On the dimension of discrete valuations of k ((X1, ..., Xn))

ON THE DIMENSION OF DISCRETE VALUATIONS OF k (( X 1 ;:::;X n )) Miguel  Angel Olalla Aosta  Faultad de Matematias. Ap do. 1160. E-41080 SEVILLA (SPAIN) olallaalgebra.us.es Novemb er 15, 2000 Abstrat Let v b e a rank-one disrete valuation of the eld k (( X 1 ;:::;X n )). We know, after [1℄, that if n = 2 then the dimension of v is 1 and if v is the usual order funtion over k (( X 1 ;:::;X n )) its dimension is n  1. In this pap er we prove that, in the general ase, the dimension of a rank-one disrete valuation an be any number b etween 1 and n  1. 1 TERMINOLOGY AND PRELIMINARIES Let k b e an algebraially losed eld of harateristi 0, R n = k [[ X 1 ;:::;X n ℄℄, M n = ( X 1 ;:::;X n ) the maximal ideal and K n = k (( X 1 ;:::;X n )) the quotient eld. Let v b e a rank-one disrete valuation of K n j k , R v its valuation ring, m v its maximal ideal and  v its residual eld of v . The enter of v in R n is m v \ R n . Throughout this pap er \disrete valuation of K n j k " will mean \rankone disrete valuation of K n j k whose enter in R n b e the maximal ideal M n ". The dimension of v is the transendene degree of  v over k . We shall supp ose that the group of v is Z . Let b K n b e the ompletion of K n with respet to v (see [2℄), b v the extension of v to b K n , R b v , m b v and  b v the ring, maximal ideal and the residual eld of b v , resp etively.  Partially supp orted by Junta de Andalua, Ayuda a grup os FQM 218. 2 THE DIMENSION OF v Fix a numb er m b etween 1 and n  1. We are featuring a onstrutive metho d in order to obtain examples of valuations with dimension m . Let us onsider the following homomorphism: ' : k [[ X 1 ;:::;X n ℄℄ ! k ( u )[[ t ℄℄ X 1 7! t X 2 7! ut X i 7! P j  1 u 1 =p j i t j where k ( u ) stands for the algebrai losure of k ( u ) and 2 < p 3 < : : : < p n are prime numb ers. Lemma 1 The homomorphism ' is one to one. Proof: Let us take the elds K 2 = k ( u ) and K i = k ( u; f u 1 =p j 3 ; j  1 g ;:::; f u 1 =p j i ; j  1 g ) for all i  3. Let us supp ose that ' is not one to one, then ker( ' ) 6 = f 0 g . So let f b e a non-zero element of M = ( X 1 ;:::;X n ) suh that f 2 ker( ' ). Let m b e the higher index suh that f 2 k [[ X 1 ;:::;X m ℄℄. If m = 1 or 2, trivially we have a ontradition. If m = 3, let us take f = f ( ' ( X 1 ) ; ' ( X 2 ) ; X 3 ) 2 K 2 [[ t; X 3 ℄℄ ; and onsider the homomorphism : K 2 [[ t; X 3 ℄℄ ! k ( u )[[ t ℄℄ t 7! t X 3 7! P j  1 u 1 =p j 3 t j : We know that f 2 ker( ) and this kernel is a prime ideal b eause is an homomorphism between integral domains. We an write f = t r g , with r  0 and t do esn't divide to g . This fores g to have some non-trivial terms in X 3 . Let s > 0 b e the minimum suh that X s 3 is one of these terms. By the Weierstrass preparation theorem we have g = U g 0 , where U ( t; X 3 ) is a unit and g 0 = X s 3 + a 1 ( t ) X s  1 3 + ::: + a s ( t ) : Sine U is a unit, g 0 2 ker( ) and ( g 0 ) = g 0 0  t; X j  1 u 1 =p j 3 t j 1 A = 0 : 2 This leads to a ontradition b eause the ro ots of g 0 are in K 2 [[ t 1 =q ℄℄, with q 2 Z , by the Puiseux theorem. If m > 3 let us take f = f ( ' ( X 1 ) ;:::;' ( X m  1 ) ; X m ) 2 K m  1 [[ t; X m ℄℄ and onsider the homomorphism : K m  1 [[ t; X m ℄℄ ! k ( u )[[ t ℄℄ t 7! t X m 7! P j  1 u 1 =p j m t j : As in the previous ase we an write f = t r h , where h 2 ker( ). So we have h = U h 0 , where U ( t; X m ) is a unit and h 0 = X r m + b 1 ( t ) X r  1 m + ::: + b r ( t ) 2 ker( ) ; so h 0 0  t; X j  1 u 1 =p j m t j 1 A = 0 : But this is again a ontradition by the Puiseux theorem: sine ker( ) is a prime ideal, we an supp ose that h 0 is an irreduible element of the ring K m  1 [[ t ℄℄[ X m ℄. In this situation the Puiseux theorem says that to obtain the o eÆients of a ro ot of h 0 = 0, like a Puiseux series in t with o eÆients in k ( u ), we have to resolve a nite number of algebrai equations of degree greater than 1 in K m  1 . Inside K m  1 we an not obtain u 1 =p m and, with a nite numb er of algebrai equations, we an obtain a nite number of p owers of u 1 =p j m but not all. So this proves the lemma. We shall extend to the quotient elds this injetive homomorphism for giving an example of a rank-one disrete valuation of k (( X 1 ;:::;X n )) of dimension 1. Lemma 2 There exists a rank-one disrete valuation of k (( X 1 ;:::;X n )) of dimension 1. Proof: We know that the homomorphism ' : k [[ X 1 ;:::;X n ℄℄ ! k ( u )[[ t ℄℄ previously dened is one to one. So we an take the valuation v =  Æ ' , where  is the usual order funtion over K ( u )(( t )) in t and ' is the natural extension to the quotient elds. Let  b e the residue X 2 =X 1 + m v 2  v . Hene, to obtain the lemma we have to prove that  = 2 k and  v is an algebrai extension of k (  ). Let us supp ose that  2 k . Then there must exist a 2 k suh that X 2 =X 1 + m v = a + m v , so X 2  aX 1 X 1 2 m v : 3 This means that v ( X 2  aX 1 ) > 1. On the other side we have ' ( X 2  aX 1 ) = ( u  a ) t; so v ( X 2  aX 1 ) = 1 and we have a ontradition. Hene  = 2 k . Let us prove that  v is an algebrai extension of k (  ). We an onsider eah element of k [[ X 1 ;:::;X n ℄℄ like a sum of forms with resp et to the usual degree. If f r is a form of degree r , then ' ( f r ) = t r P , with P a p olynomial in u and a nite numb er of elements u 1 =p i . Let us take f ; g 2 k [[ X 1 ;:::;X n ℄℄ suh that g 6 = 0 and v ( f =g ) = 0. Then ' ( f =g ) = h 0 + th 1 , where h 0 is a rational fration in u and a nite number of elements u 1 =p i . So h 0 is algebrai over k ( u ). Let us onsider P ( u; Z ) =  0 ( u ) Z m +  1 ( u ) Z m  1 + ::: +  m  1 ( u ) Z +  m ( u ) 2 k [ u ℄[ Z ℄ a p olynomial satised by h 0 , where  i ( u ) 2 k [ u ℄ for all i and  0 6 = 0. Let  b e the element  = P  X 2 X 1 ; f g  =  0  X 2 X 1   f g  m + ::: +  m  X 2 X 1  : Then we have ' (  ) =  0 ( u )( h 0 + th 1 ) m + ::: +  m ( u ) ; so v (  ) =  Æ ' (  ) > 0 and  2 m v . Subsequently, 0 + m v =  + m v = P  ; f g + m v  : This proves that f =g + m v is an algebrai element over k (  ) and, a fortiori, the lemma. Lemma 3 The dimension of a rank-one disrete valuation of k (( X 1 ;:::;X n )) is between 1 and n  1 . Proof: We know, after [1℄, that the dimension of a rank-one disrete valuation of k (( X 1 ;:::;X n )) is minor or equal than n  1. So we have to prove that there exists a transendental residue in  v . Let us supp ose that v ( X i ) = n i for all i = 1 ;:::;n . Then the value of X n 1 2 =X n 2 1 is zero, so 0 6 = ( X n 1 2 =X n 2 1 ) + m v 2  v . If this residue lies in k then there exists a 21 2 k suh that X n 1 2 X n 2 1 + m v = a 21 + m v : This implies X n 1 2 X n 2 1  a 21 = X n 1 2  a 21 X n 2 1 X n 2 1 2 m v ; 4 and then v  X n 1 2  a 21 X n 2 1 X n 2 1  > 0 : So we have v ( X n 1 2  a 21 X n 2 1 ) = m 1 > n 1 n 2 . Then v  ( X n 1 2  a 21 X n 2 1 ) n 1 X m 1 1  = 0 : If the residue of this element lies to o in k , then there must exist a 22 2 k suh that v (( X n 1 2  a 21 X n 2 1 ) n 1  a 22 X m 1 1 ) = m 2 > n 1 m 1 . We an repeat this op eration. The previous proedure is nite: if it didn't stop we would onstrut the p ower series X n 1 2  1 X i =1 b 2 i X i 1 suh that the sequene of partial sums has inreasing values. Sine  K n is a omplete eld, then this series amounts to zero in ontradition with X 1 and X 2 b eing formally independent. So the pro edure must stop and there exists a transendental element over k in  v . Theorem 4 Let m be a xed number between 1 and n  1 , then there exists a rank-one disrete valuation of k (( X 1 ;:::;X n )) of dimension m . Proof: Let us onsider the one to one (the proof of injetivity parallels that of lemma 1) homomorphism ' k [[ X 1 ;:::;X n ℄℄ ! k ( u )[[ t 1 ;:::;t m ℄℄ X 1 7! t 1 X 2 7! ut 1 X i 7! ( t i if i  m + 1 P j  1 u 1 =p j i t j 1 if i > m + 1 : We an take the valuation v :=  Æ ' , with  the usual order funtion in k ( u )[[ t 1 ;:::;t m ℄℄ and ' the natural extension to the quotient elds. We know (lemma 2) that the residue X 2 =X 1 + m v is transendental over k . Trivially the residue X i =X 1 + m v for all i = 3 ;:::;m + 1 are transendental over k ( X 2 =X 1 + m v ;:::;X i  1 + m v ) b eause t i are formally indep endent variables. Any element f =g + m v 2  v is algebrai over k ( X 2 =X 1 + m v ;:::;X m +1 + m v ) parallels that of lemma 2. So the dimension of v is m . Referenes [1℄ E. Briales, Construtive theory of valuations. , Comm. Algebra 17 (1989), no. 5, 1161{1177. [2℄ J. P. Serre, Corps loaux. , Hermann, Paris, 1968. 5