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ROCKY MOUNTAIN

JOURNAL OF MATHEMATICS

Volume 33, Number 4, Winter 2003

SOLUTION OF A PROBLEM

ABOUT SYMMETRIC FUNCTIONS

ROBERTO DVORNICICH AND UMBERTO ZANNIER

ABSTRACT. Leta>b>cbe positive integers with

(a,b,c) = 1. Then the fieldQ(X a +Y a,X b +Y b ,X c +Y c )is the field of all symmetric rational functions inX,YoverQ.

This solves a conjecture made by Mead and Stein.

LetX,Ybe independent indeterminates and, for a positive integer m,let Nm =N m (X,Y)=X m +Y m be the Newton symmetric power of orderm. In the recent paper [2], the authors calculate the degree [S:Q(N a ,Nb )], whereSis the field of all symmetric rational functions inX,Ywith rational coefficients. They also raise a few conjectures on the fieldsQ(N a ,N b ,N c ).Thepurposeof the present paper is to prove their main Conjecture 1, which we state as the following. Theorem 1.Ifa>b>care distinct positive integers with (a,b,c)=1, then the functionsNa ,N b ,N c generateSoverQ. In [2] the authors also state a conjecture (see Conjecture 4 of Sec- tion 3) about the minimal degreedof a polynomial relation satisfied byNa ,N b ,N c where, by degree of a monomialN ia N j b N kc ,theymean ai+bj+ck. At the end of the paper we shall show how our Theorem 1 implies a strong form of their conjecture, namely, Theorem 2.Assumptions being as in Theorem 1, we haved=abc/2 ifabcis even andd=(a-1)bc/2otherwise.

Proof of Theorem

1. To start with, we show that it is sufficient to

prove the analogous statement withQreplaced by its algebraic closureReceived by the editors on October 2, 2000, and in revised form on March 21,

2001.
Copyrightc?2003 Rocky Mountain Mathematics Consortium 1279

1280R. DVORNICICH AND U. ZANNIER

Q. In fact, note first that, as we shall show below, we have (1) [S:Q(N a ,N b )] = [QS:Q(N a ,N b

Then, assuming

QS=Q(N

a ,N b ,N c ) and recalling the easy fact that S/Q(N a ,N b ) is finite, we find [S:Q(N a ,N b )] = [S:Q(N a ,N b ,N c )][Q(N a ,N b ,N c ):Q(N a ,N b ≥[S:Q(N a ,N b ,N c )][QS:Q(N a ,N b =[S:Q(N a ,N b ,N c )][S:Q(N a ,N b the last equality following from (1). Therefore, [S:Q(N a ,N b ,N c )] = 1, which is the desired conclusion. To prove (1) we could appeal to the theory of regular extensions (see for instance [5]); however, it is perhaps easier to proceed directly.

Letγbe a primitive element forSoverQ(N

a ,N b )andletf? Q(N a ,N b )[X] be its minimal equation overQ(N a ,N b ). We may write f=α 1 f 1 h f h ,wheref 1

···f

h ?Q(N a ,N b )[X] are nonzero andα 1 h ?Qare linearly independent overQ. Substituting

γin place ofXwe obtain a relation 0 =α

1 f 1 h f h Nowf i (γ)?SandS=Q(N 1 ,N 2 ) is purely transcendental over

Q. Hence we must havef

i (γ)=0fori=1,...,h. Finally, [S:Q(N a ,N b X f i X f=[QS:Q(N a ,N b )]. Since the opposite inequality is trivial, this concludes the argument.

We are left with the task of proving

(2)

QS=Q(N

a ,N b ,N c

LetVbe the affine variety, over

Q, determined by the generic point

(N a ,N b ,N c ). Then the inclusionQ(N a ,N b ,N c )?QS?Q(X,Y) corresponds to a dominant rational map?:A 2 →V.Toprove(2)we have just to verify that deg?= 2. Assuming the contrary, for a pointquotesdbs_dbs19.pdfusesText_25

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