www.elsevier.com/locate/jalgebra
The centre of quantum
sl
n
at a root of unity
Rudolf Tange
School of Mathematics, University of Southampton, Highfield, SO17 1BJ, UK
Received 5 May 2005 Available online 13 February 2006 Communicated by Corrado de Concini
Summary
It is proved that the centreZof the simply connected quantised universal enveloping algebra over C,Uε,P(sln),εa primitivelth root of unity,lan odd integer>1, has a rational field of fractions.
Furthermore it is proved that iflis a power of an odd prime,Zis a unique factorisation domain.
©2005 Elsevier Inc. All rights reserved.
Introduction
In [8] de Concini, Kac and Procesi introduced the simply connected quantised universal enveloping algebraU=Uε,P(g)overCat a primitivelth root of unityεassociated to a simple finite-dimensional complex Lie algebrag. The importance of the study of the centre
ZofUand its spectrum Maxspec(Z)is pointed out in [7,8].
In this article we consider the following two conjectures concerning the centreZofU
in the caseg=sln:
(1) Zhas a rational field of fractions.
(2) Zis a unique factorisation domain (UFD).
The same conjectures can be made for the universal enveloping algebraU (g)of the Lie algebragof a reductive group over an algebraically closed field of positive characteristic.
E-mail address:[email protected].
In [16] these conjectures were proved forg=glnand forg=slnunder the condition that
nis non-zero in the field.
The second conjecture was made by Braun and Hajarnavis in [1] for the universal en-veloping algebraU (g)and suggested forU=Uε,P(g). There it was also proved thatZ is locally a UFD. In Section 3 below, this conjecture is proved forslnunder the condition that
lis a power of a prime(=2). The auxiliary results and Step 1 of the proof of Theorem 4, however, hold without extra assumptions onl.
The first conjecture was posed as a question by J. Alev for the universal enveloping algebra U (g). It can be considered as a first step towards a proof of a version of the Gelfand–Kirillov conjecture forU. Indeed the Gelfand–Kirillov conjecture forgln and
slnin positive characteristic1was proved recently by J.-M. Bois in his PhD thesis [4] using results in [16] on the centres of their universal enveloping algebras (forslnit was required thatn=0 in the field). It should be noted that the Gelfand–Kirillov conjecture forU (g)in characteristic 0 (and in positive characteristic) is still open forgnot of typeA.
As in [16], a certain semi-invariantdfor a maximal parabolic subgroup of GLnwill play an important rôle. Later we learned that (a version of) this semi-invariant already appeared before in the literature, see [10]. For quantum versions, see [12,13].
1. Preliminaries
In this section we recall some basic results, mostly from [8], that are needed to prove the main results (Theorems 3 and 4) of this article. Short proofs are added in case the results are not explicitly stated in [8].
1.1. Elementary definitions
Letgbe a simple finite-dimensional Lie algebra overCwith Cartan subalgebrah, letΦ
be its root system relative toh, let(α1, . . . , αr)be a basis ofΦand let(·|·)be the symmetric bilinear form onh∗which is invariant for the Weyl groupWand satisfies(α|α)=2 for all short rootsα. Putdi=(αi|αi)/2. The root lattice and the weight lattice ofΦ are denoted by respectivelyQandP. Note that(·|·)is integral onQ×P.
Mostly we will be in the situation whereg=sln. In this caser=n−1 and all thedi are equal to 1. We then takehthe subalgebra that consists of the diagonal matrices insln and we takeαi=A→Aii−Ai+1i+1:h→C.
Letl be an odd integer>1 and coprime to all the di, letεbe a primitivelth root of unity and letΛbe a lattice betweenQandP. LetU=Uε,Λ(g)be the quantised universal enveloping algebra ofgat the root of unityεdefined in [8] and denote the centre ofUbyZ. SinceUhas no zero divisors (see [7, 1.6–1.8]),Zis an integral domain. LetU+, U−, U0
be the subalgebras ofU generated by respectively theEi, theFi and theKλwithλ∈Λ. Then the multiplicationU−⊗U0⊗U+→U is an isomorphism of vector spaces. We
identifyU0with the group algebraCΛofΛ. Note thatW acts onU0, since it acts onΛ. LetT be the complex torus Hom(Λ,C×). ThenT can be identified with Maxspec(U0)=
HomC-Alg(U0,C)and for the action ofT onU0=C[T]by translation we havet·Kλ=
t (λ)Kλ.
Thebraid group Bacts onU by automorphisms. See [8, 0.4]. The subalgebraZ0 of
U is defined as the smallestB-stable subalgebra containing the elementsKλl,λ∈Λ, and
Eil, Fil,i=1, . . . , r. We haveZ0⊆Z. Putzλ=Kλl and letZ00 be the subalgebra ofZ0 spanned by thezλ. Then the identification of U0withCΛgives an identification of Z0 0 withClΛ. If we replaceKλbyzλin foregoing remarks, then we obtain an identification of
T with Maxspec(Z00). PutZ±0 =Z0∩U±. Then the multiplicationZ−0 ⊗Z00⊗Z0+→Z0 is an isomorphism (of algebras). See e.g. [7, 3.3].
1.2. The Harish-Chandra centreZ1and the quantum restriction theorem
LetQ∨ be the dual root lattice, that is, the Z-span of the dual root systemΦ∨. We haveQ∨∼=P∗→Λ∗. Denote the image of Q∨ under the homomorphismf →(λ→ (−1)f (λ)):Λ∗→T byQ∨2. Then the elements=1 ofQ∨2 are of order 2 and U0Q∨2 =
C(Λ∩2P ). SinceQ∨2 isW-stable, we can form the semi-direct productW˜ =WQ∨2
and thenU0W˜ =(C(Λ∩2P ))W.
Leth:U=U−⊗U0⊗U+→U0be the linear map takingx⊗u⊗ytoU(x)uU(y), whereU is the counit ofU. Thenhis a projection ofUontoU0. Furthermoreh(Z0)=
Z00=ClΛandh|Z0:Z0→Z
0
0 has a similar description ash and is a homomorphism of algebras. Define the shift automorphismγ of U0Q∨2 by settingγ (Kλ)=ε(ρ|λ)Kλ for
λ∈Λ∩2P. Hereρ is the half sum of the positive roots. Note that γ =id onZ0Q∨2
0 =
Cl(Λ∩2P ). In [8, p. 174] and [7, §2], there was constructed an injective homomorphism
¯
h:U0W˜ →Z, whose image is denoted byZ1, such thath(Z1)⊆U0Q
∨
2 and the inverse
h:Z1−→∼ U0W˜
ofh¯is equal toγ−1◦h. Note thath=honZ0∩Z1and thath|Z1 is a homomorphism
of algebras. Since Ker(h)is stable under left and right multiplication by elements ofU0
and under multiplication by elements ofZ, we can conclude that the restriction ofhto the subalgebra generated byZ0andZ1is a homomorphism of algebras.
From now on we assume thatΛ=P. LetGbe the simply connected almost simple complex algebraic group with Lie algebragand letT be a maximal torus ofG. We identify
Φ andW with the root system and the Weyl group of G relative to T. Note that the character group X(T ) of T is equal to P. In case g=sln we take T the subgroup of diagonal matrices in SLn.
1.3. Generators forC[G]GandZ1
We denote the fundamental weights corresponding to the basis (α1, . . . , αr) by
an isomorphismC[G]G−→∼ C[T]W =(CP )W, see [17, §6]. Forλ∈P denote the basis element ofCP corresponding toλ by e(λ), denote theW-orbit of λ byW ·λand put sym(λ)=μ∈W·λe(μ). Then the sym(i),i=1, . . . , r, are algebraically independent generators of(CP )W. See [3, No. VI.3.4, Théorème 1].
For a field K, we denote the vector space of alln×n matrices over K by Matn= Matn(K). Now assume thatK=C. In this section we denote the restriction to SLn of the standard coordinate functionals on Matn by ξij, 1i, j n. Furthermore, for i∈
{1, . . . , n−1},si∈C[SLn]is defined bysi(A)=tr(iA), where iAdenotes theith exterior power ofAand tr denotes the trace. Theni=(ξ11· · ·ξii)|T and therefore
sym(i)=si|T, (∗)
theith elementary symmetric function in theξjj|T. See [16, 2.4].
In the general case we use the restriction theorem for C[G] and definesi ∈C[G]G by(∗). So thens1, . . . , sr are algebraically independent generators ofC[G]G.
Identifying U0 and CP, we have U0W˜ =(C2P )W. Put ui = ¯h(sym(2i)). Then
h(ui)=sym(2i)andu1, . . . , ur are algebraically independent generators ofZ1.
1.4. The coverπand the intersectionZ0∩Z1
Let Φ+ be the set of positive roots corresponding to the basis(α1, . . . , αr)ofΦ and letU+ respectivelyU− be the maximal unipotent subgroup ofG corresponding toΦ+
respectively−Φ+. Ifg=sln, then U+ andU− consist of the upper respectively lower triangular matrices in SLnwith ones on the diagonal. PutO=U−T U+. ThenOis a non-empty open and therefore dense subset ofG. Furthermore, the group multiplication defines an isomorphismU−×T ×U+−→∼ Oof varieties. PutΩ=Maxspec(Z0).
In [7, (3.4)–(3.6)] there was constructed a groupG˜ of automorphisms ofUˆ = ˆZ0⊗Z0U,
whereZˆ0denotes the algebra of holomorphic functions on the complex analytic varietyΩ. The groupG˜ leavesZˆ0andZˆ = ˆZ0⊗Z0Zstable. In particular it acts by automorphisms on
the complex analytic varietyΩ. In [8] this action is called the “quantum coadjoint action.” In [8, §4] there was constructed an unramified coverπ:Ω→Oof degree 2r. I give a short description of the construction ofπ. PutΩ±=Maxspec(Z0±). Then we haveΩ= Ω−×T ×Ω+. Now Z:Ω →T is defined as the projection on T,X:Ω →U+ and
Y:Ω→U−as the projection onΩ±followed by some isomorphismΩ± ∼−→U±andπ
is defined asY Z2X(multiplication inG).2This means:π(x)=Y (x)Z(x)2X(x).
The following theorem says something about howG˜ andπ are related to the “Harish-Chandra centre”Z1and the conjugation action ofGonC[G]. For more precise statements see [8, 5.4, 5.5 and §6].
Theorem 1.[8, Proposition 6.3, Theorem 6.7]Considerπas a morphism toG. Then the comorphismπco:C[G] →Z0is injective and the following holds:
(i) ZG˜ =Z1.3
(ii) πcoinduces an isomorphismC[G]G−→∼ Z0G˜ =Z0∩Z1.
(iii) The monomorphism(CP )W −→∼ (CP )W obtained by combining the isomorphism in (ii)with the restriction homomorphismC[G] →C[T] =CP andh:Z1→U0=CP,
is given byx→2lx:P →P. In particularh(Z0∩Z1)=(C2lP )W.
I will give the proof of (iii). If we identify Z00 with C[T], then the homomorphism
h|Z0:Z0→Z
0
0is the comorphism of a natural embeddingT →Ω. Now we have a com-mutative diagram
G π Ω
T t→t T
2
Expressed in terms of the comorphisms this reads: (x→2x)◦resG,T =resΩ,T ◦πco, where resG,T and resΩ,T are the restrictions toT and the comorphism of the morphism between the tori is denoted by its restrictions to the character groups. Now we identifyU0
withC[T]. Composing both sides on the left withx→lxand using(x→lx)◦resΩ,T =
h|Z0:Z0→U
0=CP we obtain(x→2lx)◦resG,T =h◦πco. If we restrict both sides of this equality toC[G]G, then we can replacehbyhand we obtain the assertion.
1.5. Z0andZ1generateZ
Theorem 2.[8, Proposition 6.4, Theorem 6.4]Letu1, . . . , ur be the elements ofZ1defined
in Subsection1.3. Then the following holds:
(i) The multiplicationZ1⊗Z0∩Z1Z0→Zis an isomorphism of algebras.
(ii) Zis a freeZ0-module of ranklr with the restricted monomialsu1k1· · ·ukrr,0ki< l,
as a basis.
I give a proof of (ii). In [8, Proposition 6.4] it is proved that(CP )W is a free(ClP )W -module of ranklrwith the restricted monomials (exponents< l) in the sym(i)as a basis. The same holds then of course for(C2P )W,(C2lP )W and the sym(2i). But then the same holds forZ1,Z0∩Z1and theuiby (iii) of Theorem 1. So the result follows from (i). Recall thatΩ=Ω−×T ×Ω+and thatΩ±∼=U±. SoZ0is a polynomial algebra in dim(g)variables withrvariables inverted. In particular its Krull dimension (which coin-cides with the transcendence degree of its field of fractions) is dim(g). The same holds then forZ, since it is a finitely generatedZ0-module.
3 G˜is a group of automorphisms of the algebraUˆand does not leaveZstable. However,SG˜ can be defined in
LetZ0 be a subalgebra ofZ0containingZ1∩Z0. Then the multiplicationZ1⊗Z0∩Z1
Z0 →Z0Z1is an isomorphism of algebras by the above theorem. This gives us a way to determine generators and relations forZ0Z1: Let s1, . . . , sr be the generators of C[G]G defined in Subsection 1.3. Thenπco(s1), . . . , πco(sr)are generators ofZ0∩Z1=Z0∩Z1 by Theorem 1(ii). Now assume that we have generators and relations forZ0. We use forZ1 the generatorsu1, . . . , urdefined in Subsection 1.3. For eachi∈ {1, . . . , r}we can express
πco(si)as a polynomialgi in the generators ofZ0 and as a polynomialfi in theuj. Then the generators and relations forZ0together with theui and the relationsfi=gi form a presentation ofZ0Z1.4
The fi can be determined as follows. Write sym(li) as a polynomial fi in the sym(j). Then sym(2li) is the same polynomial in the sym(2j) and πco(si)=
fi(u1, . . . , ur)by Theorem 1(iii).
Note thatπco(C[O])=Z0−C(2lP )Z+0 and thatZ0=πco(C[O])[z1, . . . , zr].
Now assume that G=SLn. For f ∈C[SLn] denote f ◦π by f˜ and put Z˜0 =
πco(C[SLn]). Then Z˜0 is generated by the ξ˜ij; it is a copy of C[SLn] in Z0. Now O consists of the matricesA∈SLn that have an LU-decomposition (without row permu-tations), that is, whose principal minorsΔ1(A), . . . , Δn−1(A)are non-zero. So C[O] =
C[SLn][Δ−11, . . . , Δ−n−11],πco(C[O])= ˜Z0[ ˜Δ−11, . . . ,Δ˜−n−11]and
Z0= ˜Z0[z1, . . . , zn−1]
˜
Δ−11, . . . ,Δ˜−n−11.
Let prO,T be the projection of O on T. An easy computation shows that Δi|O =
(ξ11· · ·ξii)◦prO,T =i◦prO,T for i=1, . . . , n−1.5SoΔ˜i =i◦prO,T ◦π=i◦
(t→t2)◦prΩ,T =2i◦prΩ,T =z2i. In Subsection 3.3 we will determine generators and
relations forZ0Z1, whereZ0 = ˜Z0[z1, . . . , zn−1]using the method mentioned above.
2. Rationality
We use the notation of Section 1 with the following modifications. The func-tions ξij, 1i, j n, now denote the standard coordinate functionals on Matn and for i∈ {1, . . . , n}, si ∈K[Matn] is defined by si(A)=tr(
iA)
for A∈Matn. Then det(xid−A)=xn+ni=1(−1)isi(A)xn−i. This notation is in accordance with [16].
Forf ∈C[Matn]we denote its restriction to SLn byfand we denoteπco(f)byf˜. So nows1, . . . , sn−1andξij are the functionss1, . . . , sn−1andξij of Subsection 1.3 and theξ˜ij are the same.
To prove the theorem below we need to look at the expressions of the functionssi in terms of theξij. We use that those equations are linear inξ1n, ξ2n, . . . , ξnn. The treatment
4 This method was also used by Krylyuk in [14] to determine generators and relations for the centre of the universal enveloping algebraU (g)ofg. Our homomorphismπco:C[G] →Z0plays the rôle of Krylyuk’s G-equivariant isomorphismη:S(g)(1)→Zp, where we use the notation of [16].
5 For twon×nmatricesAandBwe havek(AB)=k(A)k(B). From this it follows that if eitherAis
is completely analogous to that in [16, 4.1] (we use the same symbolsR,M,d andxa) to
which we refer for more explanation. LetRbe theZ-subalgebra ofC[Matn]generated by allξij withj =n.
Let∂ijdenote differentiation with respect to the variableξijand set
M=
⎡ ⎢ ⎢ ⎣
∂1n(s1) ∂2n(s1) . . . ∂nn(s1)
∂1n(s2) ∂2n(s2) . . . ∂nn(s2)
..
. ... . .. ... ∂1n(sn) ∂2n(sn) . . . ∂nn(sn)
⎤ ⎥ ⎥
⎦, c=
⎡ ⎢ ⎢ ⎣ ξ1n ξ2n .. . ξnn ⎤ ⎥ ⎥
⎦, s=
⎡ ⎢ ⎢ ⎣ s1 s2 .. . sn ⎤ ⎥ ⎥ ⎦.
Then the matrixMhas entries inRand the following vector equation holds:
M·c=s+r, wherer∈Rn. (1)
We denote the determinant ofMbyd. Fora=(a1, . . . , an)∈Knwe set
xa= ⎡ ⎢ ⎢ ⎢ ⎢ ⎣
0 · · · 0 0 an
1 · · · 0 0 an−1
..
. . .. ... ... ...
0 · · · 1 0 a2
0 · · · 0 1 a1
⎤ ⎥ ⎥ ⎥ ⎥ ⎦.
Then the minimal polynomial ofxaequalsxn− n
i=1aixn−i, det(xa)=(−1)n−1anand
d(xa)=1 (compare Lemma 3 in [16]).
Theorem 3.Zhas a rational field of fractions.
Proof. Denote the field of fractions of Z byQ(Z). From Subsection 1.5 it is clear that
Q(Z)has transcendence degree dim(sln)=n2−1 overCand that it is generated as a field by then2+2(n−1)variablesξ˜ij,u1, . . . , un−1andz1, . . . , zn−1. To prove the assertion
we will show thatQ(Z)is generated by then2−1 elementsξ˜ij,i=j,j=n,u1, . . . , un−1 andz1, . . . , zn−1. We will first eliminate thengeneratorsξ˜1n, . . . ,ξ˜nnand then then−1
generatorsξ˜11, . . . ,ξ˜n−1n−1.
Applying the homomorphismf → ˜f =πco◦(f →f):C[Matn] →Z0to both sides of (1) we obtain the following equations in theξ˜ij ands˜1, . . . ,s˜n−1
˜
M· ˜c= ˜s+ ˜r, where˜r∈ ˜Rn. (2)
HereM,˜ c˜,˜s,r˜have the obvious meaning, except that we put the last component of˜sandr˜ equal to 0 respectively 1, andR˜ is theZ-subalgebra ofZ0generated by allξ˜ij withj=n. Choosingasuch thatan=(−1)n−1we havexa∈SLn. Sinced(xa)=1, we haved=0
and therefore det(M)˜ = ˜d=0. Furthermore, fori=1, . . . , n−1,(˜s)i= ˜si∈Z0∩Z1and
Now we will eliminate the generatorsξ˜11, . . . ,ξ˜n−1n−1. We have
z2
1= ˜Δ1= ˜ξ11
and fork=2, . . . , n−1 we have, by the Laplace expansion rule,
z2
k= ˜Δk= ˜ξkkΔ˜k−1+tk= ˜ξkkz
2
k−1+tk,
wheretk is in theZ-subalgebra ofZgenerated by theξ˜ij withi, jkand(i, j )=(k, k). It follows by induction onkthat fork=1, . . . , n−1, ξ˜11, . . . ,ξ˜kk are in the subfield of
Q(Z)generated by thezi withikand theξ˜ij withi, jkandi=j. 2
3. Unique factorisation
Recall that Nagata’s lemmaasserts the following: If x is a non-zero prime element of a Noetherian integral domain S such that S[x−1] is a UFD, then S is a UFD. See [11, Lemma 19.20]. Here an element is called prime if it generates a prime ideal. The non-zero prime elements of an integral domain are always irreducible and in a UFD the converse holds. In Theorem 4 we will see that, by Nagata’s lemma, it suffices to show that the algebraZ/(d)˜ is an integral domain in order to prove thatZis a UFD. To prove this we will show by induction that the two sequences of algebras (to be defined later):
K[SLn]/(d)∼= ¯A(K)= ¯B0,0(K)⊆ ¯B0,1(K)⊆ · · · ⊆ ¯B0,n−1(K)= ¯B0(K) in characteristicpand
¯
B0(C)⊆ ¯B1(C)⊆ · · · ⊆ ¯Bn−1(C)= ¯B(C)
over C, consist of integral domains. Lemma 2 is, among other things, needed to show thatA(K)¯ ∼=K[SLn]/(d)is an integral domain. Lemmas 3 and 4 are needed to obtain bases overZ (see Proposition ¯1), which, in turn, is needed to pass to fields of positive characteristic and to apply modpreduction (see Lemma 6).
3.1. The casen=2
In this subsection we show that the centre ofUε,P(sl2)is always a UFD, without any extra assumptions onl. The standard generators forU=Uε,P(sl2)areE, F, K andK−1. PutK=Kα=K2,z1=z =Kl ,z=zα=z21=Kl. Furthermore, following [8, 3.1], we putc=(ε−ε−1)l,x= −cz−1El,y=cFl. Thenx, yandz1are algebraically inde-pendent overCandZ0=C[x, y, z1][z−11](see [8, §3]).
˜
tr generatesZ0∩Z1. Furthermoretr˜= ¯h(z+z−1), by Theorem 1(iii). Letf ∈C[u]be the polynomial withz+z−1=f (K+K−1). Thentr˜=f (u). From the formulas in [8, 5.2] it follows thattr˜= −zxy+z+z−1.
By the construction from Subsection 1.5 (we take Z0 =Z0),Z is isomorphic to the quotient of the localised polynomial algebraC[x, y, z1, u][z1−1]by the ideal generated by
−z21xy+z21+z1−2−f (u). Clearly x, uandz1 generate the field of fractions of Z. In particular they are algebraically independent. So Z[x−1] is isomorphic to the localised polynomial algebraC[x, z1, u][z−11, x−1]and therefore a UFD. By Nagata’s lemma it suf-fices to show thatx is a prime element in Z. But Z/(x) is isomorphic to the quotient of C[y, z1, u][z1−1]by the ideal generated byz21+z−
2
1 −f (u). This ideal is also gener-ated by z41−f (u)z21+1. So it suffices to show that z41−f (u)z21+1 is irreducible in
C[y, z1, u][z−11]. From the fact thatf is of odd degreel >0 (see e.g. Lemma 4 below), one easily deduces that z14−f (u)z21+1 is irreducible in C[z1, u]and therefore also in
C[y, z1, u]. Clearlyz41−f (u)z21+1 is not invertible inC[y, z1, u][z−11], so it is also irre-ducible in this ring.
3.2. SLnand the functiond
The next lemma is needed for the proof of Theorem 4. The Jacobian matrix below con-sists of the partial derivatives of the functions in question with respect to the variablesξij.
Lemma 1.Ifn3, then there exists a matrixA∈SLn(Z)such thatd(A)=0and such that some2n-th order minor of the Jacobian matrix of(s1, . . . , sn, d, Δ1, . . . , Δn−1)is±1
atA.
Proof. The computations below are very similar to those in [16, Section 6]. We denote by X the (n×n)-matrix(ξij)and for an (n×n)-matrixB=(bij)andΛ1, Λ2⊆ {1, . . . , n} we denote byBΛ1, Λ2 the matrix(bij)i∈Λ1,j∈Λ2, where the indices are taken in the natural
order.
In the computations below we will use the following two facts: ForΛ1, Λ2⊆ {1, . . . , n}with|Λ1| = |Λ2|we have
∂ij
det(XΛ1,Λ2)
=
(−1)n1(i)+n2(j )det(X
Λ1\{i},Λ2\{j}) when(i, j )∈(Λ1×Λ2), 0 when(i, j ) /∈(Λ1×Λ2),
wheren1(i)denotes the position in whichioccurs inΛ1and similarly forn2(j ). Forknwe havesk=
Λdet(XΛ,Λ)where the sum ranges over allk-subsetsΛof
Putα=((1 1), (2 2), (2 3), . . . , (2n−1), (n n), (n−1n), . . . , (2n), (2 1), (1 2)), and let
αi denote theith component ofα. We letAbe the following (n×n)-matrix:
A=
⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣
1 0 0 · · · 0 (−1)n
0 1 0 · · · 0 0
1 1 0 · · · 0 0
0 0 1 · · · 0 0
..
. ... ... . .. ... ...
0 0 0 · · · 1 0
⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦
.
The columns of the Jacobian matrix are indexed by the pairs(i, j )with 1i, jn. Let
Mα be the 2n-square submatrix of the Jacobian matrix consisting of the columns with indices fromα. By permuting inAthe first row to the last position and interchanging the first two columns, we see that det(A)=1. We will show thatd(A)=0 and that the minor
dα:=det(Mα)of the Jacobian matrix is±1 atA.
First we consider theΔk,k∈ {1, . . . , n−1}. By inspecting the matrixAand using the fact that∂ijΔk=0 ifi > korj > k, we deduce the following facts:
(∂2iΔk)(A)=
±1 ifi=k,
0 ifi > k, fori, k∈ {1, . . . , n−1}, i=1, (∂11Δ1)(A)=1,
(∂12Δk)(A)=(∂21Δk)(A)=0 for allk∈ {1, . . . , n−1}, and
(∂inΔk)(A)=0 for allk∈ {1, . . . , n−1}and alli∈ {1, . . . , n}.
Now we consider the sk. Let i∈ {1, . . . , n} and let Λ⊆ {1, . . . , n}. Assume that
∂in(det(XΛ,Λ))is non-zero atA. Then we have:
• i, n∈Λ;
• j∈Λ⇒j−1∈Λfor allj with 4jnandj=i, since otherwise there would be a zero row (inXΛ\{i},Λ\{n}(A)=AΛ\{i},Λ\{n});
• j∈Λ⇒j+1∈Λfor alljwith 3jn−1, since otherwise there would be a zero column.
First assume thati3 and that|Λ|n−i+1. Then it follows thatΛ= {i, . . . , n}and that∂in(det(XΛ,Λ))(A)= ±1.
Next assume thati=2. Then it follows that eitherΛ= {2, . . . , n}or Λ= {1, . . . , n}. In the first case we have∂in(det(XΛ,Λ))(A)=(−1)1+n−1=(−1)n. In the second case we have∂in(det(XΛ,Λ))(A)=(−1)2+n=(−1)n.
So fori, k∈ {1, . . . , n}we have:
(∂insk)(A)=
⎧ ⎪ ⎨ ⎪ ⎩
±1 ifi3 andi+k=n+1,
0 ifi3 andi+k < n+1, (−1)n ifi∈ {1,2}andk∈ {n−1, n},
0 ifi∈ {1,2}andk < n−1.
It follows from the above equalities that inM(A)the first 2 columns are equal. Sod(A)=
det(M(A))=0.
LetΛ⊆ {1, . . . , n}. Assume that∂12(det(XΛ,Λ))is non-zero atA. Then 1,2∈Λand the first row is zero. A contradiction. So∂12(det(XΛ,Λ)) is zero atA. Now assume that
∂21(det(XΛ,Λ))is non-zero atA. Then
• 1,2∈Λ;
• n∈Λ, since otherwise the first row would be zero;
• j∈Λ⇒j−1∈Λfor allj with 4j n, since otherwise there would be a zero row.
So Λ= {1, . . . , n} and∂in(det(XΛ,Λ))(A)= ±1. Thus we have(∂12sk)(A)=0 for all
k∈ {1, . . . , n}and
(∂21sk)(A)=
±1 ifk=n,
0 otherwise.
Finally, we consider the functiond. Leti∈ {1, . . . , n}, letΛ⊆ {1, . . . , n}and assume that∂12∂in(det(XΛ,Λ))is non-zero atA. Then we have:
• 1,2, i, n∈Λandi=1;
• i=2, since otherwise the first row would be zero;
• j∈Λ⇒j−1∈Λfor allj with 4j n, since otherwise there would be a zero row.
It follows thati=2,Λ= {1, . . . , n}and∂12∂in(det(XΛ,Λ))= ±1. So fori, k∈ {1, . . . , n} we have:
(∂12∂insk)(A)=
±1 if(i, k)=(2, n),
0 if(i, k)=(2, n).
We have
d=
π∈Sn
sgn(π )∂π(1)n(s1)· · ·∂π(n)n(sn). (3)
So, by the above,
(∂12d)(A)=
sgn(π )∂π(1)n(s1)∂π(2)n(s2)· · ·∂π(n−1)n(sn−1)∂12∂2n(sn)
where the sum is over all π ∈ Sn with π(n)=2. From what we know about the
∂insk we deduce that the only permutation that survives in the above sum is given by
(π(1), . . . , π(n))=(n, n−1, . . . ,3,1,2)and that(∂12d)(A)= ±1.
If we permute the rows ofMα(A)in the order given byΔ1, . . . , Δn−1, s1, . . . , sn, dand take the columns in the order given byα, then the resulting matrix is lower triangular with
±1’s on the diagonal. So we can conclude thatdα(A)=det(Mα(A))= ±1. 2
In the remainder of this subsectionKdenotes an algebraically closed field.
Lemma 2.
(i) dis an irreducible element ofK[Matn]. (ii) K[SLn]is a UFD.
(iii) The invertible elements ofK[SLn]are the non-zero scalars.
(iv) d, Δ1, . . . , Δn−1is are mutually inequivalent irreducible elements ofK[SLn].
Proof. (i) The proof of this is completely analogous to that of Proposition 3 in [16]. One now has to work with the maximal parabolic subgroupP of GLn that consists of the in-vertible matrices(aij)withani=0 for alli < n. The elementd is then a semi-invariant of
P with the weight det·ξnn−n(the restriction of this weight to the maximal torus of diagonal matrices isnn−1).
(ii) In fact it is well known that the algebra of regular functions K[G] of a simply connected semi-simple algebraic group G over K is a UFD. See [15, the corollary to Proposition 1].
(iii) and (iv). Since Δn−1 is not everywhere non-zero on SLn, it is not invertible in
K[SLn]. From the Laplace expansion for det with respect to the last row or the last column it is clear that we can eliminate ξnn using the relation det=1, if we make Δn−1 in-vertible. So we have an isomorphismK[SLn][Δ −n−11] ∼=K[(ξij)(i,j )=(n,n)][Δ−n−11]. It maps
d, Δ1, . . . , Δn−1 to respectively d, Δ1, . . . , Δn−1, since these polynomials do not con-tain the variableξnn. The invertible elements ofK[(ξij)(i,j )=(n,n)][Δ−n−11]are the elements
αΔkn−1, α∈K\ {0},k∈Z, since Δn−1 is irreducible inK[(ξij)(i,j )=(n,n)]. So the in-vertible elements of K[SLn][Δ −n−11] are the elements αΔnk−1, α∈K\ {0},k∈Z. This shows thatΔn−1 is irreducible inK[SLn], since otherwise there would be more invert-ible elements inK[SLn][Δn −−11]. So the invertible elements of K[SLn]are the non-zero scalars. Sinced and the Δi are not scalar multiples of each other, all that remains is to show thatd andΔ1, . . . , Δn−2are irreducible. We only do this ford, the argument for theΔi is completely similar. Sinced is prime inK[(ξij)(i,j )=(n,n)]andd does not divide
Δn−1, it follows thatdis prime inK[(ξij)(i,j )=(n,n)][Δ−n−11]and therefore thatdis prime inK[SLn][Δ −n−11]. To show thatdis prime inK[SLn]it suffices to show that for every
f∈K[SLn],Δn−1f ∈(d)impliesf∈(d). So assume that
for somef, g∈K[SLn]. If we takea∈Kn such thatan=(−1)n−1, then we havexa∈
SLn,d(xa)=1 andΔn−1(xa)=0. SoΔn−1does not divided. But thenΔn−1dividesg,
sinceΔn−1is irreducible. Cancelling a factorΔn−1 on both sides of(∗), we obtain that
f ∈(d). 2
3.3. Generators and relations and aZ-form forZ˜0[z1, . . . , zn−1]Z1
For the basics about monomial orderings and Gröbner bases I refer to [5].
Lemma 3.If we give the monomials in the variablesξij the lexicographic monomial
or-dering for whichξnn> ξn n−1>· · ·> ξn1> ξn−1n>· · ·> ξn−1 1>· · ·> ξ11, thendethas
leading term±ξnn· · ·ξ22ξ11anddhas leading term±ξn nn−−11· · ·ξ322ξ21.
Proof. I leave the proof of the first assertion to the reader. For the second assertion we use the notation and the formulas of Subsection 3.2. The leading term of a non-zero polyno-mialf is denoted by LT(f ). Leti∈ {1, . . . , n}andΛ⊆ {1, . . . , n}with|Λ| =k2 and assume that∂in(det(XΛ,Λ))=0. Theni, n∈Λ. Now we use the fact that no monomial in
∂in(det(XΛ,Λ))contains a variable with row index equal toior with column index equal tonor a product of two variables which have the same row or column index.
First assume thati > n−k+1. Then
LT∂in
det(XΛ,Λ)
±ξn n−1· · ·ξi+1iξi−1i−1· · ·ξn−k+1n−k+1
with equality if and only ifΛ= {n, n−1, . . . , n−k+1}. Now assume thati=n−k+1. Then
LT∂in
det(XΛ,Λ)
±ξn n−1· · ·ξn−k+2n−k+1
with equality if and only ifΛ= {n, n−1, . . . , n−k+1}. Finally assume thati < n−k+1. Then
LT∂in
det(XΛ,Λ)
±ξn n−1· · ·ξn−k+3n−k+2ξn−k+2i with equality if and only ifΛ= {n, n−1, . . . , n−k+2, i}.
So fori, k∈ {1, . . . , n}withk2 we have:
LT(∂insk)=
±ξ
n n−1· · ·ξi+1iξi−1i−1· · ·ξn−k+1n−k+1 ifi+k > n+1,
±ξn n−1· · ·ξn−k+2n−k+1 ifi+k=n+1,
±ξn n−1· · ·ξn−k+3n−k+2ξn−k+2i ifi+k < n+1. In particular LT(∂insk)±ξn n−1· · ·ξn−k+1n−k+1 with equality if and only if i+k=
n + 1. But then, by Eq. (3), LT(d) = LT(∂nns1)LT(∂n−1ns2)· · ·LT(∂1nsn) =
±ξn nn−−11· · ·ξ322ξ21. 2
Recall that the degree reverse lexicographical ordering on the monomials uα = uα1
1 · · ·u αk
Lemma 4.Letfi∈Z[u1, . . . , un−1]be the polynomial such thatsym(li)=fi(sym(1),
. . . ,sym(n−1)). If we give the monomials in the ui the degree reverse lexicographic
monomial ordering for whichu1>· · ·> un−1, thenfihas leading termuli. Furthermore,
the monomials that appear infi−uli are of total degreeland have exponents< l.6
Proof. Letσibe theith elementary symmetric function in the variablesx1, . . . , xnand let
λi∈P =X(T )be the characterA→Aii ofT. Then sym(i)=σi(e(λ1), . . . , e(λn))for
i∈ {1, . . . , n−1}. So thefican be found as follows. Fori∈ {1, . . . , n−1}, determineFi∈
Z[u1, . . . , un]such thatσi(x1l, . . . , xnl)=Fi(σ1, . . . , σn). Thenfi =Fi(u1, . . . , un−1,1). It now suffices to show that fori∈ {1, . . . , n−1},Fi−uli is aZ-linear combination of monomials in theujthat have exponents< l, are of total degreeland that contain some
uj withj > i(the monomials that containunwill become of total degree< lwhenun is replaced by 1).
Fixi∈ {1, . . . , n−1}. Consider the following properties of a monomial in thexj:
(x1) the monomial contains at leasti+1 variables; (x2) the exponents arel;
(x3) the number of exponents equal tolisi;
and the following properties of a monomial in theuj:
(u1) the monomial contains a variableuj for somej > i; (u2) the total degree isl;
(u3) the exponents are< l.
Lethbe a symmetric polynomial in thexi and letH be the polynomial in theui such that
h=H (σ1, . . . , σn). Give the monomials in thexi the lexicographic monomial ordering for whichx1>· · ·> xn. We will show by induction on the leading monomial ofhthat if each monomial that appears inhhas property (x1) respectively property (x2) respectively prop-erties (x1), (x2) and (x3), then each monomial that appears inH has property (u1) respec-tively property (u2) respecrespec-tively properties (u1), (u2) and (u3). Letxα:=xα1
1 · · ·x αn
n be the leading monomial ofh. Thenα1α2· · ·αn. Putβ=(α1−α2, . . . , αn−1−αn, αn). Letkbe the last index for whichαk=0. Thenβ=(α1−α2, . . . , αk−1−αk, αk,0, . . . ,0). Ifxαhas property (x1), thenki+1,uβhas property (u1) and the monomials that appear inσβ have property (x1), sinceσ
kappears inσβ.
Ifxα has property (x2), thenα1l,uβ is of total degreeα1land the monomials that appear inσβhave exponentsβ1+ · · · +βk=α1l. Now assume thatxα has properties (x1), (x2) and (x3). Forj < kwe haveβj =αj−αj+1< l, sinceαj+1=0. So we have to show thatβk=αk< l. Ifαk were equal tol, then we would haveα1= · · · =αk=l, by (x2). This contradicts (x3), since we haveki+1 by (x1). Finally we show that the
6 So ourf
i are related to the polynomialsPi=xil−μdiμxμfrom the proof of Proposition 6.4 in [8] as
follows:Pi=fi(x1, . . . , xn−1)−sym(li). In particulardi0=sym(li)anddiμ∈Zfor allμ∈P\ {0}(we
monomials that appear inσβ have property (x3). Ifα1< l, then all these monomials have exponents< l. So assumeα1=l. Letj be the smallest index for whichβj =0. Then the number of exponents equal tolin a monomial that appears inσβisj. On the other hand,
α1= · · · =αj=l. So we must haveji, sincexα has property (x3).
Now we can apply the induction hypothesis toh−cσβ, wherecis the leading coefficient ofh.
The assertion about Fi −uli now follows, because the monomials that appear in
σi(x1l, . . . , xnl)−σil have the properties (x1), (x2) and (x3). 2
From now on we denotezi byzi.
7LetZ[SLn]be theZ-subalgebra ofC[SLn] gener-ated by theξij andAbe theZ-subalgebra ofZgenerated by theξ˜ij. SoA=πco(Z[SLn]). LetB be theZ-subalgebra generated by the elementsξ˜ij,u1, . . . , un−1andz1, . . . , zn−1. For a commutative ringR we putA(R)=R⊗ZAandB(R)=R⊗ZB. Clearly we can identifyA(C)withZ˜0. In the proposition below “natural homomorphism” means a homo-morphism that maps ξij toξ˜ij and, if this applies, the variablesui andzi to the equally named elements ofZ. The polynomialsfi below are the ones defined in Lemma 4.
Proposition 1.The following holds:
(i) The kernel of the natural homomorphism from the polynomial algebra
Z[(ξij)ij, u1, . . . , un−1, z1, . . . , zn−1] to B is generated by the elements det−1, f1−s1, . . . , fn−1−sn−1, z21−Δ1, . . . , z2n−1−Δn−1.
(ii) The homomorphismB(C)→Z, given by the universal property of ring transfer, is injective.
(iii) A is a free Z-module and B is a free A-module with the monomials
uk1
1 · · ·u kn−1
n−1z m1
1 · · ·z mn−1
n−1 ,0ki< l,0mi<2, as a basis.
(iv) A[z1, . . . , zn−1]∩Z1=A∩Z1=Z[˜s1, . . . ,s˜n−1]andB∩Z1is a freeA∩Z1-module
with the monomialsuk1
1 · · ·u kn−1
n−1,0ki< l, as a basis.
Proof. Let Z0 be the C-subalgebra ofZ generated by the ξ˜ij andz1, . . . , zn−1. As we have seen in Subsection 1.5, the zi satisfy the relationsz2i = ˜Δi. TheΔ˜i are part of a generating transcendence basis of the field of fractions Fr(Z˜0)ofZ˜0 by arguments very similar to those at the end of the proof of Theorem 3. This shows that the monomi-als zm1
1 · · ·z mn−1
n−1 , 0mi <2, form a basis of Fr(Z0) over Fr(Z˜0)and of Z0 over Z˜0. It follows that the kernel of the natural homomorphism from the polynomial algebra
C[(ξij)ij, z1, . . . , zn−1]toZ0 is generated by the elements det−1, z21−Δ1, . . . , z2n−1−
Δn−1. So we have generators and relations forZ0. By the construction from Subsec-tion 1.5 we then obtain that the kernelI of the natural homomorphism from the polyno-mial algebra C[(ξij)ij, u1, . . . , un−1, z1, . . . , zn−1]toZ0Z1is generated by the elements det−1, f1−s1, . . . , fn−1−sn−1, z21−Δ1, . . . , zn2−1−Δn−1.
7 In [8,9]z
Now we give the monomials in the variables(ξij)ij, u1, . . . , un−1, z1, . . . , zn−1a mono-mial ordering which is the lexicographical product of an arbitrary monomono-mial ordering on the monomials in thezi, the monomial ordering of Lemma 4 on the monomials in theui and the monomial ordering of Lemma 3 on theξij.8Then the ideal generators mentioned above have leading monomialsξnn· · ·ξ22ξ11, ul1, . . . , uln−1, z21, . . . , z2n−1 and the leading coefficients are all±1. Since the leading monomials have gcd 1, the ideal generators form a Gröbner basis; see [5, Chapter 2, §9, Theorem 3 and Proposition 4], for example. Since the leading coefficients are all±1, it follows from the division with remainder algorithm that the ideal ofZ[(ξij)ij, u1, . . . , un−1, z1, . . . , zn−1] generated by these elements con-sists of the polynomials inI that have integral coefficients and that it has theZ-span of the monomials that are not divisible by any of the above leading monomials as a direct complement. This proves (i) and (ii).
(iii) The canonical images of the above monomials form aZ-basis ofB. These monomi-als are the products of the monomimonomi-als in theξij that are not divisible byξnn· · ·ξ22ξ11and the restricted monomials mentioned in the assertion. The canonical images of the mono-mials in theξijthat are not divisible byξnn· · ·ξ22ξ11form aZ-basis ofA.
(iv) As we have seen, the monomials with exponents<2 in theziform a basis of theZ˜ 0-moduleZ0. SoA[z1, . . . , zn−1] ∩ ˜Z0=A. Therefore, by Theorem 1(ii),A[z1, . . . , zn−1] ∩
Z1 = A∩ Z1 =πco(Z[SLn]SLn). Now (ZP )W =Z[sym(1), . . . ,sym(n−1)] (see [3, No. VI.3.4, Theorem 1]) and the si are in Z[SLn], so Z[SLn]SLn =Z[s
1, . . . , sn−1] by the restriction theorem forC[SLn]. This proves the first assertion. From the proof of Theorem 2 we know that the given monomials form a basis ofZ1overZ0∩Z1and a basis ofZoverZ0. So an element ofZis inZ1if and only if its coefficients with respect to this basis are inZ0∩Z1. The second assertion now follows from (iii). 2
By (ii) of the above proposition we may identifyB(C)withZ˜0[z1, . . . , zn−1]Z1 and
B(C)[ ˜Δ−11, . . . ,Δ˜−n−11]withZ.
PutZ¯ =Z/(d)˜ . For the proof of Theorem 4 we need a version forZ¯ of Proposition 1. First we introduce some more notation. Foru∈Zwe denote the canonical image ofuinZ¯
byu¯. Forf∈C[Matn]we writef¯instead off¯˜. LetA¯be theZ-subalgebra ofZ¯generated by theξ¯ij and letB¯ be theZ-subalgebra generated by the elementsξ¯ij,u¯1, . . . ,u¯n−1and
¯
z1, . . . ,z¯n−1. For a commutative ringRwe putA(R)¯ =R⊗ZA¯andB(R)¯ =R⊗ZB¯. Proposition ¯1.The following holds:
(i) The kernel of the natural homomorphism from the polynomial algebra Z[(ξij)ij, u1, . . . , un−1, z1, . . . , zn−1] to B¯ is generated by the elements det−1, d, f1−s1, . . . , fn−1−sn−1, z21−Δ1, . . . , zn2−1−Δn−1.
(ii) The kernel of the natural homomorphismZ[Matn] → ¯Ais(det−1, d).
(iii) The homomorphism B(¯ C)→ ¯Z, given by the universal property of ring transfer, is injective.
8 So thez
(iv) A¯ is a free Z-module and B¯ is a free A¯-module with the monomials ¯
uk1
1 · · · ¯u kn−1
n−1z¯ m1
1 · · · ¯z mn−1
n−1 ,0ki< l,0mi<2, as a basis. (v) TheA¯-span of the monomialsu¯k1
1 · · · ¯u kn−1
n−1,0ki< l, is closed under multiplication. Proof. From Lemma 2(iii) we deduce that(A(C)[ ˜Δ−11, . . . ,Δ˜n−−11] ˜d)∩A(C)=A(C)d˜. From this it follows, using theA(C)-basis ofB(C), that(Zd)˜ ∩B(C), which is the kernel of the natural homomorphism B(C)→ ¯Z, equals B(C)d˜. From (i) and (ii) of Proposi-tion 1 or from its proof it now follows that the kernel of the natural homomorphism from the polynomial algebraC[(ξij)ij, u1, . . . , un−1, z1, . . . , zn−1]toZ¯ is generated by the ele-ments det−1, d, f1−s1, . . . , fn−1−sn−1, z21−Δ1, . . . , z2n−1−Δn−1.
Again using theA(C)-basis ofB(C)we obtain that(B(C)d)˜ ∩A(C)=A(C)d˜. From this it follows that the kernel of the natural homomorphismC[Matn] → ¯Zis generated by det−1 andd.
By Lemma 3 we have LT(d)= ±ξn nn−−11· · ·ξ322ξ21 which has gcd 1 with the leading monomials of the other ideal generators, so the ideal generators mentioned above form a Gröbner basis overZ. Now (i)–(iv) follow as in the proof of Proposition 1.
(v) This follows from the fact that the remainder modulo the Gröbner basis of a poly-nomial inZ[(ξij)ij, u1, . . . , un−1]is again inZ[(ξij)ij, u1, . . . , un−1]. 2
By (ii) and (iii) of the above propositionA¯andB(¯ C)[ ¯Δ1−1, . . . ,Δ¯−n−11]can be identified with respectivelyZ[Matn]/(det−1, d)andZ¯. From (iv) it follows that, for any commuta-tive ringR,A(R)¯ embeds inB(R)¯ .
3.4. The theorem
Lemma 5.LetAbe an associative algebra with1over a fieldF and letLbe an extension ofF. Assume that for every finite extensionFofF,F⊗FAhas no zero divisors. Then
the same holds forL⊗F A.
Proof. Assume that there exista, b∈L⊗FA\ {0}withab=0. Let(ei)i∈Ibe anF-basis ofAand letckij∈F be the structure constants. Writea=i∈Iαiei andb=
i∈Iβiei. LetIarespectivelyIbbe the set of indicesisuch thatαi=0 respectivelyβi =0 and let
J be the set of indicesk such thatckij =0 for some(i, j )∈Ia×Ib. ThenIa andIb are non-empty andIa,IbandJ are finite. Takeia∈Iaandib∈Ib. Sinceab=0, the following equations overF in the variablesxi,i∈Ia,yi,i∈Ib,uandvhave a solution overL:
i∈Ia, j∈Ib
ckijxiyj=0 for allk∈J,
xiau=1, yibv=1.
Lemma 6.LetRbe the valuation ring of a non-trivial discrete valuation of a fieldF and letKbe its residue class field. LetAbe an associative algebra with1overRwhich is free as anR-module and letLbe an extension ofF. Assume that for every finite extensionK
ofK,K⊗RAhas no zero divisors. Then the same holds forL⊗RA.
Proof. Assume that there exista, b∈L⊗RA\ {0}withab=0. By the above lemma we may assume thata, b∈F⊗RA\ {0}for some finite extension F ofF. Let(ei)i∈I be anR-basis of A. Letν be an extension toF of the given valuation of F, letR be the valuation ring ofν, letK be the residue class field and letδ∈Rbe a uniformiser forν. Note thatRis a local ring and a principal ideal domain (and therefore a UFD) and thatK
is a finite extension ofK(see e.g. [6, Chapter 8, Theorem 5.1]). By multiplyinga andb
by suitable integral powers ofδ we may assume that their coefficients with respect to the basis(ei)i∈I are inR and not all divisible byδ (inR). By passing to the residue class fieldKwe then obtain non-zeroa, b∈K⊗R(R⊗RA)=K⊗RAwithab=0. 2
Remark.The above lemmas also hold if we replace “zero divisors” by “non-zero nilpotent elements.”
For t ∈ {0, . . . , n−1} let B¯t be the Z-subalgebra generated by the elements ξ¯ij,
¯
u1, . . . ,u¯n−1 andz¯1, . . . ,z¯t. SoB¯n−1= ¯B. For a commutative ring R we putB¯t(R)=
R⊗ZB¯t. From (iv) and (v) of Proposition ¯1 we deduce that the monomialsu¯1k1· · · ¯uknn−−11×
¯
zm1
1 · · · ¯z mt
t , 0ki < l, 0mi <2, form a basis of B¯t overA¯. So for any commutative ringRwe have bases forB¯t(R)overA(R)¯ and overR. Note thatB¯t(R)embeds inB(R)¯ , since theZ-basis ofB¯t is part of theZ-basis ofB¯.
Modifying the terminology of [11, §16.6], we define theJacobian idealof anm-tuple of polynomialsϕ1, . . . , ϕmas the ideal generated by thek×kminors of the Jacobian matrix ofϕ1, . . . , ϕm, wherekis the height of the ideal generated by theϕi.
Theorem 4.Iflis a power of an odd primep, thenZis a unique factorisation domain.
Proof. We have seen in Subsection 3.1 that forn=2 it holds without any extra assump-tions onl, so assume thatn3. For the elimination of variables in the proof of Theorem 3 we only needed the invertibility ofd˜, soZ[ ˜d−1]is isomorphic to a localisation of a poly-nomial algebra and therefore a UFD. So, by Nagata’s lemma, it suffices to prove thatd˜is a prime element ofZ, i.e. thatZ¯=Z/(d)˜ is an integral domain. We do this in 5 steps.
Step 1.B(K)¯ is reduced for any fieldK.
2nelements.9SoB(K)¯ is Cohen–Macaulay (see [11, Proposition 18.13]). LetV be the closed subvariety of (n2+2(n−1))-dimensional affine space defined by I. Then, by [11, Corollary 18.14], V is equidimensional of dimension n2−2. By Theorem 18.15 in [11] it suffices to show that the closed subvariety ofV defined by the Jacobian ideal of det−1, d, f1−s1, . . . , fn−1−sn−1, z21−Δ1, zn2−1−Δn−1does not contain any of the irreducible components ofV. This amounts to showing that this subvariety is of codimen-sion1 inV, sinceVequidimensional.
By Lemma 2, (det−1, d) is a prime ideal of K[Matn]. So we have an embedding
K[Matn]/(det−1, d)→K[V]which is the comorphism of a finite surjective morphism of varietiesV→V (det−1, d), whereV (det−1, d)is the closed subvariety of Matnthat con-sists of the matrices of determinant 1 on whichdvanishes. This morphism maps the closed subvariety ofV defined by the Jacobian ideal of det−1, d, f1−s1, . . . , fn−1−sn−1, z21−
Δ1, . . . , z2n−1−Δn−1into the closed subvariety ofV (det−1, d)defined by the ideal gen-erated by the 2nth order minors of the Jacobian matrix of(s1, . . . , sn, d, Δ1, . . . , Δn−1) with respect to the variablesξij. This follows easily from the fact thatsn=det and that the
zj anduj do not appear in thesi andΔi. Since finite morphisms preserve dimension (see e.g. [11, Corollary 9.3]), it suffices to show that the latter variety is of codimension1 in
V (det−1, d). SinceV (det−1, d)is irreducible, this follows from Lemma 1(ii).
Step 2.B¯0(K)is an integral domain for any fieldKof characteristicp.
We may assume that K is algebraically closed. From the construction of the fi (see the proof of Lemma 4) and the additivity of the pth power map in characteristic p it follows thatfi≡uli modp. So the kernel of the natural homomorphism from the polyno-mial algebraK[(ξij)ij, u1, . . . , un−1, z1, . . . , zn−1]toB(K)¯ is generated by the elements det−1, d, ul1−s1, . . . , uln−1−sn−1 and the A(K)¯ -span of the monomials u¯k11· · · ¯uktt, 0ki < l, is closed under multiplication for eacht ∈ {0, . . . , n−1}. We show by in-duction ont thatB¯0,t(K):= ¯A(K)[ ¯u1, . . . ,u¯t]is an integral domain fort=0, . . . , n−1. Fort =0 this follows from Lemma 2 and Proposition ¯1(ii). Let t∈ {1, . . . , n−1}and assume that it holds fort−1. ClearlyB¯0,t(K)= ¯B0,t−1(K)[ ¯ut] ∼= ¯Bt−1(K)[x]/(xl− ¯st). So it suffices to prove thatxl− ¯st is irreducible over the field of fractions ofB¯0,t−1(K). By the Vahlen–Capelli criterion or a more direct argument, it suffices to show thats¯tis not apth power in the field of fractions ofB¯0,t−1(K). So assume thats¯t=(v/w)p for some
v, w∈ ¯B0,t−1(K)withw=0. Then we havevp= ¯stwp= ¯ultwp. So with l=l/p, we have(v− ¯ultw)p=0. But thenv− ¯ultw=0 by Step 1. Now recall thatv andwcan be expressed uniquely asA(K)¯ -linear combinations of monomials inu¯1, . . . ,u¯t−1with expo-nents< l. If such a monomial appears with a non-zero coefficient inw, thenu¯lt times this monomial appears with the same coefficient in the expression of 0=v− ¯ultwas anA(K)¯ -linear combination of restricted monomials inu¯1, . . . ,u¯n−1. Since this is impossible, we must havew=0. A contradiction.
Step 3.B¯0(C)is an integral domain.
This follows immediately from Step 2 and Lemma 6 applied to thep-adic valuation of
Qand withL=C.
Step 4.B¯t(C)is an integral domain fort=0, . . . , n−1.
We prove this by induction ont. Fort=0 it is the assertion of Step 3. Lett∈ {1, . . . , n−1}and assume that it holds fort−1. Clearly B¯t(C)= ¯Bt−1(C)[¯zt] ∼= ¯Bt−1(C)[x]/
(x2− ¯Δt). So it suffices to prove thatx2− ¯Δt is irreducible over the field of fractions ofB¯t−1(C). Assume thatx2− ¯Δt has a root in this field, i.e. thatΔ¯t =(v/w)2for some
v, w∈ ¯Bt−1(C)withw=0. By the same arguments as in the proof of Lemma 5 we may assume that for some finite extensionF ofQthere existv, w∈ ¯Bt−1(F )withw=0 and
w2Δ¯t=v2. Letν2be an extension toFof the 2-adic valuation ofQ, letS2be the valuation ring ofν2, letK be the residue class field and letδ∈S2be a uniformiser forν2. We may assume that the coefficients ofvandwwith respect to theZ-basis ofB¯t−1mentioned ear-lier are inS2. Assume that the coefficients ofware all divisible byδ(inS2). Thenw=0 in
¯
Bt−1(K)and thereforev2=0 inB¯t−1(K). But by Step 1,B¯t−1(K)is reduced, sov=0 in
¯
Bt−1(K)and all coefficients ofvare divisible byδ. So, by cancelling a suitable power of
δinwandv, we may assume that not all coefficients ofware divisible byδ. By passing to the residue class fieldKwe then obtainv, w∈ ¯Bt−1(K)withw=0 andw2Δ¯t=v2. But then(wz¯t−v)2=0 inB¯t(K), sincez¯2t = ¯ΔtandKis of characteristic 2. The reducedness ofB¯t(K)(Step 1) now giveswz¯t−v=0 inB¯t(K). Now recall thatv andwcan be ex-pressed uniquely asA(K)¯ -linear combinations of the monomialsu¯k1
1 · · · ¯u kn−1
n−1z¯ m1
1 · · · ¯z mt−1
t−1 , 0ki< l, 0mi <2. We then obtain a contradiction in the same way as at the end of Step 2.
Step 5.Z/(d)is an integral domain.
SinceZ¯ = ¯B(C)[ ¯Δ1−1, . . . ,Δ¯n−−11]and theΔ¯i are non-zero inA(¯ C)∼=C[SLn]/(d)by Lemma 2, this follows from Step 4. 2
Remark. To attempt a proof for arbitrary odd l >1 I have tried the filtration with deg(ξij)=2l, deg(zi)=li and deg(ui)=2i. But the main problem with this filtration is that it does not simplify the relationssi=fi(u1, . . . , un−1)enough.
References
[1] A. Braun, C.R. Hajarnavis, Smooth polynomial identity algebras with almost factorial centers, Warwick preprint: 7/2003.
[2] N. Bourbaki, Algèbre, Chapitres 1, 2 et 3, Hermann, Paris, 1970.
[3] N. Bourbaki, Groupes et Algèbres de Lie, Chapitres 4, 5 et 6, Hermann, Paris, 1968.
