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Amalgamations of Domains
Journal of Algebra 221, 90᎐101 Ž1999. Article ID jabr.1999.7944, available online at http:rrwww.idealibrary.com on
Amalgamations of Domains S. Peter Farbman* Department of Mathematics, Sul Ross State Uni¨ ersity, Alpine, Texas 79832 Communicated by Walter Feit Received November 10, 1998
The question of whether the amalgamated free product of two domains is itself a domain is considered. Some results analogous to group-theoretic approaches to the zero-divisor conjecture are proved. A condition is introduced that guarantees that the amalgam will be a domain. If, for example, R is a polynomial ring over the domain T, then only minimal restrictions on how T embeds in S guarantee that R#T S is a domain. 䊚 1999 Academic Press
1. INTRODUCTION One of the most famous open questions in group ring theory is the zero-divisor conjecture ŽZDC., which states that if G is any torsion-free group and F is any field, then the group algebra FG has no non-trivial zero-divisors. A unique-product Žu.p.. group is a group G such that if A, B are finite subsets of G, then there is some ordered pair Ž a0 , b 0 . g A = B such that if Ž a, b . g A = B and ab s a0 b 0 , then Ž a0 , b 0 . s Ž a, b .. U.p. groups were invented in the hope of settling the ZDC using a strictly group-theoretic property: if G is a u.p. group then Žfor any domain D . DG is a domain; this is easily proven, mostly by using the u.p. property and the fact that DG is free as a Žleft and right. D-module. ŽIn this paper, ‘‘domain’’ is taken to mean a not necessarily commutative nonzero unital ring with no non-trivial zero-divisors.. Many results about u.p. groups raise analogous questions about domains. The purpose of this paper is to discuss how we might achieve results analogous to the results in w1x, but in the context of more general rings. In particular, we intend to introduce some conditions under which we can be assured that the amalgamated free product of two domains is again a domain. * Current address: St. Mary’s College of Maryland, St. Mary’s City, Maryland 20686. 90 0021-8693r99 $30.00 Copyright 䊚 1999 by Academic Press All rights of reproduction in any form reserved.
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2. NOTATION All rings herein will be assumed to be associative with a Žnon-zero. multiplicative identity which is respected by subrings and homomorphisms. Fix R and S as domains and let T be a domain which embeds in both R and S. We identify T with its image in R and S so that R l S s T. In general, for rings R, S with a common subring T, the amalgam P s R#T S may not even contain copies of R, S, or T. To avoid this, and other possible complications, we assume that R, S are left and right free as T-modules. That is, we can write Rs
r T s [ Tr [ J J j
X
XX j ,
BR s r j : j g J 4 ,
BRX s r XjX : jX g J X 4 ,
Ž 1. Ss
s T s [ Ts [ K K k
X
XX k ,
BS s sk : k g K 4 ,
BSX s sXk X : kX g K X 4 .
Ž 2. We also assume in what follows that 1 g BR , BS , BRX , BSX . We call R and S factors of P. Under these assumptions, R, S, and T embed in P; furthermore, we can represent P as a right T-module which is free on a basis consisting of 1 and elements of the form c1 c 2 ⭈⭈⭈ c¨ ,
Ž 3.
with c i g Ž BS j BR . y 14 , c i , c iq1 never in the same factor. Elements in this basis will be called right monomials, and this representation will be called the right representation of P. We can also define the left representation because P is left free on the left monomials cX¨ cX¨ y1 ⭈⭈⭈ cX1 ,
Ž 4.
with cXi g Ž BSX j BRX . y 14 , cXi , cXiq1 never in the same factor. It is easy to see that P is also free as a left and right R Žresp., S . module. A basis as a free right R-module Žresp., S-module. is 1 together with the subset of right monomials as in Ž3. with c¨ g BS y 14 Žresp., c¨ g BR y 14.. We will refer to this as the right R-representation Žresp., the right S-representation .. The left R-representation Žresp., the left S-representation . is defined similarly. Note, though, that the terms ‘‘right representation’’ and ‘‘left representation’’ always refer to P as a free T-module.
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S. PETER FARBMAN
The length of an element p g P, denoted < p <, is defined to be the smallest n such that p is in the T-linear span of Ž R j S . n. For example, the length of the monomials in Ž3. and Ž4. is ¨ , and elements of T have length zero. This definition is left᎐right symmetric, and immediately yields that If p1 , p 2 g P , then < p1 p 2 < F < p1 < q < p 2 < .
Ž 5.
For p g P, we will use supp r Ž p . Žresp., supp l Ž p .. to denote the set of right Žresp., left. monomials that appear in the right Žresp., left. representation of p. The length of an element p g P is the length of the longest monomial in supp r Ž p . Žor supp l Ž p ... In w2, 3x, where most of the definitions here can be found, the word ‘‘height’’ is used instead of length. We use ‘‘length’’ to be consistent with the terminology used in w1x. Following w2x, we say that an element p g P is pure of type Ž⭈, R . if all monomials of maximal length in supp r Ž p . end with R; for example, if p is pure of type Ž⭈, R ., < p < s ¨ , and the monomial in Ž3. is in supp r Ž p ., then c¨ g BR y 14 . Similarly, if all monomials of maximal length in supp r Ž p . begin with S, then p is pure of type Ž S, ⭈ .. ŽWe will see shortly that purity does not depend on writing p in the right representation.. The other types of purity are defined analogously. This definition is extended so that, for example, if p is pure of type Ž R, ⭈ . and of type Ž⭈, S ., then we say that it is pure of type Ž R, S .. Elements of T cannot be pure.
3. LENGTH OF PRODUCTS In the context of the zero-divisor conjecture, much of the preceding is simple. For if we consider the group rings R s FG, S s FK, T s FH Žwith H F G, K ., we can take BR s BRX s G and BS s BSX s K; the monomials in Ž3. and Ž4. are just the elements of G# H K, thus we know that P s F Ž G# H K .. One of the main difficulties in extending to more general rings is that if P is not a group ring, then there could be some p g P such that cardŽsupp r Ž p .. / cardŽsupp l Ž p ... For example, the left representation of a right monomial might not be a left monomial. We first consider the simplest cases of products within P. LEMMA 1. Let R, S, T be domains as in Ž1. and Ž2., and let P s R#T S. Suppose that c s c1 c 2 ⭈⭈⭈ c¨ is a right monomial of length ¨ with each c i g BR j BS y 14 ; that d s d1 d 2 ⭈⭈⭈ d w is a right monomial of length w with each d i g BR j BS y 14 ; that t g T y 04 ; and that p s ctd g P. If c¨ g BS and d1 g BR , or if c¨ g BR and d1 g BS , then < p < s ¨ q w. If, in addition, e s e1 e2 ⭈⭈⭈ e¨ qw g supp r Ž p . with < e < s ¨ q w and each e i g
AMALGAMATIONS OF DOMAINS
93
BR j BS y 14 , then c i s e i for 1 F i F ¨ . The corresponding result holds for left monomials. Proof. Without loss of generality, suppose that c¨ g BS and d1 g BR . Certainly < p < F ¨ q w, because of Ž5.. We now proceed by induction on w. If w s 0, then d s 1, and the result is clear. If w s 1, then td1 f T because d1 f T and R is free as a left T-module. Let td1 have right representation Ý i d iŽ1. t i , with each d iŽ1. g BR . Then
Ý c1
⭈⭈⭈ c¨ d iŽ1. t i
i
is the right representation of p. Because td1 f T, at least one of the d iŽ1. is not 1; thus it must be that < p < s ¨ q 1 s ¨ q w, and every right monomial in supp r Ž p . begins with c1 ⭈⭈⭈ c¨ ; furthermore, every right monomial in supp r Ž p . of length ¨ q 1 is pure of type Ž⭈, R .. Note that this does not depend on the value of ¨ . Now let w ) 1 and define pwy 1 s ctd1 d 2 ⭈⭈⭈ d wy1. Suppose, inductively, that < pwy 1 < s ¨ q w y 1, and that every right monomial of length ¨ q w y 1 in supp r Ž pwy1 . begins with c1 ⭈⭈⭈ c¨ and is pure of type Ž⭈, X . Žwhere X is the factor containing d wy 1 .. We need to show that < p < s < pwy1 d w < s ¨ q w and that every right monomial in supp r Ž p . of length ¨ q w begins with c1 ⭈⭈⭈ c¨ and is pure of type Ž⭈, Y . Žwhere Y is the factor containing d w .. Choose some right monomial Ž e. e s c1 ⭈⭈⭈ c¨ d1Ž e. ⭈⭈⭈ d wy1 g supp r Ž pwy1 . ,
Ž 6.
with < e < s ¨ q w y 1. From the w s 1 caseᎏwhich is applicable here because e is pure of type Ž⭈, X . and d w f Xᎏwe know that there is some f g supp r Ž ed w . such that < f < s ¨ q w and f starts with c1 ⭈⭈⭈ c¨ d1Ž e. ⭈⭈⭈ Ž e. Ž . Žas d wy 1 s e. Because f starts with e, it is easy to see that f g supp r p there is nothing that can ‘‘cancel’’ f .. Therefore, < p < s ¨ q w. Finally, again using Ž5., we see that for every f g supp r Ž p . with < f < s ¨ q w, there is some e g supp r Ž pwy1 . as in Ž6. such that f g supp r Ž ed w .; hence every such f begins with c1 ⭈⭈⭈ c¨ , and p is pure of type Ž⭈, Y .. A symmetric argument yields the final statement. We can use this lemma and induction to conclude that the various concepts of purity can be given characterizations depending neither on choice of basis, nor on right versus left representation. For example, if p s x 1 y 2 x 3 ⭈⭈⭈ y¨ for arbitrary x i g R y T, yi g S y T, then < p < s ¨ and p is pure of type Ž R, S .. Although it gets a little more complicated, there are some things we can determine if equality does not hold in Ž5.:
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S. PETER FARBMAN
LEMMA 2. Let R, S, T be rings as in Ž1. and Ž2., and let P s R#T S. Let c s c1 c 2 ⭈⭈⭈ c¨ and d s d1 d 2 ⭈⭈⭈ d w be right monomials Ž with < c < s ¨ , < d < s w, and each c i , d i g BR j BS y 14., and t g T y 04 . If < ctd < s ¨ q w y 2 m or ¨ q w y Ž2 m y 1., then ¨ G m. Furthermore, if ¨ ) m, and e s e1 e2 e3 ⭈⭈⭈ g supp r Ž ctd . Ž such that e i g Ž BR j BS . y 14 for all i, and e i , e iq1 not in the same factor . with < e < s < ctd <, then c i s e i for 1 F i F ¨ y m. The corresponding result for left monomials also holds. Proof. If ¨ s 0, then c s 1, and the result is obvious; thus we may assume that ¨ G 1. Without loss of generality, let c¨ g BS y 14 . Now we proceed by induction on w. If w s 0, then d s 1 and the result is once again obvious. Suppose, inductively, that the lemma holds for w F k, and that d s d1 ⭈⭈⭈ d kq 1. If d1 g BR , then we are done by Lemma 1. So we may assume that d1 g BS y 14 . If c¨ td1 s ˜t g T, then ctd s c1 ⭈⭈⭈ c¨ y1 ˜td 2 ⭈⭈⭈ d kq1 , and we are done by induction. If, on the other hand, c¨ td1 g S y T, then by the comment after Lemma 1, < ctd < s ¨ q k s ¨ q Ž k q 1. y Ž2 ⭈ 1 y 1.. That is, m s 1 F ¨ . If ¨ ) m s 1 and c¨ td1 g S y T, then use the right representation to write c¨ td1 s Ý i d iŽ1. t i , and t i d 2 d 3 ⭈⭈⭈ d kq1 s
Ý f i j tXi j , j
where the f i j are right monomials of length no more than k; those that are of length k are of type Ž R, X . Žwhere X is the factor that contains d kq 1 .. Then ctd s
Ý Ž c1
⭈⭈⭈ c¨ y1 d iŽ1. f i j t i j . .
i, j
If e g supp r Ž ctd . with < e < s < ctd < s ¨ q k, then there is some i, j such that d iŽ1. g BS y 14 ; < f i j < s k; f i j is of type Ž R, ⭈ .; and e s Ž c1 c 2 ⭈⭈⭈ c¨ y1 d iŽ1. .Ž f i j .. In particular, e begins with c1 ⭈⭈⭈ c¨ y1 . A symmetric argument yields the final statement. We can now use these facts about monomials to examine the products of arbitrary elements of P. ŽMost of the remainder of this section consists of restatements of the results from w2, 3x that we wish to extend. They are included for completeness and to help prove Theorem 7, but most of the proofs are omitted..
AMALGAMATIONS OF DOMAINS
LEMMA 3.
95
Let R, S, T be as in Ž1. and Ž2., P s R#T S, and p, pX g P.
Ži. Ž cf. w2, Theorem 2.1x.. If < ppX < - < p < q < pX <, then either p is pure of type Ž⭈, R . and pX is pure of type Ž R, ⭈ ., or p is pure of type Ž⭈, S . and pX is pure of type Ž S, ⭈ .. Žii. Ž cf. w2, Theorem 2.3x.. Suppose that < p < s ¨ ; < pX < s w; < ppX < - ¨ qw y 1; a¨ y1 r¨ g supp r Ž p ., and rwX aXwy1 g supp l Ž pX . are monomials of maximal length Ž with r¨ g BR , rwX g BRX .; r is the coefficient of a¨ y1 in the right R-representation of p; and r X is the coefficient of aXwy 1 in the left R-representation of pX . Then rr X g T. Proof. Žii. ŽThe proof of Žii. is included as it provides motivation for Section 4.. Suppose that < ppX < - < p < q < pX < y 1. Because P is free as a left and right T-module, it follows that ¨ , w ) 0. Certainly, neither r nor r X are in T. Suppose that rr X g R y T. Then we know that < a¨ y1 rr X aXwy1 < s ¨ q w y 1 by the comment after Lemma 1. If e g supp r Ž a¨ y1 rr X aXwy1 ., with < e < s ¨ q w y 1, then e begins with a wy1 by Lemma 2. Since, by Ž5. and Lemma 2, there is no other way for e to appear in pr X aXwy 1 Žthat might cancel this appearance., it must be that e g supp r Ž pr X aXwy1 .. In particular, < pr X aXwy1 < G ¨ q w y 1. Then we haveᎏin a similar mannerᎏthat if f X g supp l Ž pr X aXwy1 . with < f X < G ¨ q w y 1, then f X ends with aXwy 1 and so it must be that f X g supp l Ž ppX .. Thus < ppX < G ¨ q w y 1, a contradiction. We also get: COROLLARY 3.1. Let R, S, T be as in Ž1. and Ž2.. If p, pX g P s R#T S such that p / 0 / pX , but ppX s 0, then < p <, < pX < ) 1. Proof. Suppose not. Because P is a free Žleft and right. T-module and T is a domain, it is certainly true that < p <, < pX < ) 0. Without loss of generality, we may assume that < p < s 1. By Lemma 3Ži., p g R or p g S. But P is free as a left R- and S-module, and both are domains, so we have reached the desired contradiction. Following w3x, we make the following definition: if A is a ring and B is a subring which is also a domain, then A is said to be an inert extension of B provided that: whenever a, aX g A y 04 such that aaX g B, there is some unit u g A such that au, uy1 aX g B. ŽNote that this implies that A is a domain.. For example, if D is a division ring and A is a D-ring which is also a domain, then A is an inert extension of D. LEMMA 4 Žcf. w3, Theorem 2.2x.. Let R, S, T be as in Ž1. and Ž2.. If p, pX g P s R#T S such that < ppX < - < p < q < pX < y 1 and S is an inert extension of T, then either p is pure of type Ž⭈, R . and pX is pure of type Ž R, ⭈ ., or there is a unit u g UŽ S . with < pu < - < p < and < uy1 pX < - < pX <.
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S. PETER FARBMAN
THEOREM 5 Žcf. w3, Corollary Following Theorem 2.2x.. Let R, S, T be rings as in Ž1. and Ž2.. Suppose that T is a domain and R, S are inert extensions of T. Then P s R#T S is an inert extension of R and S. In particular, P is a domain.
4. THE vnn PROPERTY FOR RINGS Our goal is to get results that parallel those of the preceding section in the same way that the results of w1x parallel those of w4x. Let R, S, T be domains as in Ž1. and Ž2., and let S be an inert extension of T. Suppose that p, pX g P y 04 such that ppX s 0. By Lemma 4, we may assume that p is pure of type Ž⭈, R . and pX is pure of type Ž R, ⭈ .. Taking p, pX , a¨ y1 , aXwy1 , r, r X as in Lemma 3Žii. and a¨ y1 s a¨ y3 r¨ y2 s¨ y1 , X aXwy 1 s sXwy1 rwy2 aXwy3 , we have ppX s a¨ y3 r¨ y2 s¨ y1 rr X sXwy1wy2 r X aXwy3 q other terms.
Ž 7.
As in the proof of Lemma 3Žii., for ppX to be zero Žor, more generally, for < ppX < to be zero., it must be that rr X s ˜t g T. We wish to create conditions X under which at least one term like a¨ y3 r¨ y2 s¨ y1 rr X sXwy1 rwy2 aXwy3 cannot be canceled by the ‘‘other’’ terms. If S, T are domains as in Ž2., we say BS , BSX are vnn bases provided that, for any Žnon-empty. finite subsets B s si : i g I 4 : BS and BX s sXiX : iX g I X 4 : BSX with B / 14 / BX , and elements t i, iX : i g I, iX g I X 4 : T such that t i, iX / 0 if si / 1 / sXiX , it must be that s Ý i, iX si t i, iX sXiX / 0. ŽWhat we really need is that f T, but it suffices to assume that it is non-zero. For if g T, then we can add, if necessary, si s 1 to B and sXi s 1 to BX , and alter the choice of t i, iX for si s 1 and sXiX s 1 to force s 0.. If S admits vnn bases, then we say that S is a vnn extension of T. Note that by this definition, if S is a vnn extension of T and s, sX g S such that ssX g T, then s, sX g T. In particular, if S is a vnn extension of T, then S is an inert extension of T. It is clear that if H - G are groups, then the usual bases of the group ring DG as a free DH-moduleᎏright and left coset representatives ᎏare never vnn bases; for if g g G y H, then g ⭈ 1 ⭈ gy1 q g ⭈ 0 ⭈ 1 q 1 ⭈ 0 ⭈ gy1 q 1 ⭈ Ž y1 . ⭈ 1 s 0.
Ž 8.
ŽIn fact, it is not difficult to see that because of Ž8., DG is never a vnn extension of DH.. The most obvious situation where the vnn condition holds is when S is a polynomial ring over T. For let S s T w x x and BS s x i X : i g ⺪ G 0 4 s BSX , and suppose that B s x i : i g I 4 / 14 / BX s x i : iX g I X 4 ; then
AMALGAMATIONS OF DOMAINS
97
degŽÝsi t i, iX sXiX . s i max q iXmax , regardless of the choice of the t i, iX , as long as t i ma x , iXmax / 0. The same argument shows that skew polynomial rings, or more generally, Ore extensions, are vnn extensions. It should be noted that if S is a vnn extension of T, then it is not necessarily true that all bases are vnn bases. For example, let T s ⺪ and S s ⺪w x x, the usual integer polynomial ring. Let BS s BSX s 1, x q x 2 , x 2 , x 3 , x 4 . . . 4 Žthe usual basis with x replaced by x q x 2 .. Now let B s 1, x q x 2 , x 2 4
BX s x q x 2 , x 2 4 .
Then, since Ž x q x 2 . ⭈ 1 ⭈ Ž x q x 2 . qŽ x q x 2 . ⭈ Žy1. ⭈ x 2 qx 2 ⭈ Žy1. ⭈ Ž x q x 2 . q x2 ⭈ 1 ⭈ x2 2 q1 ⭈ 0 ⭈ Ž x q x . q 1 ⭈ Žy1. ⭈ x 2 s 0, BS , BSX are not vnn bases. One immediate bonus we get from the vnn property is a strengthening of Corollary 3.1: LEMMA 6. Let R, S, T be as in Ž1. and Ž2.. If S is a ¨ nn extension of T, and p, pX g P s R#T S such that p / 0 / pX , but ppX s 0, then < p <, < pX < ) 2. Proof. We may as well suppose that pX is a right zero divisor of minimal length. Then Lemma 4 tells us that p is pure of type Ž⭈, R . and pX is pure of type Ž R, ⭈ .. Choose a¨ s a¨ y3 r¨ y2 s¨ y1 r¨ g supp r Ž p ., with r¨ y2 , r¨ g BR and s¨ y1 g BS , to be a monomial of maximal length. ŽHere, a¨ y3 or r¨ y2 might be 1, but not s¨ y1 because of Corollary 3.1.. Define Bs sigBS : there is a monomial in supp r Ž p . that starts with a¨ y3 r¨ y2 si 4 . Let ri be the coefficient of a¨ y3 r¨ y2 si in the right R-representation of p, and let r be the coefficient of a¨ . Now suppose that < pX < s 2. Define BX s sXiX g BSX : there is a monomial in supp l Ž pX . that ends with sXiX 4 . Let riXX be the coefficient of sXiX in the left R-representation of pX . Consider
Ý a¨ y3 r¨ y2 si ri riX sXi . X
i, i
X
X
Ž 9.
98
S. PETER FARBMAN
If any of the riX are in T, then the right representation of Ž9. contains monomials of length ¨ q 1 which appear in supp r Ž ppX ., a contradiction. Thus we may assume that the coefficients of the elements of BX are in R y T. Now, because of Lemma 3Žii., we know that each ri riXX g T, and so by the vnn condition, we know that Ýsi ri riXX sXiX g S y T. Therefore, in the right representation of the expression in Ž9., there is a monomial of length ¨ y 1 starting with a¨ y3 r¨ y2 . ŽIt is conceivable that a¨ y3 r¨ y2 s 1, in which case Ž9. is all of ppX .. Because of the way that B and BX have been defined, this monomial must appear in supp r Ž ppX .. Certainly, then, ppX / 0. This contradiction is brought on by the assumption that < pX < s 2. A similar argument shows that < p < ) 2. Now we can get the ring analogue to one of the main results in w1x. THEOREM 7. If S is a ¨ nn extension of T and R is a domain that is free both as a left and as a right T-module, then P s R#T S is a domain. Proof. Suppose not. Let pX g P y 04 be a right zero-divisor of minimal length, and choose some p g P y 04 such that ppX s 0. By Lemma 6, we know that < p <, < pX < ) 2; and by Lemma 4, that p is pure of type Ž⭈, R . and pX is pure of type Ž R, ⭈ .. We may assume that BS , BSX are vnn bases. First suppose that < p <, < pX < s 3. ŽWe will see that this case is all we really need to consider.. Define
ri si , j : ri g BR y 14 , si , j g BS y 14 and there is a monomial in supp r Ž p . that starts with ri si , j 4 . Define sXiX , jX riXX 4 similarly. Let ri, j be the coefficient of ri si, j in the right R-representation of p, and riXX , jX the coefficient of sXiX , jX riXX in the left R-representation of pX . ŽNote that ri g BR , si, j g BS , riXX g BRX , sXiX , jX g BSX , but the ri, j and riXX , jX can be any non-zero elements of R.. Because of Lemma 3Žii. we know that ri, j riXX , jX s t i, j, iX , jX g T y 04 for all i, iX , j, jX . Define to be the sum of all monomials of length 1 in supp r Ž p . l R Žwith their coefficients ., and similarly define X as the part of pX that lies in R. For each i, j and iX , jX , define
⭈ riXX , jX s
X
X
X
i X
ri , j ⭈ s
X
Ý ri t iŽ i , j . q Ý rk t kŽ i , j . , Ý X
i
k
t iŽXi , j. riXX
q
Ý t kŽ i , j. rkX . X
k
X
X
Here, we insist that the r k g BR , that the r k ’s be distinct from the ri ’s, and X X similarly for the r kX X ’s. We also note that it is possible that some of the t iŽ i , j . or t iŽXi, j. could be zero.
AMALGAMATIONS OF DOMAINS
99
Now consider the product
žÝ
ri si , j ri , j ⭈
/ žÝ
i, j
X
i,j
X
riXX , jX sXiX , jX riXX s
/
Ý X
i , j, i , j
s
X
ri si , j t i , j, iX , jX sXiX , jX riXX
Ý ri si , i ri . X
i, i
X
X
Because of the vnn property, each of the si, iX g S y T. This is a sum of elements of length 3. Furthermore, if we were to write it in the right Žresp. left. representation, there would be monomials of length 3 of type Ž R, R . that begin Žresp. end. with each ri Žresp., riXX .. This product is certainly not zero. Therefore, if ppX is to be zero, there must be some other way of getting from ppX a monomial of length 3, pure of type Ž R, R .. In fact, the only other way to get such a monomial is via ⭈ riXX , jX sXiX , jX riXX or ri si, j ri, j X . Thus the sum of all length 3 monomials of type Ž R, R . is included in
Ý ri si , i riX q Ý riX , j sXi , j riX q Ý ri si , j ri , j X X
i, i
X
X
X
X
i,j
s
X
X
X
X
X
i, j
Ý ri Ý si , j ⭈ t i , j, i , j ⭈ sXi , j i, i
ž
X
j, j
q Ý ri i, i
X
q
žÝ žÝ j
q Ý ri i, i
riXX
Ž 10 .
1 ⭈ t iŽ i , j . ⭈ sXiX , jX riXX
Ž 11 .
si , j ⭈ t iŽXi , j. ⭈ 1 riXX
Ž 12 .
X
X
X
X
X
X
X
/ /
/
j
X
X
X
X
X
r k t kŽ i , j . siX , jX riX
Ý X
Ž 13 .
i,j,k
q
X
Ý i , j, k
X
X
ri si , j t kŽ iX , j . r k X .
Ž 14 .
The vnn property guarantees that Ž10. ᎐ Ž12. resolves to
Ý r i i , i r i , X
i, i
X
X
Ž 15 .
where i, iX g S y T for each pair Ž i, iX .. If we write all of the terms from Ž10. to Ž14. in the right representation, then each monomial of length 3 of type Ž R, R . from Ž13. stands alone; that is, none of the other terms start with r k , and so these terms cannot be canceled. But each such term must be canceled to get ppX s 0. Thus no
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monomials of length 3 of type Ž R, R . can come from Ž13.. A similar argument tells us that there is no such monomial in Ž14.. Therefore, we must have some monomials of length 3 and type Ž R, R . in Ž15. that cannot be canceled. This means that ppX / 0 if < p < s 3 s < pX <. Now let us return to the general case where < p < s ¨ and < pX < s w can be any length Žgreater than or equal to 3 each.. As usual, choose X a¨ y3 r¨ y2 s¨ y1 r¨ g supp r Ž p ., and rwX sXwy1 rwy2 aXwy3 g supp l Ž pX .. In fact, define a s Ý pi t i , where the pi include all monomials in supp r Ž p . that start with a¨ y3 Žwith coefficients t i .. Define aX similarly Žbased on aXwy3 .. Then a s a¨ y3 ,
aX s X aXwy 3 , where , X g P have length 3 and are of type Ž R, R .. Because of Lemma 2, if there is any hope for ppX to be zero, it must be that < aaX < F ¨ q w y 3; but since and X are of the kind discussed aboveᎏthat is, < < s < X < s 3 and both are of type Ž R, R . ᎏwe know that supp r Ž X . contains monomials of type Ž R, R .. Therefore, < aaX < G ¨ q w y 3. We may, then, assume that < aaX < s ¨ q w y 3. Also, the right representation of aaX contains a monomial of length ¨ q w y 3 that starts with a¨ y3 . Define AX s
X
X
i. Ý Ž i. aŽwy 3,
i X
i. Ž . where the sum is over all aŽwy 3 of length w y 3 andX type R, ⭈ that occur as the end of an element of supp l Ž pX ., and the Ž i. are as above. Then X
X
i. aAX s a¨ y3 Ž Ý Ž i. aŽwy3 . X
X
i. s a¨ y3 Ž Ý Ž Ž i. . aŽwy3 .. X
Of course, XeachX of the Ž i. are of length 3 and are of type Ž R, R .X , so X Ž i. . Ž i. Ž i. . Ž i. Ž Ž . Ž each a wy 3 is of length w and of type R, ⭈ . Thus Ý a wy3 is of length w and of type Ž R, ⭈ ., and supp r Ž aAX . contains monomials of length ¨ q w y 3 that start with a¨ y3 . However, because of the way that a and AX have been defined, there can be no other monomials of length X ¨ q w y 3 beginning with a¨ y3 anywhere in the product of p with p . X X Therefore, < pp < G ¨ q w y 3, and in particular, pp / 0. The vnn property is rather strong, and it is not a necessary condition; it is a simple exercise to find rings R, S, T such that R#T S is a domain with neither R nor S vnn over T Ževen without inertia .. However, it is not enough merely to assume that there are free bases as in Ž1. and Ž2..
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Consider the following example: Let F be any field, and T the free algebra on the variables t 1 , t 2 , t 3 . Let R be the F-algebra generated by x, y, z, and w with the single relation xz s yw. We make T a subring of R by setting xw s t 1 , xz Žs yw . s t 2 , and yz s t 3 . Finally, let S be the F-algebra generated by a, b, c, d, t 1 , t 2 , t 3 with the two relations at 2 d s bt 2 c and at1 c s bt 3 d. It is easy to see that R and S are domains and that they are free as left and right T-modules. In P s R#T S, < ax q by < s 2 s < wc y zd <; in particular, neither of these elements is zero. However,
Ž ax q by . ⭈ Ž wc y zd . s axwc y axzd q bywc y byzd s at1 c y at 2 d q bt 2 c y bt 3 d s 0.
Ž 16 .
What goes ‘‘wrong’’ with this example is fairly straightforward. R is not inert over T since we have that xw s t 1 g T, but there is no unit u such that xu g T. Also, S is apparently not vnn over T because of Ž16.. ŽIt is, in fact, much more difficult to prove that S is not vnn over T because we would have to prove not just that the ‘‘obvious’’ bases do not have the vnn property, but that no bases can be vnn..
REFERENCES 1. S. P. Farbman, The unique product property of groups and their amalgamated free products, J. Algebra 178 Ž1995., 962᎐990. 2. P. M. Cohn, On the free product of associative rings, II, Math. Z. 73 Ž1960., 433᎐456. 3. P. M. Cohn, On the free product of associative rings, III, J. Algebra 8 Ž1968., 376᎐383. 4. A. I. Lichtman, On unique product groups, Comm. Algebra 9 Ž1981., 533᎐551.
Amalgamations of Domains S. Peter Farbman* Department of Mathematics, Sul Ross State Uni¨ ersity, Alpine, Texas 79832 Communicated by Walter Feit Received November 10, 1998
The question of whether the amalgamated free product of two domains is itself a domain is considered. Some results analogous to group-theoretic approaches to the zero-divisor conjecture are proved. A condition is introduced that guarantees that the amalgam will be a domain. If, for example, R is a polynomial ring over the domain T, then only minimal restrictions on how T embeds in S guarantee that R#T S is a domain. 䊚 1999 Academic Press
1. INTRODUCTION One of the most famous open questions in group ring theory is the zero-divisor conjecture ŽZDC., which states that if G is any torsion-free group and F is any field, then the group algebra FG has no non-trivial zero-divisors. A unique-product Žu.p.. group is a group G such that if A, B are finite subsets of G, then there is some ordered pair Ž a0 , b 0 . g A = B such that if Ž a, b . g A = B and ab s a0 b 0 , then Ž a0 , b 0 . s Ž a, b .. U.p. groups were invented in the hope of settling the ZDC using a strictly group-theoretic property: if G is a u.p. group then Žfor any domain D . DG is a domain; this is easily proven, mostly by using the u.p. property and the fact that DG is free as a Žleft and right. D-module. ŽIn this paper, ‘‘domain’’ is taken to mean a not necessarily commutative nonzero unital ring with no non-trivial zero-divisors.. Many results about u.p. groups raise analogous questions about domains. The purpose of this paper is to discuss how we might achieve results analogous to the results in w1x, but in the context of more general rings. In particular, we intend to introduce some conditions under which we can be assured that the amalgamated free product of two domains is again a domain. * Current address: St. Mary’s College of Maryland, St. Mary’s City, Maryland 20686. 90 0021-8693r99 $30.00 Copyright 䊚 1999 by Academic Press All rights of reproduction in any form reserved.
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2. NOTATION All rings herein will be assumed to be associative with a Žnon-zero. multiplicative identity which is respected by subrings and homomorphisms. Fix R and S as domains and let T be a domain which embeds in both R and S. We identify T with its image in R and S so that R l S s T. In general, for rings R, S with a common subring T, the amalgam P s R#T S may not even contain copies of R, S, or T. To avoid this, and other possible complications, we assume that R, S are left and right free as T-modules. That is, we can write Rs
r T s [ Tr [ J J j
X
XX j ,
BR s r j : j g J 4 ,
BRX s r XjX : jX g J X 4 ,
Ž 1. Ss
s T s [ Ts [ K K k
X
XX k ,
BS s sk : k g K 4 ,
BSX s sXk X : kX g K X 4 .
Ž 2. We also assume in what follows that 1 g BR , BS , BRX , BSX . We call R and S factors of P. Under these assumptions, R, S, and T embed in P; furthermore, we can represent P as a right T-module which is free on a basis consisting of 1 and elements of the form c1 c 2 ⭈⭈⭈ c¨ ,
Ž 3.
with c i g Ž BS j BR . y 14 , c i , c iq1 never in the same factor. Elements in this basis will be called right monomials, and this representation will be called the right representation of P. We can also define the left representation because P is left free on the left monomials cX¨ cX¨ y1 ⭈⭈⭈ cX1 ,
Ž 4.
with cXi g Ž BSX j BRX . y 14 , cXi , cXiq1 never in the same factor. It is easy to see that P is also free as a left and right R Žresp., S . module. A basis as a free right R-module Žresp., S-module. is 1 together with the subset of right monomials as in Ž3. with c¨ g BS y 14 Žresp., c¨ g BR y 14.. We will refer to this as the right R-representation Žresp., the right S-representation .. The left R-representation Žresp., the left S-representation . is defined similarly. Note, though, that the terms ‘‘right representation’’ and ‘‘left representation’’ always refer to P as a free T-module.
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The length of an element p g P, denoted < p <, is defined to be the smallest n such that p is in the T-linear span of Ž R j S . n. For example, the length of the monomials in Ž3. and Ž4. is ¨ , and elements of T have length zero. This definition is left᎐right symmetric, and immediately yields that If p1 , p 2 g P , then < p1 p 2 < F < p1 < q < p 2 < .
Ž 5.
For p g P, we will use supp r Ž p . Žresp., supp l Ž p .. to denote the set of right Žresp., left. monomials that appear in the right Žresp., left. representation of p. The length of an element p g P is the length of the longest monomial in supp r Ž p . Žor supp l Ž p ... In w2, 3x, where most of the definitions here can be found, the word ‘‘height’’ is used instead of length. We use ‘‘length’’ to be consistent with the terminology used in w1x. Following w2x, we say that an element p g P is pure of type Ž⭈, R . if all monomials of maximal length in supp r Ž p . end with R; for example, if p is pure of type Ž⭈, R ., < p < s ¨ , and the monomial in Ž3. is in supp r Ž p ., then c¨ g BR y 14 . Similarly, if all monomials of maximal length in supp r Ž p . begin with S, then p is pure of type Ž S, ⭈ .. ŽWe will see shortly that purity does not depend on writing p in the right representation.. The other types of purity are defined analogously. This definition is extended so that, for example, if p is pure of type Ž R, ⭈ . and of type Ž⭈, S ., then we say that it is pure of type Ž R, S .. Elements of T cannot be pure.
3. LENGTH OF PRODUCTS In the context of the zero-divisor conjecture, much of the preceding is simple. For if we consider the group rings R s FG, S s FK, T s FH Žwith H F G, K ., we can take BR s BRX s G and BS s BSX s K; the monomials in Ž3. and Ž4. are just the elements of G# H K, thus we know that P s F Ž G# H K .. One of the main difficulties in extending to more general rings is that if P is not a group ring, then there could be some p g P such that cardŽsupp r Ž p .. / cardŽsupp l Ž p ... For example, the left representation of a right monomial might not be a left monomial. We first consider the simplest cases of products within P. LEMMA 1. Let R, S, T be domains as in Ž1. and Ž2., and let P s R#T S. Suppose that c s c1 c 2 ⭈⭈⭈ c¨ is a right monomial of length ¨ with each c i g BR j BS y 14 ; that d s d1 d 2 ⭈⭈⭈ d w is a right monomial of length w with each d i g BR j BS y 14 ; that t g T y 04 ; and that p s ctd g P. If c¨ g BS and d1 g BR , or if c¨ g BR and d1 g BS , then < p < s ¨ q w. If, in addition, e s e1 e2 ⭈⭈⭈ e¨ qw g supp r Ž p . with < e < s ¨ q w and each e i g
AMALGAMATIONS OF DOMAINS
93
BR j BS y 14 , then c i s e i for 1 F i F ¨ . The corresponding result holds for left monomials. Proof. Without loss of generality, suppose that c¨ g BS and d1 g BR . Certainly < p < F ¨ q w, because of Ž5.. We now proceed by induction on w. If w s 0, then d s 1, and the result is clear. If w s 1, then td1 f T because d1 f T and R is free as a left T-module. Let td1 have right representation Ý i d iŽ1. t i , with each d iŽ1. g BR . Then
Ý c1
⭈⭈⭈ c¨ d iŽ1. t i
i
is the right representation of p. Because td1 f T, at least one of the d iŽ1. is not 1; thus it must be that < p < s ¨ q 1 s ¨ q w, and every right monomial in supp r Ž p . begins with c1 ⭈⭈⭈ c¨ ; furthermore, every right monomial in supp r Ž p . of length ¨ q 1 is pure of type Ž⭈, R .. Note that this does not depend on the value of ¨ . Now let w ) 1 and define pwy 1 s ctd1 d 2 ⭈⭈⭈ d wy1. Suppose, inductively, that < pwy 1 < s ¨ q w y 1, and that every right monomial of length ¨ q w y 1 in supp r Ž pwy1 . begins with c1 ⭈⭈⭈ c¨ and is pure of type Ž⭈, X . Žwhere X is the factor containing d wy 1 .. We need to show that < p < s < pwy1 d w < s ¨ q w and that every right monomial in supp r Ž p . of length ¨ q w begins with c1 ⭈⭈⭈ c¨ and is pure of type Ž⭈, Y . Žwhere Y is the factor containing d w .. Choose some right monomial Ž e. e s c1 ⭈⭈⭈ c¨ d1Ž e. ⭈⭈⭈ d wy1 g supp r Ž pwy1 . ,
Ž 6.
with < e < s ¨ q w y 1. From the w s 1 caseᎏwhich is applicable here because e is pure of type Ž⭈, X . and d w f Xᎏwe know that there is some f g supp r Ž ed w . such that < f < s ¨ q w and f starts with c1 ⭈⭈⭈ c¨ d1Ž e. ⭈⭈⭈ Ž e. Ž . Žas d wy 1 s e. Because f starts with e, it is easy to see that f g supp r p there is nothing that can ‘‘cancel’’ f .. Therefore, < p < s ¨ q w. Finally, again using Ž5., we see that for every f g supp r Ž p . with < f < s ¨ q w, there is some e g supp r Ž pwy1 . as in Ž6. such that f g supp r Ž ed w .; hence every such f begins with c1 ⭈⭈⭈ c¨ , and p is pure of type Ž⭈, Y .. A symmetric argument yields the final statement. We can use this lemma and induction to conclude that the various concepts of purity can be given characterizations depending neither on choice of basis, nor on right versus left representation. For example, if p s x 1 y 2 x 3 ⭈⭈⭈ y¨ for arbitrary x i g R y T, yi g S y T, then < p < s ¨ and p is pure of type Ž R, S .. Although it gets a little more complicated, there are some things we can determine if equality does not hold in Ž5.:
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LEMMA 2. Let R, S, T be rings as in Ž1. and Ž2., and let P s R#T S. Let c s c1 c 2 ⭈⭈⭈ c¨ and d s d1 d 2 ⭈⭈⭈ d w be right monomials Ž with < c < s ¨ , < d < s w, and each c i , d i g BR j BS y 14., and t g T y 04 . If < ctd < s ¨ q w y 2 m or ¨ q w y Ž2 m y 1., then ¨ G m. Furthermore, if ¨ ) m, and e s e1 e2 e3 ⭈⭈⭈ g supp r Ž ctd . Ž such that e i g Ž BR j BS . y 14 for all i, and e i , e iq1 not in the same factor . with < e < s < ctd <, then c i s e i for 1 F i F ¨ y m. The corresponding result for left monomials also holds. Proof. If ¨ s 0, then c s 1, and the result is obvious; thus we may assume that ¨ G 1. Without loss of generality, let c¨ g BS y 14 . Now we proceed by induction on w. If w s 0, then d s 1 and the result is once again obvious. Suppose, inductively, that the lemma holds for w F k, and that d s d1 ⭈⭈⭈ d kq 1. If d1 g BR , then we are done by Lemma 1. So we may assume that d1 g BS y 14 . If c¨ td1 s ˜t g T, then ctd s c1 ⭈⭈⭈ c¨ y1 ˜td 2 ⭈⭈⭈ d kq1 , and we are done by induction. If, on the other hand, c¨ td1 g S y T, then by the comment after Lemma 1, < ctd < s ¨ q k s ¨ q Ž k q 1. y Ž2 ⭈ 1 y 1.. That is, m s 1 F ¨ . If ¨ ) m s 1 and c¨ td1 g S y T, then use the right representation to write c¨ td1 s Ý i d iŽ1. t i , and t i d 2 d 3 ⭈⭈⭈ d kq1 s
Ý f i j tXi j , j
where the f i j are right monomials of length no more than k; those that are of length k are of type Ž R, X . Žwhere X is the factor that contains d kq 1 .. Then ctd s
Ý Ž c1
⭈⭈⭈ c¨ y1 d iŽ1. f i j t i j . .
i, j
If e g supp r Ž ctd . with < e < s < ctd < s ¨ q k, then there is some i, j such that d iŽ1. g BS y 14 ; < f i j < s k; f i j is of type Ž R, ⭈ .; and e s Ž c1 c 2 ⭈⭈⭈ c¨ y1 d iŽ1. .Ž f i j .. In particular, e begins with c1 ⭈⭈⭈ c¨ y1 . A symmetric argument yields the final statement. We can now use these facts about monomials to examine the products of arbitrary elements of P. ŽMost of the remainder of this section consists of restatements of the results from w2, 3x that we wish to extend. They are included for completeness and to help prove Theorem 7, but most of the proofs are omitted..
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LEMMA 3.
95
Let R, S, T be as in Ž1. and Ž2., P s R#T S, and p, pX g P.
Ži. Ž cf. w2, Theorem 2.1x.. If < ppX < - < p < q < pX <, then either p is pure of type Ž⭈, R . and pX is pure of type Ž R, ⭈ ., or p is pure of type Ž⭈, S . and pX is pure of type Ž S, ⭈ .. Žii. Ž cf. w2, Theorem 2.3x.. Suppose that < p < s ¨ ; < pX < s w; < ppX < - ¨ qw y 1; a¨ y1 r¨ g supp r Ž p ., and rwX aXwy1 g supp l Ž pX . are monomials of maximal length Ž with r¨ g BR , rwX g BRX .; r is the coefficient of a¨ y1 in the right R-representation of p; and r X is the coefficient of aXwy 1 in the left R-representation of pX . Then rr X g T. Proof. Žii. ŽThe proof of Žii. is included as it provides motivation for Section 4.. Suppose that < ppX < - < p < q < pX < y 1. Because P is free as a left and right T-module, it follows that ¨ , w ) 0. Certainly, neither r nor r X are in T. Suppose that rr X g R y T. Then we know that < a¨ y1 rr X aXwy1 < s ¨ q w y 1 by the comment after Lemma 1. If e g supp r Ž a¨ y1 rr X aXwy1 ., with < e < s ¨ q w y 1, then e begins with a wy1 by Lemma 2. Since, by Ž5. and Lemma 2, there is no other way for e to appear in pr X aXwy 1 Žthat might cancel this appearance., it must be that e g supp r Ž pr X aXwy1 .. In particular, < pr X aXwy1 < G ¨ q w y 1. Then we haveᎏin a similar mannerᎏthat if f X g supp l Ž pr X aXwy1 . with < f X < G ¨ q w y 1, then f X ends with aXwy 1 and so it must be that f X g supp l Ž ppX .. Thus < ppX < G ¨ q w y 1, a contradiction. We also get: COROLLARY 3.1. Let R, S, T be as in Ž1. and Ž2.. If p, pX g P s R#T S such that p / 0 / pX , but ppX s 0, then < p <, < pX < ) 1. Proof. Suppose not. Because P is a free Žleft and right. T-module and T is a domain, it is certainly true that < p <, < pX < ) 0. Without loss of generality, we may assume that < p < s 1. By Lemma 3Ži., p g R or p g S. But P is free as a left R- and S-module, and both are domains, so we have reached the desired contradiction. Following w3x, we make the following definition: if A is a ring and B is a subring which is also a domain, then A is said to be an inert extension of B provided that: whenever a, aX g A y 04 such that aaX g B, there is some unit u g A such that au, uy1 aX g B. ŽNote that this implies that A is a domain.. For example, if D is a division ring and A is a D-ring which is also a domain, then A is an inert extension of D. LEMMA 4 Žcf. w3, Theorem 2.2x.. Let R, S, T be as in Ž1. and Ž2.. If p, pX g P s R#T S such that < ppX < - < p < q < pX < y 1 and S is an inert extension of T, then either p is pure of type Ž⭈, R . and pX is pure of type Ž R, ⭈ ., or there is a unit u g UŽ S . with < pu < - < p < and < uy1 pX < - < pX <.
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S. PETER FARBMAN
THEOREM 5 Žcf. w3, Corollary Following Theorem 2.2x.. Let R, S, T be rings as in Ž1. and Ž2.. Suppose that T is a domain and R, S are inert extensions of T. Then P s R#T S is an inert extension of R and S. In particular, P is a domain.
4. THE vnn PROPERTY FOR RINGS Our goal is to get results that parallel those of the preceding section in the same way that the results of w1x parallel those of w4x. Let R, S, T be domains as in Ž1. and Ž2., and let S be an inert extension of T. Suppose that p, pX g P y 04 such that ppX s 0. By Lemma 4, we may assume that p is pure of type Ž⭈, R . and pX is pure of type Ž R, ⭈ .. Taking p, pX , a¨ y1 , aXwy1 , r, r X as in Lemma 3Žii. and a¨ y1 s a¨ y3 r¨ y2 s¨ y1 , X aXwy 1 s sXwy1 rwy2 aXwy3 , we have ppX s a¨ y3 r¨ y2 s¨ y1 rr X sXwy1wy2 r X aXwy3 q other terms.
Ž 7.
As in the proof of Lemma 3Žii., for ppX to be zero Žor, more generally, for < ppX < to be zero., it must be that rr X s ˜t g T. We wish to create conditions X under which at least one term like a¨ y3 r¨ y2 s¨ y1 rr X sXwy1 rwy2 aXwy3 cannot be canceled by the ‘‘other’’ terms. If S, T are domains as in Ž2., we say BS , BSX are vnn bases provided that, for any Žnon-empty. finite subsets B s si : i g I 4 : BS and BX s sXiX : iX g I X 4 : BSX with B / 14 / BX , and elements t i, iX : i g I, iX g I X 4 : T such that t i, iX / 0 if si / 1 / sXiX , it must be that s Ý i, iX si t i, iX sXiX / 0. ŽWhat we really need is that f T, but it suffices to assume that it is non-zero. For if g T, then we can add, if necessary, si s 1 to B and sXi s 1 to BX , and alter the choice of t i, iX for si s 1 and sXiX s 1 to force s 0.. If S admits vnn bases, then we say that S is a vnn extension of T. Note that by this definition, if S is a vnn extension of T and s, sX g S such that ssX g T, then s, sX g T. In particular, if S is a vnn extension of T, then S is an inert extension of T. It is clear that if H - G are groups, then the usual bases of the group ring DG as a free DH-moduleᎏright and left coset representatives ᎏare never vnn bases; for if g g G y H, then g ⭈ 1 ⭈ gy1 q g ⭈ 0 ⭈ 1 q 1 ⭈ 0 ⭈ gy1 q 1 ⭈ Ž y1 . ⭈ 1 s 0.
Ž 8.
ŽIn fact, it is not difficult to see that because of Ž8., DG is never a vnn extension of DH.. The most obvious situation where the vnn condition holds is when S is a polynomial ring over T. For let S s T w x x and BS s x i X : i g ⺪ G 0 4 s BSX , and suppose that B s x i : i g I 4 / 14 / BX s x i : iX g I X 4 ; then
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97
degŽÝsi t i, iX sXiX . s i max q iXmax , regardless of the choice of the t i, iX , as long as t i ma x , iXmax / 0. The same argument shows that skew polynomial rings, or more generally, Ore extensions, are vnn extensions. It should be noted that if S is a vnn extension of T, then it is not necessarily true that all bases are vnn bases. For example, let T s ⺪ and S s ⺪w x x, the usual integer polynomial ring. Let BS s BSX s 1, x q x 2 , x 2 , x 3 , x 4 . . . 4 Žthe usual basis with x replaced by x q x 2 .. Now let B s 1, x q x 2 , x 2 4
BX s x q x 2 , x 2 4 .
Then, since Ž x q x 2 . ⭈ 1 ⭈ Ž x q x 2 . qŽ x q x 2 . ⭈ Žy1. ⭈ x 2 qx 2 ⭈ Žy1. ⭈ Ž x q x 2 . q x2 ⭈ 1 ⭈ x2 2 q1 ⭈ 0 ⭈ Ž x q x . q 1 ⭈ Žy1. ⭈ x 2 s 0, BS , BSX are not vnn bases. One immediate bonus we get from the vnn property is a strengthening of Corollary 3.1: LEMMA 6. Let R, S, T be as in Ž1. and Ž2.. If S is a ¨ nn extension of T, and p, pX g P s R#T S such that p / 0 / pX , but ppX s 0, then < p <, < pX < ) 2. Proof. We may as well suppose that pX is a right zero divisor of minimal length. Then Lemma 4 tells us that p is pure of type Ž⭈, R . and pX is pure of type Ž R, ⭈ .. Choose a¨ s a¨ y3 r¨ y2 s¨ y1 r¨ g supp r Ž p ., with r¨ y2 , r¨ g BR and s¨ y1 g BS , to be a monomial of maximal length. ŽHere, a¨ y3 or r¨ y2 might be 1, but not s¨ y1 because of Corollary 3.1.. Define Bs sigBS : there is a monomial in supp r Ž p . that starts with a¨ y3 r¨ y2 si 4 . Let ri be the coefficient of a¨ y3 r¨ y2 si in the right R-representation of p, and let r be the coefficient of a¨ . Now suppose that < pX < s 2. Define BX s sXiX g BSX : there is a monomial in supp l Ž pX . that ends with sXiX 4 . Let riXX be the coefficient of sXiX in the left R-representation of pX . Consider
Ý a¨ y3 r¨ y2 si ri riX sXi . X
i, i
X
X
Ž 9.
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S. PETER FARBMAN
If any of the riX are in T, then the right representation of Ž9. contains monomials of length ¨ q 1 which appear in supp r Ž ppX ., a contradiction. Thus we may assume that the coefficients of the elements of BX are in R y T. Now, because of Lemma 3Žii., we know that each ri riXX g T, and so by the vnn condition, we know that Ýsi ri riXX sXiX g S y T. Therefore, in the right representation of the expression in Ž9., there is a monomial of length ¨ y 1 starting with a¨ y3 r¨ y2 . ŽIt is conceivable that a¨ y3 r¨ y2 s 1, in which case Ž9. is all of ppX .. Because of the way that B and BX have been defined, this monomial must appear in supp r Ž ppX .. Certainly, then, ppX / 0. This contradiction is brought on by the assumption that < pX < s 2. A similar argument shows that < p < ) 2. Now we can get the ring analogue to one of the main results in w1x. THEOREM 7. If S is a ¨ nn extension of T and R is a domain that is free both as a left and as a right T-module, then P s R#T S is a domain. Proof. Suppose not. Let pX g P y 04 be a right zero-divisor of minimal length, and choose some p g P y 04 such that ppX s 0. By Lemma 6, we know that < p <, < pX < ) 2; and by Lemma 4, that p is pure of type Ž⭈, R . and pX is pure of type Ž R, ⭈ .. We may assume that BS , BSX are vnn bases. First suppose that < p <, < pX < s 3. ŽWe will see that this case is all we really need to consider.. Define
ri si , j : ri g BR y 14 , si , j g BS y 14 and there is a monomial in supp r Ž p . that starts with ri si , j 4 . Define sXiX , jX riXX 4 similarly. Let ri, j be the coefficient of ri si, j in the right R-representation of p, and riXX , jX the coefficient of sXiX , jX riXX in the left R-representation of pX . ŽNote that ri g BR , si, j g BS , riXX g BRX , sXiX , jX g BSX , but the ri, j and riXX , jX can be any non-zero elements of R.. Because of Lemma 3Žii. we know that ri, j riXX , jX s t i, j, iX , jX g T y 04 for all i, iX , j, jX . Define to be the sum of all monomials of length 1 in supp r Ž p . l R Žwith their coefficients ., and similarly define X as the part of pX that lies in R. For each i, j and iX , jX , define
⭈ riXX , jX s
X
X
X
i X
ri , j ⭈ s
X
Ý ri t iŽ i , j . q Ý rk t kŽ i , j . , Ý X
i
k
t iŽXi , j. riXX
q
Ý t kŽ i , j. rkX . X
k
X
X
Here, we insist that the r k g BR , that the r k ’s be distinct from the ri ’s, and X X similarly for the r kX X ’s. We also note that it is possible that some of the t iŽ i , j . or t iŽXi, j. could be zero.
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99
Now consider the product
žÝ
ri si , j ri , j ⭈
/ žÝ
i, j
X
i,j
X
riXX , jX sXiX , jX riXX s
/
Ý X
i , j, i , j
s
X
ri si , j t i , j, iX , jX sXiX , jX riXX
Ý ri si , i ri . X
i, i
X
X
Because of the vnn property, each of the si, iX g S y T. This is a sum of elements of length 3. Furthermore, if we were to write it in the right Žresp. left. representation, there would be monomials of length 3 of type Ž R, R . that begin Žresp. end. with each ri Žresp., riXX .. This product is certainly not zero. Therefore, if ppX is to be zero, there must be some other way of getting from ppX a monomial of length 3, pure of type Ž R, R .. In fact, the only other way to get such a monomial is via ⭈ riXX , jX sXiX , jX riXX or ri si, j ri, j X . Thus the sum of all length 3 monomials of type Ž R, R . is included in
Ý ri si , i riX q Ý riX , j sXi , j riX q Ý ri si , j ri , j X X
i, i
X
X
X
X
i,j
s
X
X
X
X
X
i, j
Ý ri Ý si , j ⭈ t i , j, i , j ⭈ sXi , j i, i
ž
X
j, j
q Ý ri i, i
X
q
žÝ žÝ j
q Ý ri i, i
riXX
Ž 10 .
1 ⭈ t iŽ i , j . ⭈ sXiX , jX riXX
Ž 11 .
si , j ⭈ t iŽXi , j. ⭈ 1 riXX
Ž 12 .
X
X
X
X
X
X
X
/ /
/
j
X
X
X
X
X
r k t kŽ i , j . siX , jX riX
Ý X
Ž 13 .
i,j,k
q
X
Ý i , j, k
X
X
ri si , j t kŽ iX , j . r k X .
Ž 14 .
The vnn property guarantees that Ž10. ᎐ Ž12. resolves to
Ý r i i , i r i , X
i, i
X
X
Ž 15 .
where i, iX g S y T for each pair Ž i, iX .. If we write all of the terms from Ž10. to Ž14. in the right representation, then each monomial of length 3 of type Ž R, R . from Ž13. stands alone; that is, none of the other terms start with r k , and so these terms cannot be canceled. But each such term must be canceled to get ppX s 0. Thus no
100
S. PETER FARBMAN
monomials of length 3 of type Ž R, R . can come from Ž13.. A similar argument tells us that there is no such monomial in Ž14.. Therefore, we must have some monomials of length 3 and type Ž R, R . in Ž15. that cannot be canceled. This means that ppX / 0 if < p < s 3 s < pX <. Now let us return to the general case where < p < s ¨ and < pX < s w can be any length Žgreater than or equal to 3 each.. As usual, choose X a¨ y3 r¨ y2 s¨ y1 r¨ g supp r Ž p ., and rwX sXwy1 rwy2 aXwy3 g supp l Ž pX .. In fact, define a s Ý pi t i , where the pi include all monomials in supp r Ž p . that start with a¨ y3 Žwith coefficients t i .. Define aX similarly Žbased on aXwy3 .. Then a s a¨ y3 ,
aX s X aXwy 3 , where , X g P have length 3 and are of type Ž R, R .. Because of Lemma 2, if there is any hope for ppX to be zero, it must be that < aaX < F ¨ q w y 3; but since and X are of the kind discussed aboveᎏthat is, < < s < X < s 3 and both are of type Ž R, R . ᎏwe know that supp r Ž X . contains monomials of type Ž R, R .. Therefore, < aaX < G ¨ q w y 3. We may, then, assume that < aaX < s ¨ q w y 3. Also, the right representation of aaX contains a monomial of length ¨ q w y 3 that starts with a¨ y3 . Define AX s
X
X
i. Ý Ž i. aŽwy 3,
i X
i. Ž . where the sum is over all aŽwy 3 of length w y 3 andX type R, ⭈ that occur as the end of an element of supp l Ž pX ., and the Ž i. are as above. Then X
X
i. aAX s a¨ y3 Ž Ý Ž i. aŽwy3 . X
X
i. s a¨ y3 Ž Ý Ž Ž i. . aŽwy3 .. X
Of course, XeachX of the Ž i. are of length 3 and are of type Ž R, R .X , so X Ž i. . Ž i. Ž i. . Ž i. Ž Ž . Ž each a wy 3 is of length w and of type R, ⭈ . Thus Ý a wy3 is of length w and of type Ž R, ⭈ ., and supp r Ž aAX . contains monomials of length ¨ q w y 3 that start with a¨ y3 . However, because of the way that a and AX have been defined, there can be no other monomials of length X ¨ q w y 3 beginning with a¨ y3 anywhere in the product of p with p . X X Therefore, < pp < G ¨ q w y 3, and in particular, pp / 0. The vnn property is rather strong, and it is not a necessary condition; it is a simple exercise to find rings R, S, T such that R#T S is a domain with neither R nor S vnn over T Ževen without inertia .. However, it is not enough merely to assume that there are free bases as in Ž1. and Ž2..
AMALGAMATIONS OF DOMAINS
101
Consider the following example: Let F be any field, and T the free algebra on the variables t 1 , t 2 , t 3 . Let R be the F-algebra generated by x, y, z, and w with the single relation xz s yw. We make T a subring of R by setting xw s t 1 , xz Žs yw . s t 2 , and yz s t 3 . Finally, let S be the F-algebra generated by a, b, c, d, t 1 , t 2 , t 3 with the two relations at 2 d s bt 2 c and at1 c s bt 3 d. It is easy to see that R and S are domains and that they are free as left and right T-modules. In P s R#T S, < ax q by < s 2 s < wc y zd <; in particular, neither of these elements is zero. However,
Ž ax q by . ⭈ Ž wc y zd . s axwc y axzd q bywc y byzd s at1 c y at 2 d q bt 2 c y bt 3 d s 0.
Ž 16 .
What goes ‘‘wrong’’ with this example is fairly straightforward. R is not inert over T since we have that xw s t 1 g T, but there is no unit u such that xu g T. Also, S is apparently not vnn over T because of Ž16.. ŽIt is, in fact, much more difficult to prove that S is not vnn over T because we would have to prove not just that the ‘‘obvious’’ bases do not have the vnn property, but that no bases can be vnn..
REFERENCES 1. S. P. Farbman, The unique product property of groups and their amalgamated free products, J. Algebra 178 Ž1995., 962᎐990. 2. P. M. Cohn, On the free product of associative rings, II, Math. Z. 73 Ž1960., 433᎐456. 3. P. M. Cohn, On the free product of associative rings, III, J. Algebra 8 Ž1968., 376᎐383. 4. A. I. Lichtman, On unique product groups, Comm. Algebra 9 Ž1981., 533᎐551.