Algèbre non commutative groupes quantiques et invariants: Septième Contact Franco-Belge, Reims

285 Pages·1997·3.34 MB·English

by  Alev

Preface
The seventh meeting of the Contact Franco-Belge was held in Reims, June 26th
to 30th, 1995. This conference, followed with great enthusiasm, was the occasion to
numerous discussions among specialists with close mathematical interests. The goal
of the meeting was the presentation of recent advances in several related subjects, in
which, a non commutative algebraic approach, often turns out to be necessary.
The organization committee consisting of J. Alev, G. Cauchon, J. Calais, M. P.
Malliavin, J.P. Tignol and F. Van Oystaeyen, encouraged the presentation of both
original results and surveys. Thus, these proceedings contain three parts including
sixteen papers: thirteen on recent results and three surveys.
In the first part, five papers on recent developments concerning classical problems
are presented: S. Abhyankar’s contribution on Hilbert’s thirteenth problem, the
survey of B. Keller on tilting theory, the paper of A. Melnikov and A.A. Kirillov
on a remarkable sequence of polynomials, M. Van den Bergh’s paper on a class of
division algebras and A. Van den Essen’s survey on the jacobian conjecture.
In the second part, one can find five papers on Hopf algebras, quantum groups
and their representations, corresponding to the talks by S. Caenepeel, D. Gurevich,
S. Khoroshkin, C. Ringel and L. Willaert. These articles deal with different aspects
of the algebraic properties of quantum groups.
In the third part, there are six papers on invariant theory and representations,
correspondingtothetalksbyM.Brion,A.Joseph,H.Kraft,L.Lebruyn,O.Mathieu
et A. Melnikov. Invariant theory still continues to motivate numerous interesting
results in various directions.
On behalf of the organizing committee, we would like to thank the following
institutions for their financial support: the U.R.A. 1870 C.N.R.S., the University
and the City of Reims, the Région of Champagne-Ardenne, the Ministère de
l’Education Nationale and the European Union.
J. Alev, G. Cauchon
Société Mathématiquede France
Hilbert’s Thirteenth Problem
Shreeram S. ABHYANKAR∗
Abstract
Some progress is made in Hilbert’s Thirteenth problem.
Résumé
Un certain progrès est réalisé dansle treizième problème deHilbert.
1 Introduction
Amongst the 23 problems which Hilbert formulated at the turn of the last century
[Hi1],the13thproblemasksifeveryfunctionofnvariablesiscomposedoffunctions
of n−1 variables, with the expectation that this is not so for any n≥2.
Hilbert’s continued fascination with the 13th problem is clear from the fact that
in his last mathematical paper [Hi2], published in 1927, where he reported on the
status of his problems, Hilbert devoted 5 pages to the 13th problem and only 3
pages to the remaining 22 problems. In [Hi2], in support of the n = 2 case of the
13thproblem,Hilbertformulatedhissexticconjecturewhichsaysthat,althoughthe
solution of a generalequation of degree 6 can be reduced to the situation when the
coefficients depend on 2 variables, this cannot be cut down to 1 variable.
In the 1955 paper [A01] which represents the failure part of his Ph.D. Thesis,
AbhyankarshowedthatJung’smethodofresolvingsingularitiesofcomplexalgebraic
surfacesdoesnotcarryovertononzerocharacteristic;hedidthisbyconstructinga6
degree surface coveringwith nonsolvablelocal Galois groupabovea simple point of
thebranchlocus.Inhis1957paper[A04],bytakingasectionofthissurfacecovering,
Abhyankar was led to write down several explicit families of bivariate polynomials
f(X,Y) giving unramified coverings of the affine line in nonzero characteristic and
to suggest that their Galois groups be computed. It turned out that these Galois
groups include all the alternating and symmetric groups Alt and Sym where
N N
N > 1 is any integer, all the Mathieu groups M , M , M , M and M , the
11 12 22 23 24
linear groups SL(N,q) and PSL(N,q) where N >1 is any integer and q >1 is any
AMS1980Mathematics Subject Classification (1985 Revision):12F10,14H30,20D06,20E22
∗Mathematics Department, Purdue University, West Lafayette, IN 47907, USA — This work
waspartlysupportedbyNSFgrantDMS91–01424 andNSAgrantMDA904–92–H–3035.
Société Mathématiquede France
2 Shreeram S. ABHYANKAR
prime power, the unitary groups SU(2N −1,q) and PSU(2N −1,q) where N > 1
is any integer and q > 1 is any prime power, the symplectic groups Sp(2N,q) and
PSp(2N,q) where N > 2 is any integer and q > 1 is any prime power, and the
orthogonalgroupsΩ−(2N,q)andPΩ−(2N,q)whereN >3isanyintegerandq >1
is any odd prime power; see Abhyankar [A06] to [A12].
In the 1956 paper [A02] which represents the success part of his Ph.D. Thesis,
Abhyankarresolvedsurfacesingularitiesinnonzerocharacteristicandobservedthat
this completes the solution of Zariski’s version of Hilbert’s 14th problem in the
2 dimensional case, and shows the birational invariance of arithmetic genus for
2 dimensional varieties; later in his 1966 monograph [A05], Abhyankar resolved
singularities of 3 dimensional varieties in nonzero characteristic and observed that
this shows the birational invariance of arithmetic genus for 3 dimensional varieties.
Remarkably, it became apparent after 40 years that the above cited 6 degree
surface covering constructed in Abhyankar’s failure paper [A01] precisely solves
Hilbert’s sextic conjecture, and hence settles the n = 2 case of his 13th problem,
by showing that the algebraic closure k(X,Y)∗ of the bivariate rational function
field k(X,Y) over a field k is strictly bigger than the compositum of the algebraic
closuresk(f)∗ofk(f)withf varyingoverallelementsofthepolynomialringk[X,Y].
Likewise,Galoistheorytogetherwithideasfromresolutionofsingularitiesofhigher
dimensionalvarieties leads to a weak formof the 13thproblemfor general n,which
saysthatthealgebraicclosurek(Z ,...,Z )∗ ofthen-variablerationalfunctionfield
1 n
k(Z ,...,Z ) is strictly biggerthan the compositum of the algebraic closuresk(g)∗
1 n
of k(g) as g varies over all (n−1)-tuples g1,...,gn−1 of elements of k[Z1,...,Zn]
whose linear parts are linearly independent.
In Section 4 we shall prove the stronger version of the n = 2 case of the 13th
problemwhichsaysthat,foranyn>1,theintegralclosureB ofA =k[Z ,...,Z ]
n n 1 n
inthe algebraicclosureL =k(Z ,...,Z )∗ ofthen-variablerationalfunctionfield
n 1 n
K = k(Z ,...,Z ) over a field k is strictly bigger than the integral closure of
n 1 n
A in the compositum L(1) of the algebraic closures k(f)∗ of k(f) (in L ) with f
n n,1 n
varying over all elements of A . Actually, we shall prove more. Namely, let L(2)
n n,1
be the compositum of the algebraic closures k(f(1))∗ of k(f(1)) with f(1) varying
over all elements of L(1) which are integral over A , let L(3) be the compositum of
n n n
the algebraic closures k(f(2))∗ of k(f(2)) with f(2) varying over all elements of L(2)
n
which are integral over A , and so on. Let L = L(1) ∪L(2) ∪L(3) ∪... and let
n n,1 n,1 n,1 n,1
(cid:1)
B be the integral closure of A in L . Let A = the formal power series ring
n,1 n n,1 n
k∗[[Z ,...,Z ]] over the algebraic closure k∗ of k, let K(cid:1) = the meromorphicseries
1 n n
field k∗((Z ,...,Z ))= the quotient field of A(cid:1) , and let B(cid:1) be the integralclosure
1 n n n
(cid:1) (cid:1) (cid:1) (cid:1)
ofA inthe algebraicclosureL ofK ,wherewesupposethatL is anoverfieldof
n n n n
(cid:1) (cid:1) (cid:1) (cid:1)
L . Finally, let Ksol be the maximal solvable extension of K (in L ), i.e., Ksol is
n n n n n
SéminairesetCongrès 2
Hilbert’s Thirteenth Problem 3
(cid:1) (cid:1)
the maximal normal extension of K (in L ) such that the Galois groups of all the
n n
intermediate finite normal extensions are solvable (where we note that the Galois
group of a finite normal extension coincides with the Galois group of the maximal
(cid:1)
separable subextension); alternatively, Ksol may be defined to be the compositum
n
(cid:1)
of all the finite normal extensions of K with solvable Galois groups. In Section
n
2 we shall show that then L ⊂ K(cid:1)sol. In Section 3 we shall indicate how the
n,1 n
unsolvable 6 degree surface covering of [A01] solves Hilbert’s sextic conjecture. By
putting together the results of Sections 2 and 3, in Section 4 we shall show that B
n
isstrictlybiggerthanB ;wecallthisthepresingleton versionofthe13thproblem.
n,1
To statethe correspondingversionofthe generalcase ofthe 13thproblem,given
any n>m≥1, let L(1) be the compositum of the algebraic closures k(g)∗ of k(g)
n,m
with g varying over all m-tuples of elements of A , let L(2) be the compositum
n n,m
of the algebraic closures k(g(1))∗ of k(g(1)) with g(1) varying over all m-tuples of
elements of L(1) which are integral over A , let L(3) be the compositum of the
n,m n n,m
algebraic closures k(g(2))∗ of k(g(2)) with g(2) varying over all m-tuples of elements
ofL(2) whichareintegraloverA ,andsoon.LetL =L(1) ∪L(2) ∪L(3) ∪...,
n,m n n,m n,m n,m n,m
andletB betheintegralclosureofA inL .Thenthesaidversionconjectures
n,m n n,m
that B is strictly bigger than B ; we call this the general version of the 13th
n n,m
problem.InSection2weshallformulateaversionwhichisstrongerthanthegeneral
version and call it the analytic version of the 13th problem.
InSection5weshallsettleaweakversionofthegeneralcaseofthe13thproblem
byprovingthat,whenevern>m≥1,B isstrictlybiggerthantheintegralclosure
n
B(cid:3) of A in the compositum L(cid:3) of K and the algebraicclosures k(g)∗ of k(g)
n,m n n,m n
as g varies over all m-tuples g ,...,g of elements of A whose linear parts (i.e.,
1 m n
terms ofdegree1)arelinearlyindependent overk;we callthis the prelinear version
of the 13th problem.
InSection6we shallproveanextremelyweakversionofthe 13thproblemwhich
says that, for any partition n +···+n = n of n into positive integers n ,...,n
1 t 1 t
with t > 1, B is strictly bigger than the integral closure B(cid:3)(cid:3) of A in the
n n1,...,nt n
compositum L(cid:3)n(cid:3)1,...,nt of Kn and the algebraic closures k({Zj : n1 +···+ni−1 <
j ≤ n1+···+ni})∗ of k({Zj :n1+···+ni−1 < j ≤ n1+···+ni}) for 1 ≤i ≤t;
we call this the prepartition version of the 13th problem. It may be noted that the
n=2 case ofthis canbe found in Abhyankar’s1956paper [A03] which waswritten
to answer a question of Igusa.
In Sections 4, 5 and 6 we shall actually prove the analytic, and hence
stronger,forms ofthe presingleton,prelinearandprepartitionversionsandwe shall
respectively call these the singleton, linear and partition versions.
In his discussionof the 13th problem, Hilbert did not make it clear what kind of
functions he had in mind. We have interpreted them as integral functions. In their
Société Mathématiquede France
4 Shreeram S. ABHYANKAR
1976 reformulation, Arnold-Shimura [ArS] took them to be algebraic functions. In
their1963articles,Arnold[Ar]andKolmogorov[Kol]thoughtofthemascontinuous
functions.
It is a pleasure to thank Jim Madden for stimulating conversations concerning
the Hilbert 13th problem.
2 Analytic version and solvability
Givenanyfieldkandintegersn>m≥1,letA ,B ,K ,L ,k∗,A(cid:1) ,B(cid:1) ,K(cid:1) ,L(cid:1) ,K(cid:1)sol
n n n n n n n n n
and L(1) ,L(2) ,L(3) ,...,L ,B be as in Section 1. Let L(cid:2)(1) be the composi-
n,m n,m n,m n,m n,m n,m
tum of the algebraic closures k∗(g)∗ of k∗(g) with g varying over all m-tuples of
elements of A(cid:1) . Let L(cid:2)(2) be the compositum of the algebraic closures k∗(g(1))∗ of
n n,m
k∗(g(1))withg(1) varyingoverallm-tuplesofelementsofL(cid:2)(1) ∩B(cid:1) ,letL(cid:2)(3) bethe
n,m n n,m
compositum of the algebraic closures k∗(g(2))∗ of k∗(g(2)) with g(2) varying over all
m-tuplesofelementsofL(cid:2)(2) ∩B(cid:1) ,andsoon.LetL(cid:2) =L(cid:2)(1) ∪L(cid:2)(2) ∪L(cid:2)(3) ∪...,
n,m n n,m n,m n,m n,m
(cid:2) (cid:1) (cid:2)
and let B be the integral closure of A in L . Now obviously:
n,m n n,m
Remark 2.1. L ⊂L(cid:2) and hence B ⊂B(cid:2) .
n,m n,m n,m n,m
Therefore if we conjecture that B (cid:10)⊂ B(cid:2) and call this the preanalytic version
n n,m
of the 13th problem, then clearly:
Remark 2.2. The preanalytic version for k,n,m implies the general version for
k,n,m.
(cid:1)
For any finite sequence r = (r ,...,r ) of elements in B , by basic properties
1 u n
(cid:1)
of complete local rings, as given in Chapter VIII of [ZS2], we see that A [r] is an
n
n-dimensional complete local domain and k∗ is a coefficient field of A(cid:1) [r], i.e., k∗
n
(cid:1) (cid:1)
is mapped bijectively onto the residue field A [r]/M(A [r]) by the residue class
n n
epimorphism µ : A(cid:1) [r] (cid:11)→ A(cid:1) [r]/M(A(cid:1) [r]) where M(A(cid:1) [r]) is the maximal ideal
r n n n n
(cid:1) (cid:1)
in A [r]. Given any finite sequence of elements s = (s ,...,s ) in A [r], we put
n 1 v n
s¯= (s¯ ,...,s¯ ) = (s −s˜ ,...,s −s˜ ), where s˜ ,...,s˜ are the unique elements
1 v 1 1 v v 1 v
in k∗ such that µ (s ) = µ (s˜ ),...,µ (s ) = µ (s˜ ), and by k∗[[s]] we denote the
r 1 r 1 r v r v
closureofk∗[s¯]inA(cid:1) [r]withrespecttoitsKrulltopology.Notethatthenk∗[[s]]isa
n
complete localdomain of dimension at most v and k∗ is a coefficient field of k∗[[s]];
by k∗((s)) we denote the quotientfield of k∗[[s]]; likewise by k∗((s))∗ we denote the
algebraic closure of k∗((s)) (in L(cid:1) ). If r(cid:3) =(r(cid:3),...,r(cid:3) ) is any other finite sequence
n 1 u(cid:1)
in B(cid:1) suchthatthe elements s ,...,s belong to A(cid:1) [r(cid:3)]then bypassingto A(cid:1) [r,r(cid:3)]
n 1 v n n
weseethat(foranyfinitesequencesinB(cid:1) )theabovedefinitionsofs¯,k∗[[s]],k∗((s))
n
and k∗((s))∗ are independent of r (for instance we can take r = s). Note that if s
is a singleton, i.e., if v =1, then either k∗[[s]]=k∗ or k∗[[s]] is a complete discrete
valuation ring, and hence in both the cases (by generalized Newton’s Theorem)
k∗((s))∗ is a solvable extension of k∗((s)), i.e., k∗((s))∗ is a normal extension of
SéminairesetCongrès 2
Hilbert’s Thirteenth Problem 5
k∗((s)) such that the Galois groups of all the finite normal intermediate extensions
are solvable. [The said generalized Newton’s Theorem says that the Galois group
of a finite Galois extension of a field which is complete with respect to a discrete
valuationwithalgebraicallyclosedresiduefieldisalwayssolvable;inviewofHensel’s
Lemma(seeChapterVIII of[ZS2]),thisfollowsfromthefactthattheinertiagroup
of a discrete valuation is always solvable (see Chapter V of [ZS1])].
Let L(cid:1)(1) be the compositum of the fields k∗((g))∗ with g varying over all m-
n,m
tuples ofelements of A(cid:1) . Let L(cid:1)(2) be the compositum ofthe fields k∗((g(1)))∗ with
n n,m
g(1) varyingoverallm-tuplesofelementsofL(cid:1)(1) ∩B(cid:1) ,letL(cid:1)(3) bethecompositum
n,m n n,m
ofthefieldsk∗((g(2)))∗ withg(2) varyingoverallm-tuplesofelementsofL(cid:1)(2) ∩B(cid:1) ,
n,m n
andsoon.LetL(cid:1) =L(cid:1)(1) ∪L(cid:1)(2) ∪L(cid:1)(3) ∪...,andletB(cid:1) betheintegralclosure
n,m n,m n,m n,m n,m
(cid:1) (cid:1)
of A in L . Now obviously:
n n,m
Remark 2.3. L(cid:2) ⊂L(cid:1) and hence B(cid:2) ⊂B(cid:1) .
n,m n,m n,m n,m
Therefore if we conjecture that B (cid:10)⊂ B(cid:1) and call this the analytic version of
n n,m
the 13th problem, then clearly:
Remark 2.4. The analytic version for k,n,m implies the preanalytic version for
k,n,m.
By induction on i we shall show that L(cid:1)(i) ⊂K(cid:1)sol for all i≥0 where L(cid:1)(0) =K(cid:1) .
n,1 n n,1 n
Obviously L(cid:1)(0) ⊂ K(cid:1)sol. So let i > 0 and assume that L(cid:1)(i−1) ⊂ K(cid:1)sol. Given any
n,1 n n,1 n
h ∈ L(cid:1)(i), we can find a finite sequence r = (r ,...,r ) of elements in L(cid:1)(i−1)∩B(cid:1)
n,1 1 u n,1 n
such that h is algebraic over the compositum D of k∗((r )),...,k∗((r )). Clearly
1 u
D is the quotient field of the compositum C of k∗[[r ]],...,k∗[[r ]], and we have
1 u
C ⊂ A(cid:1) [r]. By the induction hypothesis A(cid:1) [r] ⊂ K(cid:1)sol and hence D ⊂ K(cid:1)sol. As
n n n n
noted above, k∗((r ))∗ is a solvable extension of k∗((r )). This being so for every
j j
j we see that D(k∗((r ))∗,...,k∗((r ))∗) is a solvable extension of D. Therefore
1 u
D(k∗((r ))∗,...,k∗((r ))∗) ⊂ K(cid:1)sol and hence h ∈ K(cid:1)sol. Consequently L(cid:1)(i) ⊂ K(cid:1)sol.
1 u n n n,1 n
This completes the induction. Thus, in view of 2.1 and 2.3, we have proved that:
Theorem2.5 — L(cid:1) ⊂K(cid:1)sol and hence in particular L ⊂K(cid:1)sol.
n,1 n n,1 n
3 Unsolvable coverings
Given any field k and integer n>1, let A ,B ,K ,L ,k∗,A(cid:1) ,B(cid:1) ,K(cid:1) ,L(cid:1) ,K(cid:1)sol be
n n n n n n n n n
as in Section 1. Let
F =F(Y)=YQ+ZRY +ZS ∈A [Y]⊂A(cid:1) [Y]
2 1 n n
whereRandS arepositiveintegersandQ>1isanintegerwithGCD(Q−1,R)=1.
Bythe calculationoftheY-discriminantDisc (F)ofF onpage105of[A06]wesee
Y
Société Mathématiquede France
6 Shreeram S. ABHYANKAR
that Disc (F)(cid:10)=0 and hence we can talk about the Galois group Gal(F,K(cid:1) ) of F
Y n
(cid:1) (cid:1)
over K as a subgroup of Sym . Let G= Gal(F,K ).
n Q n
In Example 5 of [A01] we have concluded that if char k (= characteristic of k)
is a prime number p, n = 2, Q = p+1, R = p−1 and S = p+1, then G is a
large complicatedsubgroupof Sym because its order is divisible by p(p+1). By
p+1
using the MTR (= Method of Throwing away Roots) technique of [A06] and by
paraphrasingaproofgiventhere weshallshowthatif p(cid:10)=7 andthe integersRand
S have suitable divisibility properties then actually G= PSL(2,p).
Moreoverweshallshowthat,withouttheaboveassumptionsofthesaidExample
5, most of the time (especially when char k is zero) G is unsolvable.
More precisely we shall prove 3.1 to 3.5:
Lemma3.1 — G is doubly transitive.
Lemma3.2 — If char k = p > 0 and Q = q +1 where q > 1 is a power of p,
and in case of p = 2 we have GCD(q−1,S)= 1 whereas in case of p > 2 we have
GCD(q−1,S) = 2, then G = PSL(2,q) except that in case of q = p = 7 we may
have G= PSL(2,7) or AΓL(1,8).
Lemma3.3 — If Q is not a prime power then G is unsolvable.
Theorem3.4(Aformofthesexticconjecture) — If Q=6 then G is unsolvable.
Corollary3.5 — B (cid:10)⊂K(cid:1)sol.
n n
To prove 3.1 we first note that obviouslyF is an irreducible monic distinguished
polynomial in Z over k∗[[Y,Z ,...,Z ]] and hence by a Gauss Lemma type
1 2 n
argument using the Weierstrass Preparation Theorem we see that F is irreducible
(cid:1)
as a polynomial in Y over K . Therefore G is transitive. Let V be the real discrete
n
(cid:1) (cid:1)
valuationofK whosevaluationringisthelocalizationofA attheprincipalprime
n n
ideal generated by Z . Now the coefficients of F have nonnegative V-value and by
1
reducing them modulo the maximal ideal of the valuation ring of V we get the
polynomial H = YQ +ZRY. Clearly H factors as H = Y(YQ−1 +ZR) into two
2 2
coprime irreducible factors over the residue field k∗((Z ,...,Z )) of V. Therefore
2 n
by Hensel’s Lemma, F factors into two coprime monic irreducible polynomials of
degrees 1 and Q−1 in Y overthe V-completion k∗((Z ,...,Z ))((Z )) of K(cid:1) , and
2 n 1 n
hence upon letting β to be arootofF(Y) wesee thatV hasexactly twoextensions
W and W(cid:3) to K(cid:1) (β) and after labelling them suitably we have W(β)>0=W(cid:3)(β)
n
and then the ramificationexponents of W and W(cid:3) are both 1 whereas their residue
degreesare1andQ−1respectively.FromthisitfollowsthatGisdoublytransitive,
which proves 3.1.
By Burnside’s Theorem (see page 89 of [A06] including footnotes 37 to 40), a
doubly transitive permutation group contains a unique minimal normal subgroup,
and the said subgroupis either elementary abelian or nonabeliansimple; moreover,
SéminairesetCongrès 2
Hilbert’s Thirteenth Problem 7
the firstcaseoccursifandonlyifthe unique minimalnormalsubgroupisregularas
a permutation group; hence in the first case the degree of the group = the degree
of the said subgroup = the order of the said subgroup = a prime power. Therefore
3.1 implies 3.3.
Noting that 6 is (the smallest integer which is) not a prime power, 3.3 implies
3.4. Now β ∈B and, taking Q=6, by 3.4 we get β (cid:10)∈K(cid:1)sol, which proves 3.5.
n n
To prove 3.2, assume that char k = p > 0 and Q = q + 1 where q > 1 is a
power of p. Let F(cid:3)(Y) ∈ K(cid:1) (β)[Y] be obtained by throwing away the root β of
n
F(Y). Then F(cid:3)(Y)=(1/Y)[F(Y +β)−F(β)]=Yq+βYq−1−(ZS/β). Let F(cid:2)(Y)
1
be obtained from F(cid:3)(Y) by reciprocation. Then F(cid:2)(Y) = (−β/ZS)YqF(cid:3)(1/Y) =
1
Yq−(β2/ZS)Y −(β/ZS).LetF(cid:2)(cid:3)(Y)∈K(cid:1) (β,γ)[Y]beobtainedbythrowingawaya
1 1 n
rootγ ofF(cid:2)(Y).ThenF(cid:2)(cid:3)(Y)=(1/Y)[F(cid:2)(Y +γ)−F(cid:2)(γ)]=Yq−1−(β2/ZS).Hence if
1
p>2andS ≡0 (mod 2)thenF(cid:2)(cid:3)(Y)=[Y(q−1)/2+(β/ZS/2)][Y(q−1)/2−(β/ZS/2)].
1 1
In view of the relations F(β)=0 and W(β)>0 we have β =ZSβ˜ with W(β˜)=0.
1
NowinviewoftheequationF(β)=0weseethatW(β˜+Z−R)>0.Consequentlyin
2
(cid:2) (cid:2)
view ofthe equationF(γ)=0 we see thatW has a unique extension U to K (β,γ)
n
and for this extension the ramification exponent is 1 and the residue degree is q. It
followsthatifp=2andGCD(q−1,S)=1thenthepolynomialF(cid:2)(cid:3)(Y)isirreducible
over K(cid:2) (β,γ), whereas if p > 2 and GCD(q − 1,S) = 2 then the polynomials
n
Y(q−1)/2+(β/ZS/2)andY(q−1)/2−(β/ZS/2)areirreducibleoverK(cid:2) (β,γ).Therefore
1 1 n
as on page 114 of [A06], as a consequence of the Zassenhaus-Feit-Suzuki Theorem,
we get 3.2.
4 Singleton version
Given any field k and integer n > 1, let A ,B ,K ,L ,k∗,A(cid:1) ,B(cid:1) ,K(cid:1) ,L(cid:1) ,K(cid:1)sol
n n n n n n n n n
andB ,L ,B(cid:1) ,L(cid:1) be as inSection1.Let us callthe assertionB (cid:10)⊂B(cid:1) the
n,1 n,1 n,1 n,1 n n,1
singleton version of the 13th problem. Then by 2.5 and 3.5 we get the following:
Theorem4.1 — The singleton version is true, i.e., B (cid:10)⊂ B(cid:1) . In particular, the
n n,1
presingleton version is true, i.e., B (cid:10)⊂B .
n n,1
5 Linear version
Given any field k and integers n > m ≥ 1, let A ,B ,K ,L ,k∗,A(cid:1) ,B(cid:1) ,K(cid:1) ,L(cid:1)
n n n n n n n n
and B(cid:3) ,L(cid:3) be as in Section 1. Let L(cid:1)(cid:3) be the compositum of K(cid:1) and the
n,m n,m n,m n
algebraicclosuresk∗((g))∗ ofk∗((g))with g varyingoverallm-tuplesofelementsof
(cid:1)
A whose constant terms are zero and whose linear parts are linearly independent
n
over k∗. Let B(cid:1)(cid:3) be the integral closure of A(cid:1) in L(cid:1)(cid:3) . Now obviously:
n,m n n,m
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8 Shreeram S. ABHYANKAR
Remark 5.1. L(cid:3) ⊂L(cid:1)(cid:3) and hence B(cid:3) ⊂B(cid:1)(cid:3) .
n,m n,m n,m n,m
ThereforeifweassertthatB (cid:10)⊂B(cid:1)(cid:3) andcallthisthe linear version ofthe 13th
n n,m
problem, then clearly:
Remark 5.2. The linear version for k,n,m implies the prelinear version for k,n,m.
We shall now prove the following:
Lemma5.3 — Let ∆ be a nonzero homogeneous polynomial of degree e > 1 in
Z ,...,Z with coefficients in k∗ such that (0,...,0) is the only point in k∗n at
1 n
which ∆=0 =∆ for 1 ≤i≤ n where ∆ is the partial derivative of ∆ relative to
i i
Z . Then (I) for every set of linearly independent homogeneous linear polynomials
i
z1,...,zn in Z1,...,Zn with coefficients in k∗ we have ∆ (cid:10)∈ k∗[z1,...,zn−1] (thus,
in the sense of Hironaka’s desingularization paper [Hir], for the singularity of the
hypersurface ∆=0 at the origin we have ν =e and τ =n).
Moreover (II) if n > 2 then ∆ is irreducible in k∗[Z ,...,Z ]. Now let Θ ∈ A(cid:1)
1 n n
be such that Θ−∆∈M(A(cid:1) )e+1 where M(A(cid:1) ) is the maximal ideal in A(cid:1) , let d>1
n n n
(cid:1)
be an integer which is nondivisible by char k, and let Θ1/d be a dth root of Θ in L ,
n
(cid:1)
i.e., an element of L whose dth power is Θ.
n
Then (III) assuming n > 2 we have Θ1/d (cid:10)∈ B(cid:1)(cid:3) (in particular, by taking
n,m
Θ = ∆ = Ze + ··· + Ze where e > 1 is an integer nondivisible by char k, we
1 n
get a concrete element Θ∈A which has the desired properties and hence for which
n
we have Θ1/d ∈B but Θ1/d (cid:10)∈B(cid:1)(cid:3) ).
n n,m
In view of the last parenthetical observation, 5.3 implies the linear version for
n>2; for n=2, the linear versionfollows from the singletonversionprovedin 4.1.
To prove (I), let δ be the expression of ∆ as a polynomial in z ,...,z with
1 n
coefficients in k∗, and let δ be the partial derivative of δ with respect to z . Now
i i
the condition that (0,...,0) is the only point of k∗n at which ∆ = 0 = ∆ for
i
1 ≤ i ≤ n is equivalent to the condition that (0,...,0) is the only point of k∗n
at which δ = 0 = δi for 1 ≤ i ≤ n. If ∆ ∈ k∗[z1,...,zn−1] then we would have
δ = 0 = δ for 1 ≤ i ≤ n at (0,...,a ) for every a ∈ k∗ which would be a
i n n
contradiction. Therefore we must have ∆(cid:10)∈k∗[z1,...,zn−1]. This proves (I).
If ∆ = ∆(cid:3)∆(cid:3)(cid:3) with nonconstant polynomials ∆(cid:3) and ∆(cid:3)(cid:3) then ∆(cid:3) and ∆(cid:3)(cid:3) must
be homogeneous, ∆(cid:3) =0=∆(cid:3)(cid:3) for an (n−2)-dimensional algebraic set in k∗n, and
every point of ∆(cid:3) =0=∆(cid:3)(cid:3) is singular for ∆=0. This proves (II).
Toprove(III)assumethatΘ1/c ∈B(cid:1)(cid:3) wherecisapositiveintegernondivisible
n,m
(cid:1)
by char k. Then Θ1/c is separable over K . Therefore we can find a finite number
n
of triples (g(j),h(j),P(j))1≤j≤u such that, for 1 ≤ j ≤ u, g(j) is an m-tuple of
(cid:1)
elements of A whose constant terms are zero and whose linear parts are linearly
n
independent over k∗, h(j) ∈ k∗((g(j)))∗, and P(j) = P(j)(Y) is a univariate monic
polynomial over k∗[[g(j)]] whose Y-discriminant Disc (P(j)) is a nonzero element
Y
SéminairesetCongrès 2
Hilbert’s Thirteenth Problem 9
of k∗[[g(j)]] and for which P(j)(h(j)) = 0, and such that Θ1/d ∈ K(cid:1) (h(1),...,h(u)).
n
(cid:1)
Now assume that n>2. Then by (II) we see that Θ is irreducible in A , and hence
n
(cid:1)
we get a real discrete valuation Ω of K whose valuation ring is the localization
n
(cid:1)
of A at the principal prime ideal generated by Θ. For any j, by (I) we see that
n
(cid:1) (cid:1)
Disc (P(j)) is nondivisible by Θ in A andhence Ω is unramifiedin K (h(j)). This
Y n n
beingsofor1≤j ≤u,weconcludethatΩisunramifiedinK(cid:1) (h(1),...,h(u)).Since
n
Θ1/d ∈K(cid:1) (h(1),...,h(u)), we must have c=1. This proves (III).
n
As said above, as a consequence of 4.1, 5.2 and 5.3 we get:
Theorem5.4 — The linear version is true, i.e., B (cid:10)⊂ B(cid:1)(cid:3) . In particular, the
n n,m
prelinear version is true, i.e., B (cid:10)⊂B(cid:3) .
n n,m
6 Partition version
Given any field k and integers n +···+n = n with n > 0,...,n > 0,t > 1
1 t 1 t
let A ,B ,K ,L ,k∗,A(cid:1) ,B(cid:1) ,K(cid:1) ,L(cid:1) and B(cid:3)(cid:3) ,L(cid:3)(cid:3) be as in Section 1.
Let Ln(cid:3)(cid:3) n nbe nthe conmponsitumn onf K(cid:1) annd1,..t.,hnet alng1e,.b..r,natic closures k∗(({Z :
n1,...,nt n j
n1+···+ni−1 <j ≤n1+···+ni}))∗ofk∗(({Zj :n1+···+ni−1 <j ≤n1+···+ni}))
for 1≤i≤t.LetB(cid:3)(cid:3) be the integralclosureofA(cid:1) inL(cid:3)(cid:3) .Now obviously:
n1,...,nt n n1,...,nt
Remark 6.1. L(cid:3)(cid:3) ⊂L(cid:1)(cid:3)(cid:3) and hence B(cid:3)(cid:3) ⊂B(cid:1)(cid:3)(cid:3) .
n1,...,nt n1,...,nt n1,...,nt n1,...,nt
Therefore if we assert that B (cid:10)⊂ B(cid:1)(cid:3)(cid:3) and call this the partition version of
n n1,...,nt
the 13th problem, then clearly:
Remark 6.2. The partition version for k,n ,...,n implies the prepartition version
1 t
for k,n ,...,n .
1 t
Also clearly:
Remark 6.3. The partition version obviously follows from the linear version 5.4.
Alternatively:
Remark 6.4. Upon letting λ=k∗((Z2,...,Zn−1))∗ and Λ = the integral closure of
λ[[Z ,Z ]] in the compositum of λ((Z ))∗, λ((Z ))∗ and λ((Z ,Z )), by the two
1 n 1 n 1 n
proofs sketched in [A03] we see that for any g(Z ) ∈ λ[[Z ]] and h(Z ) ∈ λ[[Z ]]
1 1 n n
with g(0) = 0 (cid:10)= g(Z ) and h(0) = 0 (cid:10)= h(Z ) and any integer E > 1 nondivisible
1 n
by char k we have [g(Z ) + h(Z )]1/E (cid:10)∈ Λ. Clearly B(cid:1)(cid:3)(cid:3) ⊂ Λ. By taking
1 n n1,...,nt
g(Z ) ∈ k[Z ] and h(Z ) ∈ k[Z ] (for instance g(Z ) = Z and h(Z ) = Z ))
1 1 n n 1 1 n n
we also get [g(Z )+h(Z )]1/E ∈ B . Thus the partition version also follows from
1 n n
[A03].
In view of 6.2, by 6.3 or 6.4 we get:
Theorem6.5 — The partition version is true, i.e., B (cid:10)⊂ B(cid:1)(cid:3)(cid:3) . In particular,
n n1,...,nt
the prepartition version is true, i.e., B (cid:10)⊂B(cid:3)(cid:3) .
n n1,...,nt
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