McAdam Swan, Unique Comaximal Factorization 2004

a

e

le

nit

ts, of

, welasilynon-

).

Journal of Algebra 276 (2004) 180–192

www.elsevier.com/locate/jalgebr

Unique comaximal factorization

Stephen McAdama,∗ and Richard G. Swanb

a Department of Mathematics, The University of Texas at Austin, Austin, TX 78712-1082, USAb Department of Mathematics, The University of Chicago, Chicago, IL 60637, USA

Received 9 December 2002

Communicated by Susan Montgomery

Introduction

We study an analogue to unique factorization domains. In a domainR, we consider thefollowing concepts.

Definitions.

(i) Let b be a nonzero non-unit element ofR. We will say b is a pseudo-irreduciblelement if it is impossible to factorb as b = cd with c and d comaximal (i.e.,(c, d) = R) non-units.

(ii) Let b be a nonzero non-unit element ofR. We will call b = b1b2 · · ·bm a completecomaximal factorization ofb if the bi are pairwise comaximal pseudo-irreducibelements.

(iii) We will call R a comaximal factorization domain (CFD) if any nonzero non-uelementb of R has a complete comaximal factorization.

(iv) We will call R a unique comaximal factorization domain (UCFD) ifR is a CFDin which complete comaximal factorizations are unique (up to order and unicourse).

The main result of Section 1 (Theorem 1.7) is thatR is a UCFD if and only ifR is a CFD in which every 2-generated invertible ideal is principal. In Section 2prove that the polynomial domainR[X] is a UCFD if and only ifR is a semi-normaUCFD. Section 3 shows the class of UCFD’s is fairly large by showing they are eproduced via pullbacks. Section 4 gives an example of a UCFD in which there is a

* Corresponding author.E-mail addresses:[email protected] (S. McAdam), [email protected] (R.G. Swan

0021-8693/$ – see front matter 2004 Elsevier Inc. All rights reserved.doi:10.1016/j.jalgebra.2004.02.007

S. McAdam, R.G. Swan / Journal of Algebra 276 (2004) 180–192 181

ened).result5

al

deals

ent

ze

, and

ais

, butis a

principal invertible ideal (showing the statement of Theorem 1.7 cannot be strengthSection 5 studies an ideal-theoretic version of comaximal factorizations, extending aof E. Noether. Throughout this work,R is a commutative domain, except in SectionwhereR will be a commutative ring.

1. Unique comaximal factorization domains

We begin with a lemma showing that CFD’s are common.

Lemma 1.1.

(i) If every nonzero non-unit element ofR has only finitely many prime ideals minimover it, thenR is a CFD.

(ii) If every nonzero non-unit element is contained in only finitely many maximal ithenR is a CFD.

(iii) If R is Noetherian, thenR is a CFD.

Proof. Suppose the hypothesis of (i) holds, and for any nonzero non-unit elemaof R, let min(a) be the nonempty finite set of prime ideals minimal overa. SupposeRis not a CFD. Then there is a nonzero non-unit elementa of R which does not havea complete comaximal factorization. Among all sucha, consider one such that the siof min(a) is minimal. If a cannot be factored into two comaximal non-units, thena is acomplete comaximal factorization of itself, a contradiction. Thus writea = bc with b andc

comaximal non-units. The comaximality ofb andc shows that min(a) is the disjoint unionof min(b) and min(c). As b andc are non-units, both these latter sets are nonemptyso both are proper subsets of min(a), and hence are both smaller than min(a). Therefore,both b and c have complete comaximal factorizations, and their concatenation formscomplete comaximal factorization ofa, again giving a contradiction. The proof of (ii)similar. Finally, (iii) follows from (i). �

Aside from Lemma 1.1, we will not concern ourselves with the existence of CFD’swill only be concerned when a given CFD is in fact a UCFD. The following conceptmajor tool in that endeavor.

Definition. A nonzero idealI of R is called anS-ideal if there are elementsa andc suchthatI = (a, c) = (a2, c).

Lemma 1.2. Let a and c be in R. The following are equivalent.

(i) (a, c) = (a2, c).(ii) There is an elementb with (a, b) = R and withc dividingab.

182 S. McAdam, R.G. Swan / Journal of Algebra 276 (2004) 180–192

nt

me

at

Proof. Suppose (i) holds, and writea = a2r + cd . Let b = 1 − ar, and noteab = cd .Conversely, if (ii) holds, thena ∈ aR = a(a, b) = (a2, ab) ⊆ (a2, c). Thus (a, c) =(a2, c). �Definition. Let c be a nonzero non-unit element ofR. We sayc is a pseudo-prime elemeif for every pair of elementsa andb in R, if c dividesab and if (a, b) = R, then eithercdividesa or c dividesb.

Lemma 1.3. Concerning the following three statements,(i) implies (ii) and (ii) implies(iii) .

(i) EveryS-ideal ofR is principal.(ii) Every pseudo-irreducible element ofR is a pseudo-prime element.(iii) If x1x2 · · ·xn = y1y2 · · ·ym are two complete comaximal factorizations of so

element ofR, then these factorizations are identical up to order and units.

Proof. (i) ⇒ (ii). Suppose (i) holds. Letc be pseudo-irreducible and supposec dividesab, with (a, b) = R. By Lemma 1.2,(a, c) = (a2, c). As c �= 0, (a, c) is anS-ideal. Bysymmetry,(b, c) is also anS-ideal. By (i), we may write(a, c) = (x) and(b, c) = (y). As(a, b) = R, (x) and(y) are comaximal, so that(c) ⊆ (x) ∩ (y) = (x)(y) ⊆ (c) (usingc

dividesab). Thusc equals a unit timesxy. As c is pseudo-irreducible, we must have thone ofx or y is a unit, so that c is comaximal to one ofa or b. Sincec dividesab and iscomaximal to one of them, it divides the other.

(ii) ⇒ (iii). Suppose (ii) holds, and letx1x2 · · ·xn = y1y2 · · ·ym be two completecomaximal factorizations of some element ofR. By (ii), x1 divides someyj with 1 �j � m. Re-ordering, we may assumex1 dividesy1. By symmetry,y1 divides somexi

with 1 � i � n. As x1 dividesy1, which in turn dividesxi , we see thatx1 andxi are notcomaximal. It follows thati must equal 1. We now see thatx1 andy1 differ by a unit.Cancellation and induction now show that (iii) holds.�Lemma 1.4. SupposeI = (e, c) = (e2, c). Then there are elementsa, b, andd such thatI = (a, c) = (a2, c), (a, b) = R = (c, d), andab = cd .

Proof. Let a = e + cz with z to be determined, and note thatI = (a, c) = (a2, c). Writee = e2r + cs, and letb = 1 − ar. Clearly (a, b) = R, and ab = a − a2r = cd whered = s + z − 2erz − rcz2. We will now selectz so as to make(c, d) = R. Note thatd ≡ s+z(1−2er)modc. Aser = (er)2+csr ≡ (er)2 modc, we see(1−2er)2 ≡ 1 modc.Specifyingz = (1− s)(1− 2er), we seed ≡ 1 modc, and it follows that(c, d) = R. �Lemma 1.5.

(a) S-ideals are invertible.(b) Any2-generated invertible ideal is isomorphic to anS-ideal.


e

ince

nown

rse,

Proof. (a) Let I = (a, c) = (a2, c) �= (0) be anS-ideal. If a = 0, clearlyI is invertible.Otherwise, writea = a2r + cd , and note thatI (a, d) = (a, c)(a, d) = (a).

(b) If J = (x, y) is invertible, then there existχ,ψ ∈ J−1 with χx + ψy = 1. We mayassumeχ �= 0. As χx = (χx)2 + ψx(χy), we see thatχJ = (χx,χy) = ((χx)2, χy) isanS-ideal. �Lemma 1.6. Suppose(x, y) = R. For c ∈ R, (xy, c) = (x, c)(y, c).

Proof. Clearly (x, c)(y, c) ⊆ (xy, c). Since(x, c) and(y, c), are comaximal,c ∈ (x, c) ∩(y, c) = (x, c)(y, c), and the lemma follows. �Theorem 1.7. LetR be a CFD. The following are equivalent.

(a) R is a UCFD.(b) Every pseudo-irreducible element is a pseudo-prime element.(c) Every2-generated invertible ideal is principal.(d) EveryS-ideal is principal.

Proof. (d) ⇔ (c). This follows from Lemma 1.5.(d) ⇒ (b) ⇒ (a). This is given by Lemma 1.3 and the fact thatR is a CFD.(a) ⇒ (d). SupposeR is a UCFD and letI be anS-ideal. By Lemma 1.4, there ar

elementsa, b, c andd , with I = (a, c), with ab = cd , and with(a, b) = R = (c, d). If a

or c is zero or a unit, thenI is principal, and we are done. Ifb or d is zero, thena or c is aunit, and we are done. If eitherb or d is a unit, then eitherc dividesa or a dividesc, andwe are done. Thus assumea, b, c, andd are nonzero non-units. Leta = ∏

ai , b = ∏bj ,

c = ∏ch, andd = ∏

dk be the complete comaximal factorizations of these elements. S(∏

ai)(∏

bj ) and(∏

ch)(∏

dk) are both complete comaximal factorizations ofab = cd ,we see eachai either dividesc or dividesd (and is comaximal toc). By Lemma 1.6,I = ∏

(ai, c), and each factor in this product equals either(ai) or R. �Remark. The equivalence of (a) and (b) in Theorem 1.7 is our analogue of the well kfact thatR is a UFD if and only if every nonzero non-unit element ofR is a product ofirreducible elements (i.e.,R is atomic), and every irreducible element is prime. Of couCFD’s are our analogue of atomic domains.

Corollary 1.8. A UFD is an UCFD.

Proof. Lemma 1.1(i) showsR is a CFD. As every invertible ideal ofR is principal,Theorem 1.7 showsR is a UCFD. �Corollary 1.9. Let R be a Prüfer domain which is a CFD. ThenR is a UCFD if and onlyif R is a Bezout domain.


l. Inws

ercipal.tion

y

ly

heubsets..1.7.only

al.

oimal

t

r

l

Proof. If R is a Bezout domain, then every finitely generated ideal is principaparticular, every 2-generated invertible ideal is principal, and so Theorem 1.7 shoR

is a UCFD.Conversely, supposeR is a UCFD. Consider a finitely generated idealI = (a1, a2, a3,

. . . , an). We must showI is principal. Now(a1, a2) is a finitely generated ideal in a Prüfdomain, and so is invertible. Being a 2-generated invertible ideal in a UCFD, it is prinTherefore the two generatorsa1 anda2 can be replaced by a single generator. IterashowsI is principal. �Corollary 1.10. Suppose every nonzero ideal ofR is contained in only finitely manmaximal ideals( for instance, ifR is a 1-dimensional Noetherian domain). ThenR is aUCFD if and only if every invertible ideal is principal. In particular, a domain with onfinitely many maximal ideals is a UCFD.

Proof. LetmspecR be the set of maximal ideals ofR with the standard spec topology. Thypothesis implies that the closed sets are the whole space together with all finite sBy the Forster–Swan theorem [8, Theorem 1], every invertible ideal ofR is 2-generatedSince Lemma 1.1(ii) tells usR is a CFD, the stated equivalence follows from TheoremThe final statement is now immediate from the well known fact that in a domain withfinitely many maximal ideals, all invertible ideals are principal.�

In Section 4, we give an example of a UCFD having a non-principal invertible ideFor further results related toS-ideals, we refer the reader to [5].

2. Polynomial rings

In this section we explore when a polynomial ring overR is a UCFD.R continues to bea domain.

Definition. We will say R is a bounded CFD ifR is a CFD in which for every nonzernon-unit elementb of R, there is an upper bound to the lengths of all complete comaxfactorizations ofb.

Lemma 2.1. LetR be a bounded CFD. ThenR[X] is a bounded CFD.

Proof. As R is a bounded CFD, for a nonzero non-unit elementb let L(b) be the leasupper bound of the lengths of all complete comaximal factorizations ofb. Also letL(b) = 0whenb = 0 orb is a unit.

Let g(X) be a nonzero non-unit element ofR[X]. We claim that ifL(g(0)) = 0 andif g(X) = ch(X) with c ∈ R and with(c,h(X)) = R[X], thenc must be a unit. (Clearlyc �= 0.) Sinceg(0) = ch(0), we see that the claim is true ifg(0) is a unit. On the othehand, ifg(0) = 0, thenh(0) = 0. Since(c,h(0)) = R, we again see thatc is a unit.

We now claim that for any nonzero non-unitg(x) ∈ R[X], the length of any comaximafactorization ofg(X) does not exceed degg(X) + L(g(0)). This will show thatR[X] is a


s.

ve

at

w

t

w

ining

bounded CFD, since among all possible comaximal factorizations ofg(X), our bound willassure there is one of maximal length, and that one will be complete.

Consider a complete comaximal factorizationg(X) = d1d2 · · ·drk1(X)k2(X) · · ·ks(X)

where thedi are constant (non-unit) factors and thekj (X) are nonconstant factorObviously s � degg(X), and so it will suffice to showr � L(g(0)). Letting k(X) =∏

kj (X) andd = ∏di (or d = 1 if r = 0), we haveg(X) = dk(X). SupposeL(g(0)) = 0.

The first claim shows thatd is a unit. Since anydi which exist are non-units, we must har = 0 = L(g(0)). Next, supposeL(g(0)) > 0. Asg(0) = dk(0), and(d, k(0)) = R, we seeL(g(0)) � L(d) + L(k(0)). Sinced1d2 · · ·dr is a complete comaximal factorization ofd ,we haver � L(d) � L(g(0)), as desired. �Lemma 2.2. Let R ⊆ T be domains. Suppose there are elementsα, β , δ andρ in T suchthatαβ �= 0, δρ, αδ, andβρ are all in R. Furthermore, supposeαδ andβρ are comaximalin R. Finally, suppose no associate ofα in T lies inR. ThenR is not a UCFD.

Proof. As (αδ,βρ)R = R, and asαβ divides (αδ)(βρ) in R, Lemma 1.2 showsI =(αδ,αβ)R is anS-ideal. We now showI is not principal. Suppose to the contrary thx ∈ R with I = xR. ThenxT = IT = (αT )((δ,β)T ). We claim(δ,β)T = T . If not, thenthere is a maximal idealN of T containing bothδ andβ . ThusN ∩ R contains bothαδ

andβρ, contradicting that these two elements are comaximal inR. The claim now showsxT = αT . Thusx andα are associates inT , contradicting the hypothesis. SinceI is nowseen to be a non-principalS-ideal ofR, Theorem 1.7 shows thatR is not a UCFD. �Theorem 2.3. Let X1,X2, . . . ,Xn (n � 1) be indeterminates overR. ThenR[X1,X2,

. . . ,Xn] is a UCFD if and only ifR is a semi-normal UCFD.

Proof. If R[X] is a UCFD, it is straightfoward to verify thatR is also a UCFD.Furthermore, ifR is semi-normal, it is easily seen that so isR[X]. These comments shoit will suffice to do the casen = 1.

First, assumeR[X] is a UCFD. We already noted thatR will be a UCFD, which we musshow is semi-normal. Suppose that is false. Then by definition, ifT is the integral closureof R there is ab ∈ T − R with b2 andb3 both inR. (We now modify a construction from[4, Theorem 1.5].)

Let α = bX + 1, δ = bX − 1, andβ = ρ = b2 in T [X]. It is trivial to see thatαβ �= 0,δρ, αδ, andβρ are all inR[X] andαδ andβρ are comaximal inR[X]. Now supposeu isa unit inT [X] (and hence inT ) such thatuα ∈ R[X]. Thenu(bX + 1) ∈ R[X] impliesu

andub are both inR. As T is integral overR, and asu ∈ R is a unit inT , u is also a unitin R. Thusub ∈ R impliesb ∈ R, which is a contradiction. Therefore, no associate ofα inT [X] lies inR[X]. By Lemma 2.2,R[X] is not a UCFD. This contradiction showsR mustbe semi-normal.

Conversely, supposeR is a semi-normal UCFD. Since Lemma 2.1 showsR[X] is aCFD, in order to showR[X] is a UCFD Theorem 1.7 tells us it will suffice to shoevery 2-generatedinvertible ideal ofR[X] is principal. Now for any domainR, there isan injection of Pic(R) (the ideal class group ofR) into Pic(R[X]), as follows. Given aclass in Pic(R), select an idealI in that class, and send that class to the class conta


to.

.

s, by

d

l

t

IR[X] in Pic(R[X]). Furthermore, ifR is semi-normal [4] shows this map is also onTherefore, ifH is a 2-generated invertible ideal ofR[X], then for some invertible idealIof R, H andIR[X] are in the same class in Pic(R[X]). In particular, this meansIR[X] isalso 2-generated. SayIR[X] = (f (X),g(X)). SinceI = IR[X] ∩ R, we easily see thatIis generated by the constant coefficients off (X) andg(X), showingI is also 2-generatedSinceR is a UCFD, Theorem 1.7 showsI is actually principal, so thatIR[X] andH areprincipal. �

3. Pullbacks

In this section we show that UCFD’s constitute a fairly large class of domainshowing that they are easily produced via pullbacks.

We review the standard pullback construction. Let(S,P ) be a quasi-local domain, anlet T be a sub-domain ofS/P . Let R be the inverse image ofT in S (i.e., the union ofthose cosets ofP in S which lie inT ). ObviouslyR is a domain,R ⊆ S, P is a prime idealof R, andR/P = T . We will show thatR is a UCFD if and only ifT is.

We need the following fact. LetI be any ideal ofR and suppose there is ab ∈ I − P .Now b is clearly a unit inS. Thus forx ∈ P , we haveb−1x ∈ P , so thatx = b(b−1x) ∈bR ⊆ I . Therefore,P ⊆ I . In other words,P is comparable to every ideal. We will phraseTheorem 3.6 using that as our hypothesis.

Lemma 3.1. Let b be a nonzero non-unit element of a domainR. Thenb is pseudo-primeif and only ifR/(b) is connected(i.e., its only idempotents are0 and1).

Proof. Supposeb is pseudo-prime, and (with overbars denoting images inR/(b)), let e bean idempotent ofR. Thene2 = e impliese(1−e) ∈ (b). Sinceb is pseudo-prime,b divideseithere or 1− e, so thate equals either0 or 1.

Conversely, supposeb is not pseudo-prime. Then there are elementsc and d with(c, d) = R and withb dividing cd , such thatb divides neitherc nor d . Write ab = cd ,and also writecx + dy = 1. Multiplying this last bycx, we getcx = c2x2 + cdxy =c2x2 + abxy, so thatcx(1 − cx) = abxy. Thuscx is idempotent inR. We claim it isa nontrivial idempotent. Ifcx = 0 thenb divides cx, showingb divides c2x + aby =c2x + cdy = c(cx + dy) = c, which is a contradiction. On the other hand, ifcx = 1, thenb divides 1− cx = dy. That impliesb dividesabx +d2y = cdx +d2y = d , which is againa contradiction. �Lemma 3.2. Suppose a prime idealP �= 0 of a domainR is contained in every maximaideal. Suppose0 �= b ∈ P . Thenb is pseudo-prime.

Proof. By Lemma 3.1, it will suffice to showR∗ = R/(b) is connected. Supposee∗ isan idempotent inR∗. Thene∗(1∗ − e∗) = 0∗ ∈ P ∗. Supposee∗ ∈ P ∗, so e ∈ P . As P iscontained in every maximal ideal ofR, 1− e is a unit. Thus 1∗ − e∗ is a unit idempotenin R∗ and so equals 1∗, showinge∗ = 0∗. The other case givese∗ = 1∗. �


l. If

two

,.

et

If

its,

n

say

Lemma 3.3. Suppose a primeP of a domain is comparable to every other prime ideab ∈ R − P , thenb + P is pseudo-prime inR/P if and only ifb is pseudo-prime inR.

Proof. As pseudo-primes need to be nonzero non-units, we mention the followingfacts. Clearlyb and b + P are both nonzero elements. SinceP is contained in everymaximal ideal ofR, we see thatb is a non-unit inR if and only if b + P is a non-unitin R/P .

By Lemma 3.1, it will suffice to showR/(b) is connected if and only if(R/P)/(b +P)

is connected. Sinceb /∈ P , we have that every prime minimal overb containsP .Therefore,(P, b)/(b) is a nil ideal inR/(b). Therefore, it follows from [1, Chapter IIIProposition 2.10] thatR/(b) is connected if and only if[R/(b)]/[(P, b)/(b)] is connectedHowever, this last ring is isomorphic toR/(P,b) [R/P ]/[(P, b)/P ] = (R/P)/

(b + P). �Lemma 3.4. Suppose a prime idealP of a domainR is comparable to every ideal, and lb ∈ R − P . If c ≡ b modP , then there is a unitu of R with c = ub.

Proof. Supposec ≡ b modP , and writec = b + p with p ∈ P . As b /∈ P , we havep ∈ P ⊆ (b). Write p = bq , so thatc = b + bq = b(1 + q). As bq = p ∈ P , we haveq ∈ P . Thusq is contained in every maximal ideal, showing 1+ q is a unit, which we takeas ouru. �Lemma 3.5. Suppose a prime idealP of a domainR is comparable to every ideal.b ∈ R − P , thenb + P is pseudo-irreducible inR/P if and only ifb is pseudo-irreduciblein R.

Proof. We first note that sinceP is contained in all maximal ideals, anyr ∈ R is a non-unitof R if and only if r + P is a non-unit ofR/P .

Supposeb is pseudo-irreducible inR, but that inR/P we can factorb + P intotwo comaximal non-unit factors(c + P)(d + P). As b ≡ cd modP , Lemma 3.4 showsthere is a unitu ∈ R with b = (uc)d . As these two factors are comaximal non-unwe have a contradiction. Thus ifb is pseudo-irreducible, so isb + P . The converse isstraightforward. �Theorem 3.6. Suppose a prime idealP of a domainR is comparable to every ideal. TheR is a CFD if and only ifR/P is a CFD,R is a bounded CFD if and only ifR/P is abounded CFD, andR is a UCFD if and only ifR/P is a UCFD.

Proof. SupposeR is a CFD, and letb + P be a nonzero non-unit element ofR/P . Thenb is a nonzero non-unit element ofR, and so has a complete comaximal factorization,b = ∏

bi . Now b + P = ∏(bi + P) is clearly a comaximal factorization. As eachbi is

pseudo-irreducible inR, Lemma 3.5 shows eachbi +P is pseudo-irreducible inR/P , andso we have a complete comaximal factorization ofb + P .

Conversely, supposeR/P is a CFD, and letb be a nonzero non-unit element ofR. Ifb ∈ P , then Lemma 3.2 showsb has a complete comaximal factorization. Now sayb /∈ P ,


CFD’sgoing

n

he

al

silar

t

so thatb + P has a complete comaximal factorization inR/P , sayb + P = ∏(bi + P).

By Lemma 3.4, there is a unitu ∈ R with b = u∏

bi . By Lemma 3.5, eachbi is pseudo-irreducible inR, and so we have a complete comaximal factorization ofb.

We have now proven the fact concerning CFD’s, and the fact concerning boundedis done similarly, merely noting that the sizes of our factorizations do not change infrom R to R/P and vice-versa (elements inP causing no concern by Lemma 3.2).

Next, supposeR is a UCFD. Then it is a CFD, and so by the foregoing,R/P is a CFD.By Theorem 1.7(b)⇒ (a), in order to showR/P is a UCFD it will suffice to show anypseudo-irreducible element ofR/P is pseudo-prime. Letb + P be pseudo-irreducible iR/P . By Lemma 3.5,b is pseudo-irreducible inR. As R is a UCFD,b is pseudo-primein R. By Lemma 3.3,b + P is pseudo-prime inR/P .

Finally, supposeR/P is a UCFD, so that we already knowR is a CFD. Letb bea pseudo-irreducible element ofR. We must showb is pseudo-prime. Ifb ∈ P , thenLemma 3.2 showsb is pseudo-prime. Ifb ∈ R − P , the argument is the reverse of tpreceding argument.�

4. A UCFD with nonzero Pic

In response to a question of the referee we give examples of UCFD’sAn for which notall invertible ideals are principal, i.e., Pic(An) �= 0. Let Bn = R[x0, . . . , xn]/(x2

0 + · · · +x2n − 1) be the ring of real valued polynomial functions on then-sphereSn. It is a domain

for n � 1 since∑

x2i − 1 is irreducible. LetAn be the subring ofBn consisting of all even

functions. This ring is generated by the elementsxixj and is the coordinate ring of the reprojective spacePn. SinceA0 = R, An is a domain for alln � 0.

Define a map

R[y1, . . . , yn][(

1+∑

y2i

)−1] → An

[(x2

0

)−1] (1)

by sendingyi to x0xi/x20 = xi/x0. This is onto since 1+ ∑

y2i maps to 1/x2

0 andyiyj (1+ ∑

y2i )−1 maps toxixj . Since both rings aren-dimensional domains it follow

that the map (1) is an isomorphism. We will identify the two rings involved. Since a simresult holds for eachAn[(x2

i )−1] and∑

x2i = 1, we see thatAn is a regular domain.

Let Pn = (x20, x0x1, . . . , x0xn) = x0Bn ∩ An. ThenAn/Pn ⊂ Bn/x0Bn = Bn−1 (up to a

re-indexing of thexi ) and the image inBn−1 is An−1 so thatAn/Pn = An−1 showing thatPn is prime. This also implies thatxixj /∈ Pn for i, j �= 0 sincexixj �= 0 in Bn/x0Bn. SinceP 2

n is generated by allx0xix0xj and∑

x2i = 1 we see thatP 2

n = (x20) showing thatPn is

invertible and also thatPn = Rad(x20).

Lemma 4.1. For n � 1, Pic(An) = Z/2Z generated byPn.

Proof. SinceAn is regular, Pic(An) = C(An), the class group ofAn. If R is a normalNoetherian domain andS is a multiplicatively closed subset ofR we have an exac


g

for

,

r

tion.

localization sequence [2]

1 → U(R) → U(RS) → D(R,S) → C(R) → C(RS) → 0 (2)

whereU(R) is the group of units ofR andD(R,S) is the group of divisors meetingS,i.e., the free abelian group on the primes of height 1 meetingS. The mapU(RS) →D(R,S) sendsu to

∑ordP (u)P over primesP of height 1 meetingS. Here ord is the

additive valuation associated to the valuation ringRP . Applying this toR[y1, . . . , yn][(1+∑y2i )−1] and using (1) we see thatC(An[(x2

0)−1]) = 0.Now apply (2) toR = An and S = {(x2

0)m | m � 0}. This shows thatC(An) is thecokernel ofU(RS) → D(R,S). SincePn = Rad(x2

0), Pn is the only height 1 prime meetinS soD(R,S) = Z generated byPn. It is clear from (1) thatU(RS) = R

∗ × Z where theZ factor is generated byx2

0. To compute ordP (x20) we localizeR to RP = V say. InV all

xixj with i, j �= 0 are units and it follows easily thatPP = (x0x1). Since(x20) = (x0x1)

2 inV we find ordP (x2

0) = 2 and deduce thatC(An) = Z/2Z generated byPn. �Theorem 4.2. If n � 2, An is a UCFD withPic(An) �= 0.

Proof. It follows from Lemma 4.1 that any invertible ideal ofAn is either principal orisomorphic toPn. It was shown in [7] that the minimal number of generators ofPn isn + 1. Therefore ifn � 2 all 2-generator invertible ideals ofAn are principal. SinceAn isNoetherian, Theorem 1.7 shows thatAn is a UCFD. �Remark. The fact that the minimalnumber of generators ofPn is n + 1 was provedin a much simpler way by Gilmer [3]. His ring was different but his proof worksAn with no essential change. The referee suggested that Gilmer’s ringDn with n � 3might be an example of a UCFD with Pic�= 0. This is indeed the case. LetCn =R[x1, . . . , xn][(∑x2

i )−1] and letDn be the subring ofCn consisting of all even functionsi.e., the subring generated by allxixj and(

∑x2i )−1. The above proof applies toDn with

the following changes. In place of (1) we consider the map

R[y1, . . . , yn][y−1

1 ,(∑

y2i

)−1] → Dn

[(x2

1

)−1]

sendingyi to x1xi , and in the proof that Pic(Dn) = Z/2Z we now have, withR = Dn andS = {(x2

1)m}, thatU(RS) = R∗ × Z × Z where the twoZ factors are generated byx2

1 and∑x2i . The factor

∑x2i is in the image ofU(R) and maps to 0 inD(R,S). As above we

compute ordP (x21) = 2 and deduce thatC(An) = Z/2Z generated byPn. The remainde

of the proof is the same as above.

5. Comaximal factorization of ideals

In this section, we will discuss the ideal-theoretic version of comaximal factorizaHere,I will be a proper ideal in an arbitrary commutative ringR.


ornotn

unique.) We

l,

o

s

e

ny

h

Definitions. We define a comaximal factorization ofI to beI = I1I2 · · · In with each factora proper ideal and withIi + Ij = R for i �= j . This factorization is complete if each factis pseudo-irreducible, where the proper idealJ being pseudo-irreducible means one canwrite J = HL with H + L = R andH , L �= R. Equivalently, the comaximal factorizatioI = ∏

Ii is complete if it cannot be refined into a longer such factorization.

In her classic paper [6], Noether showed that an ideal in a Noetherian ring has acomplete comaximal factorization. (That work is also discussed in [9, Section 89]extend that work.

Theorem 5.1. If I has a complete comaximal factorization, it is unique.

Proof. SupposeI = I1I2 · · · In andI = J1J2 · · ·Jm are two comaximal factorizations ofI .For 1� i � n and 1� j � m, let Hij = Ii + Jj . The variousHij are pairwise comaximaso for a fixedIi we haveIi ⊆ Hi1 ∩ Hi2 ∩ · · · ∩ Him = Hi1Hi2 · · ·Him ⊆ Ii , showingIi = Hi1Hi2 · · ·Him. Therefore, deleting thoseHij which equalR, we have thatI = ∏

Hij

(Hij �= R) is a comaximal factorization ofI which refinesI = ∏Ii , and by symmetry, als

refinesI = ∏Jj .

Now if I = ∏Ii andI = ∏

Jj are complete comaximal factorizations, then neither haa proper refinement, and so they are identical (up to order).�

We thank the referee for pointing out the next lemma and theorem.

Lemma 5.2. There is a bijection betweencomaximal factorizations ofI and finite directsum decompositions ofR/I given by mappingI = ∏

Ii to R/I ⊕R/Ii .

Proof. We may assumeI = (0). Suppose(0) = ∏Ii is a comaximal factorization. Th

Chinese remainder theorem shows that the image ofIi underR ⊕R/Ii is

⊕R/Ij ,

j �= i. Therefore, the inverse to our map sends the finite decompositionR ⊕Ri to

the comaximal factorization(0) = ∏Ii , whereIi is taken as the image inR of

⊕Rj ,

j �= i. �Theorem 5.3. I is pseudo-irreducible if and only ifR/I is connected.I has a completecomaximal factorization if and only ifR/I can be written as a direct sum of finitely maconnected rings.

Proof. This follows easily from Lemma 5.2.�The hypothesis of the next result clearly holds whenR is Noetherian. Combined wit

Theorem 5.1, that recaptures Noether’s result.

Theorem 5.4. Suppose there is a finite listP1,P2, . . . ,Pm of prime ideals containingI ,such that for any prime idealP minimal overI , P + Ph �= R for somePh in the list. ThenI has a complete comaximal factorization.


aster

of

atingthat

ns

quetap in

ll

s

shows

Proof. It is not hard to deduce this from Theorem 5.3. However, it is perhaps fto prove it as follows. SupposeI = I1I2 · · · In is a comaximal factorization ofI . ThenV (I) = ⋃

V (Ii) is disjoint. If n > m, then someV (Ii) does not contain anyPh. Pick P

minimal over thatIi , and so also minimal overI . EachPh is in someV (Ij ) with j �= i,and soP + Ph = R for all h, which is a contradiction. Thereforem is an upper boundto the lengths of all comaximal factorizations ofI . There is a comaximal factorizationmaximal length, and it is complete.�

The next lemma can be deduced from Lemma 5.2, but we find it more illuminto do it directly. It is essentially the ideal-theoretic version of the well known factidempotents can be lifted modulo a nil ideal.

Lemma 5.5. Let I ⊆ H ⊆ Rad(I). There is a bijection between comaximal factorizatioof I and comaximal factorizations ofH , mappingI = ∏

Ii to H = ∏(H + Ii).

Proof. Let H = DF be a comaximal factorization. We claim there exists a unicomaximal factorizationI = JK such thatD = H + J andF = H + K. (Observe thaJ ⊆ D ⊆ Rad(J ). Therefore, inductive use of the claim produces the inverse to the mthe lemma.)

SinceR/H = D/H ⊕ F/H , there are elementsd andf , both idempotents moduloH ,with D = (H,d) andF = (H,f ). We may assumed andf are idempotents moduloI[1, Chapter III, Proposition 2.10]. Asdf ∈ DF = H ⊆ Rad(I), and as we may replacedandf by powers, we may assumedf ∈ I . LetJ = (I, d) andK = (I, f ). The parentheticaobservation shows thatJ + K = R, from which we seeI = JK is a comaximafactorization, as isH = (H + J )(H + K). SinceH + J ⊆ D andH + K ⊆ F , an easyexercise using comaximality shows thatH + J = D and H + K = F . Thus I = JK

satisfies the claim.For uniqueness, supposeI = J ′K ′ also satisfies it. Thend ∈ D ⊆ Rad(J ′). Replacing

d by a power showsJ ⊆ J ′, and similarly,K ⊆ K ′. The easy exercise now showsJ = J ′andK = K ′. �

There is an ideal-theoretic analogue to pseudo-prime elements. We sayI (in anycommutative ring) is a pseudo-prime ideal if wheneverJ and K are comaximal idealwith JK ⊆ I , then eitherJ ⊆ I or K ⊆ I .

Pseudo-irreducible elements in a domain need not be pseudo-prime. Lemma 5.6the ideal-theoretic analogue is better behaved.

Lemma 5.6. The following are equivalent.

(i) I is pseudo-irreducible.(ii) I is pseudo-prime.(iii) If x, y ∈ R with xy ∈ I and(x, y) = R, thenx ∈ I or y ∈ I .

Proof. (i) ⇒ (ii). SupposeJK ⊆ I with J +K = R. ThenI = (I +J )(I +K). If (i) holds,then we may assumeI + K = R, showingJ ⊆ I .


en by

ch

ed the

.

(ii) ⇒ (iii). This is immediate.(iii) ⇒ (i). SupposeI = JK with J + K = R, and writej + k = 1. Sincejk ∈ I , by

(iii) we may assumej ∈ I ⊆ K. ThusK = R. �The referee points out that the equivalence of (i) and (ii) in Lemma 5.6 can be se

noting both are equivalent to havingV (I) connected.It is not hard to use Lemma 5.6(i)⇒ (ii) to give an alternate proof to Theorem 5.1 whi

mimics the standard proof of unique factorization in a UFD.

Acknowledgment

The authors thank the referee for extensive suggestions which greatly improvpresentation of this work.

References

[1] H. Bass, AlgebraicK-Theory, Benjamin, New York, 1968.[2] R. Fossum, The Divisor Class Group of a Krull Domain, in: Ergeb. Math. Grenzgeb., vol. 74, Springer-Verlag,

1978.[3] R. Gilmer, A note on the generating sets for invertible ideals, Proc. Amer. Math. Soc. 22 (1969) 426–427[4] R. Gilmer, R. Heitmann, On Pic(R[X]) for R semi-normal, J. Pure Appl. Algebra 16 (1980) 251–257.[5] S. McAdam, R.G. Swan, A special type of invertible ideal, manuscript.[6] E. Noether, Idealtheorie in Ringbereichen, Math. Ann. 83 (1921) 24–66.[7] R.G. Swan, Vector bundles and projective modules, Trans. Amer. Math. Soc. 105 (1962) 264–277.[8] R.G. Swan, The number of generators of a module, Math. Z. 102 (1967) 318–322.[9] B.L. van der Waerden, Modern Algebra, vol. II, Frederick Ungar, New York, 1950.

McAdam Swan, Unique Comaximal Factorization 2004

Documents

Transcript of McAdam Swan, Unique Comaximal Factorization 2004