is prime and that y R J n ) we see that Z , . . . , Z n P J n (and x P J n ), so ht J n ą n `
1, a contradiction. So Z R J n . Consider the k -homomorphism π : R n Ñ k r y, Z s defined by π p x q “ “ π p Z q “ ¨ ¨ ¨ π p Z n ` q , π p y q “ y and π p Z q “ Z . If f is a nonzero element of k r y, Z s X J n , write f “ Z m g where m P N , g P k r y, Z s and Z ∤ g ; since J n is prime and Z R J n , we have g P J n ; write g “ xg ` ř ni “ p y s p i ` q´ Z s p i ` q i ´ Z i ´ q g i ( g , . . . , g n P R n ) and note that g “ π p g q “ ´ Z π p g q P Z k r y, Z s , a contradiction. This shows that f does not exist, i.e., k r y, Z s X J n “ p q . So (17) is true.Write B { xB “ k r y, z , z , . . . s where z i is the canonical image of Z i in R { J “ B { xB . Then y s p i ` q´ z s p i ` q i ´ z i ´ “ B { xB for all i ě
1. Since y ‰ B { xB , it follows that z i is algebraicover k p y, z , . . . , z i ´ q for all i ě
1. So frac p B { xB q is an algebraic extension of k p y, z q . We have k p y, z q “ k p q by (17), so tr . deg k p B { xB q “ (cid:3) Lemma 6.5. B { yB is a domain and frac p B { yB q “ k p q .Proof. We use
Lemma R “ A r Z , Z , . . . s and, for each n ě R n “ A r Z , . . . , Z n ` s and J n “ p y q ` p xZ i ` ` y s p i ` q´ Z s p i ` q i ´ Z i ´ q ni “ .Let J “ Ť i ě J n , then B { yB “ R { J . Fix n ě
1. Then J n “ p y q ` p xZ i ` ´ Z i ´ q ni “ . Ifwe define V i ´ “ xZ i ` ´ Z i ´ for i “ , . . . , n , then R n “ k r x, y, V , . . . , V n ´ , Z n , Z n ` s and J n “ p y, V , . . . , V n ´ q , so R n { J n “ k r s and J n is a prime ideal of R n of height n `
1. By
Lemma J is a prime ideal of R , so B { yB is a domain. Note that x ‰ R n { J n (otherwise wewould have x P J n , so x, y, Z , . . . , Z n ´ P J n , so ht J n ą n `
1, a contradiction). Write R n { J n “ k r x, z , . . . , z n ` s where z i is the canonical image of Z i in R n { J n . Since xz n ` ´ z n ´ “ x ‰ R n { J n , we have z n ` P k p x, z , . . . , z n q , so condition (ii ) of Lemma p B { yB q “ frac p R n { J n q “ k p q . (cid:3) Lemma 6.6. B is a UFD, x, y are prime elements of B such that x ∤ y , and m : “ xB ` yB is amaximal ideal of B . roof. Define L “ k p y q , S “ k r y szt u , A L “ S ´ A “ L r x s , and B L “ S ´ B . Then B L “ L r x sr Z , Z , Z , . . . s{p xZ i ` ` y s p i ` q´ Z s p i ` q i ´ Z i ´ q i ě . According to [5] the ring R : “ L r x sr Z , Z , Z , . . . s{p xZ i ` ` Z s p i ` q i ´ Z i ´ q i ě is a UFD. Let φ be the L r x s -automorphism of L r x sr Z , Z , Z , . . . s defined by φ p Z i q “ yZ i for all i ě
0. Then φ maps the ideal p xZ i ` ` Z s p i ` q i ´ Z i ´ q i ě of L r x sr Z , Z , Z , . . . s onto the ideal p xZ i ` ` y s p i ` q´ Z s p i ` q i ´ Z i ´ q i ě of L r x sr Z , Z , Z , . . . s . So B L – L R and B L is a UFD.Write B “ k r x, y, z , z , . . . s where z i is the canonical image of Z i in B . Then xz i ` ` y s p i ` q´ z s p i ` q i ´ z i ´ “ B for all i ě
1; since x ‰ B (because A Ñ B is injective), it follows that B x “ A x r z , z s .Since frac B “ k p q by Lemma z , z must be algebraically independent over A . So B x “ A r s x is a UFD.Next, we claim that B “ B x X B L . Indeed, define B n “ A r z n , z n ` s for all n ě
0. Since xz i ` ` y s p i ` q´ z s p i ` q i ´ z i ´ “ i ě
1, we have z n P B n ` and hence B n Ă B n ` for all n ě
0. This gives the filtration B “ Ť n ě B n . Since z , z are algebraically independent over A , and since z , z P B n “ A r z n , z n ` s for all n ě z n , z n ` are algebraically independent over A and hence B n “ A r s for all n ě
0. So B x “ ď n ě A x r z n , z n ` s and B L “ ď n ě A L r z n , z n ` s where A x r z n , z n ` s “ A r s x and A L r z n , z n ` s “ A r s L for all n ě
0. Given f P B x X B L , choose n ě f P A x r z n , z n ` s X A L r z n , z n ` s and write: f “ ř i,j α ij z in z jn ` “ ř i,j β ij z in z jn ` p α ij P A x , β ij P A L q . By uniqueness of coefficients in frac p A qr z n , z n ` s – frac p A q r s , we see that α ij “ β ij for each pair i, j ě
0, so these coefficients belong to A x X A L “ A . So f P B n Ă B , and B “ B x X B L is proved.Since B is the intersection of two UFDs, it is a Krull domain. Since x is prime in B and B x is aUFD, it follows by Nagata’s Criterion that B is a UFD.Since (by Lemma
Lemma . deg k p B { xB q “ . deg k p B { yB q “
3, the primeideals xB and yB are nonzero and distinct, so x, y are prime elements of B and x ∤ y . We have B { m “ k r x, y, Z , Z , . . . s{ J where J “ p x, y q ` p xZ i ` ` y s p i ` q´ Z s p i ` q i ´ Z i ´ q i ě “ p x, y, Z , Z , . . . q , so B { m “ k and m is a maximal ideal of B . (cid:3) We continue to use the notation B “ k r x, y, z , z , . . . s where z i is the canonical image of Z i in B . Lemma 6.7.
We have ‰ z P Ş n ě m n . Moreover, B is not noetherian.Proof. For each i ě
1, we have xz i ` ` y s p i ` q´ z s p i ` q i ´ z i ´ “ z i ´ P m z i ` m z i ` .So z P ř j ě m z j and, for each n ě ÿ j ě m n z j Ă ÿ j ě m n p m z j ` ` m z j ` q Ă ÿ j ě m n ` z j It follows that z P ř j ě m n z j for all n ě
1, so z P Ş n ě m n . Note that z ‰
0, because frac p B q “ k p x, y, z , z q – k p q . Since m is a proper ideal of the domain B , the fact that Ş n ě m n ‰ p q impliesthat B is non-noetherian, by the Krull Intersection Theorem ([7], Corollary 5.4). (cid:3) emma 6.8. Let B “ B r T s{p xT ´ y q , where B r T s – B r s . Then B – B r y { x s Ă frac p B q and B does not satisfy the ascending chain condition on principal ideals. In particular, B is not a UFD.Proof. By Lemma B is a UFD and x, y P B are nonzero relatively prime elements. By Samuel’sCriterion ( Theorem B is an integral domain isomorphic to B r y { x s . Since y P xB , Lemma ‰ z P č n ě x n B . Using that m is a proper ideal of B and that x, y are relatively prime in B , we find that x is not aunit of B . Consequently, B does not satisfy the ascending chain condition on principal ideals, so B is not a UFD. (cid:3) Acknowledgments.
The work of the third author was supported by JSPS Overseas ChallengeProgram for Young Researchers (No. 201880243) and JSPS KAKENHI Grant Numbers JP18J10420and JP20K22317. The authors gratefully acknowledge the helpful advice of S. Bhatwadekar andXiasong Sun in preparing this article. The third author wishes to express his gratitude to membersof the Department of Mathematics at Western Michigan University, which he visited for periods in2018 and 2019.
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