symplectic modules over overrings of polynomial rings

20 Pages·2012·0.14 MB·English

IndianJ.PureAppl. Math.,43(4): 371-390,August2012
(cid:176)c IndianNationalScienceAcademy
SYMPLECTICMODULESOVEROVERRINGSOFPOLYNOMIALRINGS
AlpeshM.Dhorajia
DepartmentofMathematics,IITMumbai,Mumbai400076,India
e-mail: [email protected]
(Received13January2011;afterfinalrevision9June2012;
accepted20June2012)
Let B be a commutative Noetherian ring of dimension d and let S be a set
of all monic polynomials in B[X]. Let A be a subring of S−1B[X] which
contains B[X]. Let P be a symplectic A-module of rank 2n ≥ d, n > 0.
ThenweprovethatESp(A2 ⊥P,(cid:104),(cid:105))actstransitivelyonUm(A2⊕P).
Keywords: Projectivemodule;unimodularelement;cancellationproblem.
1. INTRODUCTION
LetAbeacommutativeNoetherianring. AsymplecticA-moduleisapair(P,(cid:104),(cid:105)),
where P is a finitely generated projective A-module and (cid:104),(cid:105) : P ×P → A is a
non-degeneratealternatingbilinearform.
Let A be a ring of dimension d and P be a symplectic A-module of rank 2n
with2n ≥ d. ThenSwan[11]hasprovedthatESp(A2 ⊥ P,(cid:104),(cid:105))actstransitively
on Um(A2 ⊕ P). Further, the above result of Swan is extended to polynomial
372 ALPESHM.DHORAJIA
rings by Bhatwadekar and Laurent polynomial rings by Keshari in [2] and [6] re-
spectively.
In this paper we prove the following result see (3.9). This generalizes a result
ofBhatwadekar([2],Theorem4.8).
Theorem A — Let B be a ring of dimension d and let S be a multiplicative
closed subset of B[X] containing monic polynomials. Let A be a ring such that
B[X] ⊆ A ⊆ S−1B[X]. LetP beasymplecticA-moduleofrank2nwith2n ≥ d,
n > 0. ThenESp(A2 ⊥ P,(cid:104),(cid:105))actstransitivelyonUm(A2⊕P).
Weprovethefollowingresult(3.13),whichgeneralizesaresultofKeshari([6],
TheoremA.7).
TheoremB—LetB bearingofdimensiondandA = B[X ,...,X ,Y, 1],
1 m g
whereg ∈ B[Y]ismonic. LetP beasymplecticA-moduleofrank2n ≥ d. Then
ESp(A2 ⊥ P,(cid:104),(cid:105))actstransitivelyonUm(A2⊕P).
Thefollowingresult(4.3),generalizesaresultofKeshari([6],TheoremA.8).
TheoremC—LetB bearingofdimension2andA = B[X ,...,X ,Y, 1],
1 n g
whereg isamonicpolynomialinB[Y]. LetP beaprojectiveA-moduleofrank2
withtrivialdeterminant. IfA2 iscancellative,thenP iscancellative.
2. PRELIMINARIES
All rings considered in this paper are assumed to be commutative Noetherian and
allmodulesarefinitelygenerated.
Let A be a ring and let P be a projective A-module. Recall that p ∈ P is
called a unimodular element if there exists a φ ∈ P∗ = Hom (P,A), such that
A
φ(p) = 1. ThesetofallunimodularelementsofP isdenotedbyUm(P). Arow
(a ,...,a ) ∈ An is called a unimodular row if there exists (b ,...,b ) ∈ An
1 n 1 n
such that a b +...+a b = 1. The set of all unimodular rows of length n is
1 1 n n
denotedbyUm (A).
n
SYMPLECTICMODULESOVEROVERRINGS 373
Let n be a positive integer. Let GL (A) be the set of all n × n invertible
n
matricesoverA. SL (A) := {M ∈ GL (A)|det(M) = 1}. LetI beanidealin
n n
A, we define SL (A,I) to be kernel of the natural map SL (A) → SL (A/I).
n n n
WhenI = (a),wewriteSL (A,a)forSL (A,I).
n n
Werecallsomepreliminaryfactsaboutsymplecticmodules. Fornotationsand
terminologywefollow[2]. Formoredetail,see([2],Section4).
Let A be a ring and let P be a finitely generated projective A-module. A
bilinear form (cid:104),(cid:105) : P ×P → A is called alternating if (cid:104)p,p(cid:105) = 0 ∀p ∈ P. An
alternating bilinear form (cid:104),(cid:105) induces a homomorphism α : P → P∗ (defined as
α(p)(q) = (cid:104)p,q(cid:105)) such that α+α∗ = 0. An alternating form (cid:104),(cid:105) on P is called
non-degenerateiftheinducedhomomorphismfromP toP∗ isanisomorphism.
A symplectic A-module is a pair (P,(cid:104),(cid:105)), where P is a finitely generated
projectiveA-moduleand(cid:104),(cid:105) : P ×P → Aisanon-degeneratealternatingbilinear
form. If (P,(cid:104),(cid:105)) is a symplectic A-module then the rank of P is even and P has
trivialdeterminant.
If (P,(cid:104),(cid:105)) and (Q,(cid:104),(cid:105)) are two symplectic modules, then the non-degenerate
alternating bilinear forms on P and Q will give rise (in a canonical manner) to a
non-degenerate alternating bilinear form onP ⊕Q and we denote the symplectic
module thus obtained by (P ⊥ Q,(cid:104),(cid:105)). The standard alternating form on A2 (in
thenaturalbasis{e ,e })isgivenby(0 1).
1 2 1 0
Twosymplecticmodules(P,(cid:104),(cid:105))and(Q,(cid:104),(cid:105))areisomorphicifthereexistsan
isomorphismτ : P → Qsuchthat
(cid:104)p ,p (cid:105) = (cid:104)τ(p ),τ(p )(cid:105),∀p ,p ∈ P.
1 2 1 2 1 2
Anisometryofthesymplecticmodule(P,(cid:104),(cid:105))isanautomorphismof(P,(cid:104),(cid:105)).
We denote by Sp(P,(cid:104),(cid:105)) the group of isometries of (P,(cid:104),(cid:105)). It is easy to see that
SL (A) ⊆ Sp(A2,(cid:104),(cid:105)).
2
Let(P,(cid:104),(cid:105))beasymplecticA-moduleandletu,v ∈ P besuchthat(cid:104)u,v(cid:105) = 0.
374 ALPESHM.DHORAJIA
Leta ∈ Aandletτ : P → P beamapdefinedby
(a,u,v)
τ (p) = p+(cid:104)p,v(cid:105)u+(cid:104)p,u(cid:105)v+a(cid:104)p,u(cid:105)u.
(a,u,v)
Thenitiseasytoseethatτ ∈ Sp(P,(cid:104),(cid:105)). Anisometryτ iscalleda
(a,u,v) (a,u,v)
symplectic transvectionifeitheruorvisunimodular. WedenotebyESp(P,(cid:104),(cid:105))
the subgroup of Sp(P,(cid:104),(cid:105)), generated by symplectic transvections. ESp(P,(cid:104),(cid:105))
isanormalsubgroupofSp(P,(cid:104),(cid:105)).
Let (P,(cid:104),(cid:105)) be a symplectic A-module. Let c,d ∈ A, q ∈ P. If u = (0,1,0)
and v = (0,c,q) ∈ A2 ⊕P, then τ is a symplectic transvection of (A2 ⊥
(−c,u,v)
P,(cid:104),(cid:105))suchthat
τ ((a,b,p)) = (a,b+ca+(cid:104)p,q(cid:105),p+aq).
(−c,u,v)
Similarly,ifu = (1,0,0)andv = (−d,0,−q),then
τ ((a,b,p)) = (a+bd+(cid:104)q,p(cid:105),b,p+bq).
(d,u,v)
E(A2 ⊥ P,(cid:104),(cid:105))denotesthesubgroupofSp(A2 ⊥ P,(cid:104),(cid:105))generatedbyΘ and
(c,q)
σ forc,d ∈ Aandq ∈ P,whereΘ ,σ aredefinedasfollows:
(d,q) (c,q) (d,q)
Θ (a,b,p) = (a,b+ca+(cid:104)p,q(cid:105),p+aq)
(c,q)
σ (a,b,p) = (a+bd+(cid:104)q,p(cid:105),b,p+bq)
(d,q)
for(a,b,p) ∈ A2⊕P.
Let B be a ring and let A = B[X]. Let F be a free A-module with a basis
(cid:80)
{e ,...,e }. Let p(X) = r γ (X)e ∈ F, where γ (X) ∈ A for 1 ≤ i ≤ r.
1 r i=1 i i (cid:80) i
Then,forb ∈ B,wedenotebyp(bX)theelement r γ (bX)e ofF.
i=1 i i
WestatethefollowingresultwhichisduetoBhatwadekar([2],Theorem4.8).
Theorem 2 — Let R be a ring of dimension d. Let A be a polynomial ring in
r (≥ 0)variablesoverR. Let(P,(cid:104),(cid:105))beasymplecticA-moduleofrank2n > 0.
If2n ≥ dthenESp(A2 ⊥ P,(cid:104),(cid:105))actstransitivelyonUm(A2⊕P).
SYMPLECTICMODULESOVEROVERRINGS 375
Remark 2.2 : We have observed that in the above theorem we can replace
ESp(A2 ⊥ P,(cid:104),(cid:105)) by E(A2 ⊥ P,(cid:104),(cid:105)). Let I be an ideal in A, then clearly the
naturalmapE(A2 ⊥ P,(cid:104),(cid:105)) → E((A/I)2 ⊥ P/IP,(cid:104),(cid:105))issurjective.
The following result is due to Keshari ([6], Theorem A.7), which extends the
aboveresultofBhatwadekartoLaurentpolynomialring.
Theorem 3 — Let B be a ring of dimension d and A = B[Y ,...,Y ,X±1,
1 r(cid:48) 1
...,X±1]. Let(P,(cid:104),(cid:105))beasymplecticA-moduleofrank2n ≥ d,thenESp(A2 ⊥
r
P,(cid:104),(cid:105))actstransitivelyonUm(A2⊕P).
3. MAIN THEOREM
WebeginthissectionbyprovingthefollowingresultofBhatwadekarandRoy([3],
Lemma4.1).
Lemma 3.1 — Let B ⊂ C be rings of dimension d and x ∈ B such that
B = C . Then
x x
(i)B/(1+xb) = C/(1+xb)forallb ∈ B.
(ii)IfI isanidealofC suchthatht(I) ≥ dandI+xC = C,thenthereexists
anelementb ∈ B suchthat1+xb ∈ I.
PROOF: (i)SinceBx = Cx,bygoingmodulotheideal(1+xb)Bxwithb ∈ B,
theprooffollows.
(ii) Since ht(I) = d, without loss of generality we may assume that I is a
maximal ideal of height d. By hypothesis, localizing C at x(1 + xB), we have
B = C . SincedimB < dimB,wehavedimC <
x(1+xB) x(1+xB) x(1+xB) x(1+xB)
d. SinceI ismaximalidealofheightd,wegetIC = C . Therefore
x(1+xB) x(1+xB)
I contains an element of the form xm(1 + xa)n for some a ∈ B and for some
positiveintegersmandn. SinceI+xC = C andI isaprimeideal,I containsan
elementoftheform1+xbforsomeb ∈ B. Thiscompletestheproofof(ii).
ThefollowingresultisaconsequenceofatheoremofEisenbud-Evansasstated
376 ALPESHM.DHORAJIA
in([9],p.1420). (cid:50)
Lemma 3.2 — Let A be a ring and let P be a projective A-module of rank r.
Let(α,a) ∈ (P∗⊕A). Thenthereexistsanelementβ ∈ P∗ suchthatht(I ) ≥ r,
a
whereI = (α+aβ)(P). Inparticular,iftheideal(α(P),a)hasheight≥ r,then
ht(I) ≥ r. Further,if(α(P),a)isanidealofheight≥ r andI isaproperidealof
R,thenht(I) = r.
ThefollowinglemmaisduetoSuslin([10],Lemma2.1).
Lemma 3.3 — Let B be a ring and let A = B[X, F1,..., Fr], where F ∈
g g i
B[X] for 1 ≤ i ≤ r and g ∈ B[X] is monic. Suppose f ,f ∈ B[X] and
1 2
c ∈ (f ,f )B[X] ∩ B. Then for any ideal I of A and g,h ∈ A ∩ B[X] with
1 2
h−g ∈ cI,thereexists∆ ∈ SL (A,I)suchthat[f (g),f (g)]∆ = [f (h),f (h)].
2 1 2 1 2
Wenowprovethefollowingextensionlemma,whichforpolynomialring(i.e.
A = B[X])isprovedin[7]. Ourprooffollowsusingthesameargumentasinthe
proof([7],Lemma1.1,Chap.3).
Lemma 3.4 — Let B be a ring and let A = B[X, F1,..., Fr], where F ∈
g g i
B[X] for 1 ≤ i ≤ r and g ∈ B[X] is monic. If I is an ideal of A containing
1+gH for some monic polynomial H ∈ B[X] and J is an ideal of B such that
I +JA = A,thenB∩I +J = B.
PROOF: LetR = A/I ⊇ B/(B∩I)andletJ¯betheimageofJ inB/(B∩I).
Byhypothesis,wehaveJR = R. SinceI containsapolynomial1+gH,itiseasy
to seethatR isintegraloverB/(B ∩I). Therefore the“Going-Up” Theorem for
integral extensions ([8], p.34) implies that J = B/(B ∩ I), therefore B ∩ I +
J = B. (cid:50)
Definition3.5—LetAbearingandsbeanon-zerodivisorinA. Let(P,(cid:104),(cid:105))
beasymplecticA-moduleofrank2n. Aset{e ,...,e ,f ,...,f } ⊂ P iscalled
1 n 1 n
ans-symplecticbasisofP ifthefollowingholds:
(i)(cid:104)e ,e (cid:105) = 0 = (cid:104)f ,f (cid:105)for1 ≤ i,j ≤ n.
i j i j
SYMPLECTICMODULESOVEROVERRINGS 377
(ii)(cid:104)e ,f (cid:105) = sfor1 ≤ i ≤ nand(cid:104)e ,f (cid:105) = 0foreveryi (cid:54)= j.
i i i j
Remark 3.6 : Let P be a symplectic A-module of rank 2n. If
{e ,...,e ,f ,...,f } ⊆ P is s-symplectic basis of P then the module F :=
1 n 1 n
(cid:80) (cid:80)
Ae + Af isafreeA-submoduleofP andsP ⊆ F. Fortheproofsee([2],
i j
Lemma4.2).
Lemma3.7—LetB beareducedringofdimensiondandletg ∈ B[X]bea
monicpolynomial. Assumethateither
(i)A = B[X, f1,..., fn],wheref ∈ B[X]fori = 1,...,nor
g g i
(ii)A = B[X ,...,X ,X, 1].
1 m g
Let (P,(cid:104),(cid:105)) be a symplectic A-module of rank 2r ≥ d, r > 0. Then P has
an s-symplectic basis {e ,...,e ,f ,...,f } such that sP ⊆ F, where F :=
1 n 1 n
(cid:80) (cid:80)
Ae + Af isafreeA-submoduleofP ofrank2r.
i j
PROOF : Let S be the set of all non-zerodivisors in B. Since rank of P is
constant, S−1A is a PID if A is as in (i) else S−1A = k[X ,...,X ,X, 1] for
1 m g
some field k. Using [5], in both the cases S−1P is a free S−1A-module of rank
2r. Let p˜, q˜ ∈ S−1P such that (cid:104)p˜,q˜(cid:105) = 1, where (cid:104),(cid:105) is a induced form on
1 1 1 1
S−1P. WriteS−1P = p˜S−1A⊕q˜S−1A⊕Q,whereQisaS−1A-submoduleof
1 1
S−1P ofrank2r−2. ApplythesameargumenttoQ,byinductively,threreexists
p˜,...,p˜ and q˜,...,q˜ such that (cid:104)p˜,q˜(cid:105) = 1 and (cid:104)p˜,q˜(cid:105) = 0 for i (cid:54)= j. Also,
2 r 2 r i i i j
wehave(cid:104)p˜,q˜(cid:105) = 0and(cid:104)p˜,q˜(cid:105) = 0for2 ≤ i ≤ r.
1 i i 1
Chooset ∈ S suchthatp˜ = ei,q˜ = fi,wheree ,f ∈ P for1 ≤ i ≤ r. Let
i t i t i i
s = t2.
Claim: {e ,...,e ,f ,...,f }isans-symplecticbasisofP.
1 r 1 r
ProofofClaim : Weshowthat(i)(cid:104)e ,e (cid:105) = 0 = (cid:104)f ,f (cid:105)for1 ≤ i,j ≤ nand
i j i j
(ii)(cid:104)e ,f (cid:105) = sand(cid:104)e ,f (cid:105) = 0foreveryi (cid:54)= j.
i i i j
Foranyi,j for1 ≤ i,j ≤ r,wehave(cid:104)e ,e (cid:105) = (cid:104)tp˜,tp˜(cid:105). Sinces(cid:104)p˜,p˜(cid:105) = 0
i j i j i j
in S−1A, we have (cid:104)e ,e (cid:105) = 0. Similarly, we can show that (cid:104)f ,f (cid:105) = 0 for
i j i j
378 ALPESHM.DHORAJIA
1 ≤ j ≤ r and(cid:104)e ,f (cid:105) = 0foreveryi (cid:54)= j.
i j
Foranyi,1 ≤ i ≤ r,wehave
e f
(cid:104)e ,f (cid:105) = (cid:104)t i,t i(cid:105) = t2(cid:104)p˜,p˜(cid:105) = s(cid:104)p˜,p˜(cid:105) = s.
i i i j i j
t t
Note that the above expression is computed in S−1A. This proves the claim.
Nowby(3.6),theprooffollows. (cid:50)
Now we prove the following result which is inspired by and the proof closely
follows,Bhatwadekar’sresult([2],Proposition4.7).
Proposition 3.8 — Let B be a ring and let s ∈ B be a non-zerodivisor. Let
A = B[X, F1,...,Fr], where F ∈ B[X] and g ∈ B[X] is monic and let
g g i
(P,(cid:104),(cid:105)) be a symplectic A-module of rank 2n. Let {e ,...,e ,f ,...,f } ⊆ P
1 n 1 n
be an s-symplectic basis of P. Let (α(X),β(X),p(X)) ∈ Um(A2 ⊕P), where
(cid:80) (cid:80)
α(X),β(X) ∈ B[X], p(X) ∈ n B[X]e + n B[X]f with α(X) ≡ 1
i=1 i j=1 j
modulo (sXg) and β(X) = 1+gH with H ∈ B[X] is monic. Let b,b(cid:48) ∈ B be
suchthatb−b(cid:48) ∈ sB. ThenthereexistsΨ ∈ SL (A,sX)ESp(A2 ⊥ P,(cid:104),(cid:105))such
2
thatΨ(α(bX),β(bX),p(bX)) = (α(b(cid:48)X),β(b(cid:48)X),p(b(cid:48)x)).
PROOF : Since ESp(A2 ⊥ P,(cid:104),(cid:105)) is a normal subgroup ofSp(A2 ⊥ P,(cid:104),(cid:105)),
G = SL (A,sX)ESp(A2 ⊥ P,(cid:104),(cid:105))isagroup. LetJ bethesetofelementsc ∈ B
2
havingfollowingproperty, b−b(cid:48) ∈ csB ⇒ ∃Φ ∈ G suchthatΦ(α(bX),β(bX),
p(bX)) = (α(b(cid:48)X),β(b(cid:48)X),p(b(cid:48)x)). ClearlyJ isnonempty. SinceGisagroup,
J isanidealofB. WeprovethatJ = B.
Let t ∈ B ∩(α(X)2,β(X)2)A and let b,b(cid:48) ∈ B be such that b−b(cid:48) ∈ tsB.
(cid:80) (cid:80)
Sincep(X) ∈ F = B[X]e + B[X]f ,p(bX) = p(b(cid:48)X)−tsq(X)forsome
i j
q(X) ∈ F.
Claim: B∩(α(X)2,β(X)2)A = B∩(α(X)2,β(X)2)B[X].
ProofofClaim: ItisenoughtoproveB∩(α(X)2,β(X)2)A ⊂ B∩(α(X)2,
β(X)2)B[X]. Lett ∈ B∩(α(X)2,β(X)2)Athent = aα(X)2+bβ(X)2,where
SYMPLECTICMODULESOVEROVERRINGS 379
a,b ∈ Aandβ(X)2 = 1+gK. MultiplyingbygK toexpressionoft,wegetthe
following
gKt = agKα(X)2+bgKβ(X)2.
Addingbothsides−tandsimplifying,weget
−t = −t(1+gK)+agKα(X)2+bgKβ(X)2.
Sinceβ(X)2 = 1+gK,weget
t = −agKα(X)2+(t−bgK)β(X)2.
Applyingthisargumentfurtherifneeded,weseethattbelongstotheidealof
B[X]generatedbyα(X)2 andβ(X)2. Thisprovestheclaim.
Nowsincet ∈ (α(X)2,β(X)2)∩B,talsobelongstotheidealofB[X]gen-
eratedbyα(bX)2 andβ(bX)2. By([2],Lemma4.3),thereexistsψ ∈ ESp(A2 ⊥
P,(cid:104),(cid:105)) such that ψ(α(bX),β(bX),p(bX)) = (α(bX),β(bX),p(b(cid:48)X)). Since
bX − b(cid:48)X ∈ tsXB[X], by (3.3), the element (α(bX),β(bX),p(b(cid:48)X)) can be
transformedto(α(b(cid:48)X),β(b(cid:48)X),p(b(cid:48)X)),byanelementoftheSL (A,sX),hence
2
t ∈ J.
Sinceβ(X) = 1+gH withH ∈ B[X]monic,B/(B∩(α(X)2,β(X)2)) (cid:44)→
A/(α(X)2,β(X)2)) is an integral extension. Since (α(X)2,β(X)2)+sA = A,
by(3.4),wehaveB∩(α(X)2,β(X)2)+sB = B. Thereforetheaboveargument
showsthatJ +sB = B.
Let m be a maximal ideal of B. If s ∈ m then m + J = B. Now as-
sume that s ∈/ m. To complete the proof, it is enough to show that m + J =
B. Since α(X) ≡ 1 modulo (sXg) and (α(X),β(X),p(X)) ∈ Um(A2 ⊕
P), we have (α(X),β(X),sXp(X)) ∈ Um(A2 ⊕ P). Let “−” denote reduc-
tion modulo the ideal (mA,β(X)). Note that (α(X),sXp(X)) ∈ Um(A¯ ⊕
P¯). Then it is easy to see that g ≡ 1 modulo (mA,β(X)), where g is of
1 1
the form α(X) + (cid:104)p(cid:48),sXp(X)(cid:105) for some p(cid:48) ∈ P. This shows that the ide-
1 1
als (α(X) + (cid:104)p(cid:48),sXp(X)(cid:105),β(X)) and mA are comaximal. Further α(X) +
1
380 ALPESHM.DHORAJIA
(cid:104)p(cid:48),sXp(X)(cid:105) + dβ(X) ∈ B[X] for some suitable d ∈ A. Clearly the ideals
1
(α(X)+(cid:104)p(cid:48),sXp(X)(cid:105)+dβ(X),β(X))and(α(X)+(cid:104)p(cid:48),sXp(X)(cid:105),β(X))are
1 1
equal. Therefore we can assume that g = α(X) + (cid:104)p(cid:48),sXp(X)(cid:105) ∈ B[X] for
1
some p(cid:48) ∈ P with (cid:104)p(cid:48),sXp(X)(cid:105) ∈ B[X]. Also we can take q(X) = sXp(cid:48) ∈
(cid:80) (cid:80)
B[X]e + B[X]f .
i j
Letp (X) = p(X)+β(X)q(X)andη(X) = α(X)+(cid:104)q(X),p(X)(cid:105). Clearly
1
(cid:80) (cid:80)
η(X) ∈ B[X] and p (X) ∈ B[X]e + B[X]f . Then η(X) ≡ 1 modulo
1 i j
(sX)andmA+(η(X),β(X)) = A. Moreover,forb ∈ B
σ (α(bX),β(bX),p(bX)) = (α(bX)+(cid:104)q(bX),p(bX)(cid:105),β(X),
(0,q(bX))
p(bX)+β(bX)q(bX))
= (η(bX),β(bX),p (bX)).
1
LetJ = B∩(η(X)2,β(X)2). SincemA+(η(X)2,β(X)2) = Aandβ(X)2
1
ismonic,asbeforewegetm+(η(X)2,β(X)2)∩B = B. Thereforem+J = B.
1
Let t ∈ J . As before we see that b,b(cid:48) ∈ B such that b−b(cid:48) ∈ st B then there
1 1 1
existsΦ ∈ Gsuch thatΦ(η(bX),β(bX),p (bX)) = (η(b(cid:48)X),β(b(cid:48)X),p (b(cid:48)X)).
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Therefore
σ −1◦Φ◦σ (α(bX),β(bX),p(bX)) = (α(b(cid:48)X),β(b(cid:48)X),p(b(cid:48)X)).
(0,q(b(cid:48)X)) (0,q(bX))
Hence J ⊂ J and thus m+J = B. Therefore J = B. This completes the
1
proof. (cid:164)
Let B be a commutative Noetherian ring of dimension d. Let A be a ring
such that B[X] ⊆ A ⊆ S−1B[X], where S denotes the multiplicatively closed
set of monic polynomials in B[X]. Let P be any finitely generated projective A-
module then P is regarded as a finitely generated projective D-module, where D
is a subring of A generated by B[X] and some finite number of elements of A.
Moreover,wemayassumethatD ⊆ B[X, 1]forsomeg ∈ S. Thereforetoprove
g
Theorem A, we may assume that (P,(cid:104),(cid:105)) is a symplectic A = B[X, f1,..., fl]-
g g
module.