t j i, and then can be naturally extended to K(cf. [S-V]). The Hasse derivatives satisfy the following: is defined on k(t) by D (i)

Transcription

1 A NOTE ON WRONSKIANS AND THE ABC THEOREM IN FUNCTION FIELDS OF RIME CHARACTERISTIC Julie Tzu-Yueh Wang Insiue of Mahemaics Academia Sinica Nankang, Taipei Taiwan, R.O.C. May 14, 1998 Absrac. We provide a sharp bound for he order sequence of Wronskians. We also give anoher proof of he runcaed second main heorem over funcion fields which is a generalizaion of he ABC heorem due o Mason, Voloch, Brownawell and Masser, Noguchi and he auhor. 1. Inroducion Le C be an irreducible nonsingular projecive algebraic curve of genus g defined over an algebraically closed field k of characerisic p. Le K be he funcion field of C. Suppose ha is a local parameer of a poin of C, i.e. v () = 1. K is hen a finie separable field exension of k().(cf. [Si], Chaper II, roposiion 1.4) Recall ha he Hasse derivaives D (i) is defined on k() by D (i) ( j ) = ( j i) j i, and hen can be naurally exended o K(cf. [S-V]). The Hasse derivaives saisfy he following: roposiion 1.1. (a) D (i) (zw) = i (b) D (i) D (j) = ( ) i+j i D i+j. D (j) (z)d (i j) (w), z, w K. Le now p > 0. Following Garcia and Voloch s noaion in [G-V], we le K m = {x K D (i) (x) = 0 for 1 i < p m } and le K = {x K D (i) (x) = 0 for i 1}. roposiion 1.2. For all m 1, K m is a field. Moreover, D (pm ) on K m and K m+1 = {x K m D (pm ) (x) = 0}. is a derivaion roof. See [G-V], roposiion 1. 1 Typese by AMS-TEX

2 2 JULIE TZU-YUEH WANG Remark. K m = kk pm and K = k (See [G-V]). Throughou he paper x 0,...,x n will denoe elemens of K. Garcia and Voloch also proved he following (cf. [G-V]) Theorem A. x 0,...,x n are linearly independen over K m if and only if here exis ( ) inegers 0 = ɛ 0 < < ɛ n < p m wih de D (ɛi) (x j ) 0 In [Wa1], he auhor gave an upper bound for each ɛ i by showing he following: Theorem B. If x 0,..., x n are linearly independen over K m, hen here exis inegers 0 = ɛ 0 < ɛ 1 <... < ɛ n wih ɛ i ip m 1 such ha de(d (ɛ i) x j ) 0 i,j n 0. This bound is he bes possible when n < p. For example, Example. Le K = k() and p > n. Le x i = ip(m 1), 0 i n, where m is a posiive ineger. Then x 0,..., x n are linearly independen over K pm, and ɛ i = ip (m 1). When n p, his bound is clearly no he bes possible. We will see laer ha he field exension degree [K : K pα ] = p α. Therefore, when p α n < p α+1 i only makes sense o consider he case where x 0,..., x n are linearly independen over K pm for m α + 1. In Secion 2, we will locae he bes possible bound for each ɛ i in he general case. The bound on ɛ i is closely relaed o he heory of Weiersrass poins in posiive characerisic.(cf. [G-V], [S-V], ec.) In [Wa1], he auhor used he previous heorem and some resuls of [S-V] on Weiersrass poins o prove he runcaed second main heorem of funcion fields of any characerisic which is a generalizaion of he ABC Theorem due o Mason, Voloch, and Brownawell and Masser. The runcaed second main heorem has many applicaions in funcion field Diophanine geomery. For example, i was used by he auhor o sudy he S-inegral poins on projecive spaces minus hyperplanes.(cf. [Wa1], [Wa2], [Wa3], and [Wa4]) In Secion 3, we will include anoher proof which is in he flavor of Nevanlinna heory and only involves Wronskians. We should also menion here ha Noguchi(cf. [No]) also used echniques from Nevanlinna heory o show a runcaed second main heorem for funcion fields of higher dimension wih characerisic zero.

3 WRONSKIANS AND ABC THEOREM 3 Ackowlegemens. This noe is moivaed by some quesions abou he ABC heorem and he heory of Weiersrass poins ha arose during he ABC workshop of he 1998 Arizona Winer School. The auhor would like o hank rofessor W. McCallum for suggesions on he proof of Lemma 2.1. The auhor also wishes o hank rofessor J. F. Voloch for some helpful commens. 2. Wronskians We firs formulae some resuls o undersand he srucure of he vecor space spanned by {x 0,..., x n } over K pγ, γ 1. Lemma 2.1. Le u and m be nonnegaive inegers. Then he field exension degree of [K pm, K pm+u ] is p u. Furhermore, { ipm 0 i p u 1} is a se of basis for he vecor space K pm over K pm+u. roof. To show he field exension degree of [K pm, K pm+u ] is p u, we firs prove ha [K : K p ] = p. I hen follows similarly ha [K pm [K pm, K pm+u ] = p u. : K pm+1 ] = p. Therefore Since he funcion field K is a finie separable field exension over k(), here exiss an elemen y K such ha K = k(, y). Furhermore, we may assume ha y saisfies a monic minimal polynomial f(y ) k()[y ]. Then we have [k(, y) : k()][k() : k( p )] = [k(, y)] : k( p )] = [k(, y) : k( p, y p )][k( p, y p ) : k( p )]. Because ha [k() : k( p )] = p, i remains o show ha [k(, y) : k()] = [k( p, y p ) : k( p )]. Suppose ha f(y ) = Y r + a n 1 ()Y r a 0 () = f(, Y ), where a i () k(). We raise each coefficien of a i () o is p h power and denoe i by a p i (). Le f p (, y) = Y r + a p n 1 ()Y r a p 0 (). Then f p ( p, y p ) = ( f(, y)) p = 0. Therefore, [k( p, y p ) : k( p )] r. Since k is an algebraically closed field, k = k p. Therefore if f p ( p, Y p ) is reducible over k( p ), hen f(, Y ) will be reducible in k()[y ] which gives a conradicion. Hence [k( p, y p ) : k( p )] = r = [k(, y) : k()]. Similarly, one can show ha [K pm : K pm+1 ] = p. Therefore [K pm, K pm+u ] = p u. ( One can easily see ha de D (ipm ) jpm) 0, so he second par of 0 i,j p u 1 he asserion follows direcly from he Theorem A and he firs asserion.

4 4 JULIE TZU-YUEH WANG Lemma 2.2. Suppose p α n < p α+1 for some nonnegaive ineger α, and le β be a nonnegaive ineger less han or equal o α. If x 0,..., x n are linearly independen over K pα+s for some posiive ineger s, hen he dimension of he vecor space spanned by x 0,..., x n over K pβ+s is sricly greaer han p β. roof. Suppose r + 1 is he dimension of he vecor space spanned by x 0,..., x n over K pβ+s. Wihou loss of generaliy, we can assume ha x 0,..., x r are linearly independen over K pβ+s. By Lemma 2.1, { ipβ+s 0 i p α β 1} is a se of basis for he vecor space K pβ+s over K pα+s. Then he se is linearly independen over K pα+s. Since x j, r + 1 j n, is a linear combinaion of x 0,..., x r { ipβ+s x j 0 i p α β 1, 0 j r} over K pβ+s, he dimension of he vecor space spanned by { ipβ+s x j 0 i p α β 1, 0 j r} over K pα+s dimension of he vecor space spanned by {x 0,..., x n } over K pα+s. p α β (r + 1) n + 1 > p α. Hence r + 1 > p β. is no less han he Therefore, We will formulae an upper bound on ɛ i according o he dimension of each vecor space spanned by {x 0,..., x n } over K pγ for γ 0. Laer, we will offer some examples o show ha his formulaion is necessary. Theorem 1. Suppose ha p α n < p α+1 and m α + 1. Le he dimension of he vecor space spanned by x 0,..., x n over K pγ be l γ + 1 for 0 γ m. Le 0 = γ 0 < γ 1 < < γ u m be a sequence of inegers beween 0 and m such ha 0 = l γ0 < l γ1 < < l γu = n and l γδ = l γδ +1 = = l γδ+1 1. Then x 0,..., x n are linearly independen over K pm if and only if here exis inegers 0 = ɛ 0 < ɛ 1 < ɛ n wih p γ δ 1 + s 1 ɛ lγδ 1 +s min{sp γ δ 1, p γ δ 1}, 1 s l γδ l γδ 1, such ha de ( D (ɛi) x j )0 i,j n 0. Remark 2.1. By Lemma 2.2, l m α+i p i. Remark 2.2. I is easy o see ha in general one has ɛ i ip m. Therefore Theorem 1 is sronger han he Theorem B. Remark 2.3. When he characerisic of k is 0, ɛ i = i.

5 WRONSKIANS AND ABC THEOREM 5 Corollary. ɛ lγδ 1 +1 = p γ δ 1, for δ = 1, 2,..., u. roof. I follows easily from he inequaliies in Theorem 1. roof of Theorem 1. The if par comes from Theorem A. Therefore, we only need o show he oher direcion. Wihou loss of generaliy, we may assume ha x 0,..., x lγδ are linearly independen over K pγ δ. By definiion of γ δ, x 0,..., x lγδ 1 +1 are linearly dependen over K pγ δ 1. Therefore, by Theorem A we have ɛ lγδ 1 +1 p γ δ 1. Hence, ɛ lγδ 1 +s > ɛ lγδ s 1 p γ δ 1 + s 1. This proves one side of he inequaliy. For he oher side of he inequaliy, we will prove i by inducion. Since x 0,..., x lγ1 are linearly independen over K pγ 1, from Theorem B here exis 0 = ɛ 0 < ɛ 1 < ( wih ɛ i ip γ1 1 such ha de D (ɛi) x j 0. Togeher wih Theorem )0 i,j l γ1 ɛ lγ1 A, we conclude ɛ i min{ip γ1 1, p γ1 1}, for 1 i l γ1. The proof will be compleed by wo inducion seps. Firs we show ha if he heorem is rue for n = l γβ, 0 β < u, hen i is rue for n = l γβ + 1. We hen show ha if he heorem is rue for n = l γβ + s, hen i is rue for n = l γβ + s + 1, where 1 s < l γβ+1 l γβ. Now suppose ha he heorem is rue for n = l γβ heorem does no hold for n = l γβ + 1. Then he vecors and he conclusion of he (x j, D (1) x j, D (2) x j,..., D (pγ β+1 1) x j ), 0 j l γβ + 1 are linearly dependen over K. Then here exis a 0,..., a lγβ +1 K such ha Since x 0,..., x lγβ l γβ +1 a j D (i) x j = 0 for 0 i p γ β+1 1. (2.1) are linearly independen over K pγ β, by he inducion hypohesis we have ha he l γβ + 1 vecors above, for 0 j l γβ, are linearly independen over K. Hence a lγβ +1 0 and, wihou loss of generaliy, we can assume ha a lγβ +1 = 1. We will show ha a j K pγ β+1, for 0 j l γβ. Then we have x 0,..., x lγβ +1 are linearly dependen over K pγ β+1 ha complees his sep of inducion proof. which will give a conradicion

6 6 JULIE TZU-YUEH WANG To prove ha a j K pγ β+1, by roposiion 1.2, i suffices o show ha D (r) a j = 0 for r = 1, p, p 2,..., p γ β+1 1. We will prove i by inducion. For r = 1, applying D o (2.1), we have (i + 1) I follows from (2.1) ha l γβ +1 l γβ a j D (i+1) x j + D a j D (i) x j = 0. l γβ D a j D (i) x j = 0, for 0 i p γ β (2.2) Since he l γβ + 1 vecors above (i.e. for 0 j l γβ ) are linearly independen over K we ge D a j = 0 for 0 j l γβ. Assume ha D (pν ) a j = 0 for ν = 0, 1,..., r 1 < γ β+1 1 and for 0 j l γβ. Then we have a j K pr. Therefore, D (b) a j = 0 for 1 b p r 1 and for 0 j l γβ. Apply D (b), 1 b p r 1, o (2.1) for i = p γβ+1 1, i.e. l γβ +1 a j D (pγβ+1 1 ) x j = 0. Then we have ( p γ β b ( ) p γ β+1 Since 1 + b 0 mod p, b l γβ +1 b ) l γβ +1 a j D (pγ β+1 1 +b) x j = 0. a j D (i) x j = 0, for 1 i p γ β p r 1. (2.3) Applying he operaor D pr o (2.1) for 0 i p γ β+1 1 1, we hen have l β D (pr ) a j D (i) x j = 0, for 1 i p γβ Since he l γβ + 1 vecors above (i.e. for 0 j l γβ ) are linearly independen over K we ge D (pr ) a j = 0 for j = 0, 1,..., l γβ. This complees he firs sep of he inducion proof. Nex, we will show ha if he heorem is rue for n = l γβ + s, hen i is rue for n = l γβ + s + 1, 1 s < l γβ+1 l γβ. If (s + 1)p γβ+1 1 p γ β+1 1, hen by Theorem A he heorem holds. Therefore, we may assume ha (s + 1)p γ β+1 1 <

7 WRONSKIANS AND ABC THEOREM 7 p γ β+1 1. Suppose ha he heorem is rue for n = l γβ + s, and i does no hold for n = l γβ + s + 1, 1 s < l β+1 l β. Similar o he previous argumen, we have l γβ +s+1 a j D (b) x j = 0 for 0 b (s + 1)p γ β+1 1. (2.4) Similarly, we can assume ha a lγβ +s+1 = 1. I hen suffices o show ha D (pν ) a j = 0 for ν = 1, 2, 3,..., γ β+1 1. We will do i by inducion. Suppose D (ν) a j = 0 for ν = 1, 2, 3,..., r 1 < γ β+1 1. Applying he operaor D (pr ) 0 b sp γ β+1 1, we have o (2.4) for l γβ +s D (pr ) a j D (b) x j = 0 for 0 b sp γβ+1 1. (2.5) By he inducion hypohesis, he l γβ + s + 1 vecors above (i.e. 0 j l γβ + s) are linearly independen over K. Therefore, D (pr ) a j = 0 for j = 0, 1,..., l γβ. Inducively, we have D (ν) a j = 0 for ν = 1, 2, 3,..., p γβ+1 1. This complees he second sep of he inducion proof. Therefore he proof for his heorem is compleed. We provide he following examples o show ha he resuls is indeed he bes possible and he decomposiion on each K pγ is necessary. Example 1. Le K = k(), p = 3, and n = 10. Le x i = i for 0 i 9, and le x 10 = 18. Then x 0,..., x 10 are linearly independen over K 27. In his case, α = 2, m = 3, l 0 = 0, l 1 = 2, l 2 = 8, and l 3 = 9; γ 0 = 0, γ 1 = 1, γ 2 = 2, and γ 3 = 3. We have ɛ i = i for 0 i 9, and ɛ 10 = 18. Example 2. Le K = k(), p = 3, and n = 4. Le x 0 = 1, x 1 = 3, x 2 = 6, x 3 = 9, x 4 = 18. Then x 0,..., x 4 are linearly independen over K 27. In his case, α = 1, m = 3, l 0 = 0, l 1 = 0, l 2 = 2, and l 3 = 4; γ 0 = 0, γ 1 = 2, and γ 2 = 3. We have ɛ 1 = 3, ɛ 2 = 6, ɛ 3 = 9, ɛ 4 = 18. Example 3. Le K = k(), p = 3, and n = 3. Le x 0 = 1, x 1 = 3, x 2 = 9, x 3 = 18. Then x 0,..., x 3 are linearly independen over K 27. In his case, α = 1, m = 3, l 0 = 0, l 1 = 0, l 2 = 1, and l 3 = 3; γ 0 = 0, γ 1 = 2, and γ 2 = 3. We have ɛ 1 = 3, ɛ 2 = 9, ɛ 3 = 18.

8 8 JULIE TZU-YUEH WANG 3. The roof of he Truncaed Second Main Theorem In his secion we will give anoher proof for he runcaed second main heorem in he flavor of Nevanlinna heory. The mehod which we are going o use here is basically he argumen from [B-M] and [La](p.220), bu we replace he ordinary higher derivaives by he Hasse derivaives in order o deal wih funcion fields of posiive characerisic. We will need some basic proposiions. roposiion 3.1. (a) If g i = a ij x j wih (a ij ) GL n+1 (k), hen de(d ɛi g j ) = de(a ij )de(d ɛi x j ). (b) If h K, hen de(d ɛ i hx j ) = h n+1 de(d ɛ i x j ). (c) If x is anoher separaing variable, hen roof. See [S-V], roposiion 1.4. de(d ɛi x x j ) = ( d dx )ɛ0+ɛ1...+ɛn de(d ɛ i x j ). roposiion 3.2. Le be a local parameer of a poin C. If x be a nonzero elemen in K, hen { v ( Di x i, if x ) v (x) 0 0, if v (x) = 0. roof. If v (x) = 0, we have v (D x) 0. Therefore v ( Di x x ) 0 if v (x) = 0. If v (f) = m 0, hen here exiss η K wih v (η) = 0 such ha x = m η. By roposiion 1.1.a D i x x Therefore v ( Di x x ) i. = 1 x = i η i ( ) m j i m j D i j η ( ) m i j j Di j η.

9 WRONSKIANS AND ABC THEOREM 9 roposiion 3.3. Le L 1,..., L q be linear forms in n+1 variables wih coefficiens in k and in general posiion. Le x 0,..., x n be elemens of K such ha L i (x 0,..., x n ) 0. Denoe by l i = L i (x 0,..., x n ). If v (l 1 ) v (l 2 )... v (l q ), hen v (l 1 ) = v (l 2 ) =... = v (l q n ) = min 0 i n {v (x i )}. roof. See [Wa], roposiion 4.2. Theorem 2. Suppose ha x 0,..., x n are elemens of K, and are linearly independen over kk pm. Suppose ha L 1,..., L q are linear forms in n+1 variables wih coefficiens in k, and are in general posiion, i.e. any n+1 elemens of {L i } are linearly independen over k. Then here exiss a sequence of numbers 0 = ɛ 0 < ɛ 1 < ɛ n, such ha (q n 1)h(x 0,..., x n ) q min {ɛ n, v (L i (x 0,..., x n )) min {v (x j )}} 1 i q 0 j n i=1 / S + (ɛ ɛ n ) max{0, 2g 2 + S }. (3.1) Furhermore, when he characerisic of p = 0, ɛ i = i; when p > 0, he upper bound for ɛ i is described in Theorem 1. roof. If x 0,..., x n are linearly independen over kk pm (p 0), hen by Theorem 1 and is remarks here exiss a sequnce of bounded numbers 0 = ɛ 0 < < ɛ n such ha Le L i (x 0,..., x n ) = n de(d ɛi x j ) 0. a ij x j := l i for 1 i q. Le I = {i 0,..., i n } be an index subse of {1,..., q} and {u 1,..., u q n 1 } be he complemen of I. Le By roposiion 3.1.a G = l 1... l q de(d ɛ i x j ). (3.2) Then we have he following formula: l u1... l uq n 1 de(d ɛi l ij ) = de(a ij )de(d ɛi x j ). = c I G de(dɛ i l ij ) l i0... l in = c I G de( Dɛ i l ij n i=0 l i j )( d d ) n i=0 ɛi, (3.3)

10 10 JULIE TZU-YUEH WANG where c I is a nonzero consan in k. Le e 3.1.b we also have = min 0 i n {v (x i )}. By roposiion v (de( Dɛ i l ij l ij )) = v (de( Dɛi L ij ( e f 0,..., e f n ) n i=0 L i j ( e f 0,..., e )). (3.4) f n ) Assume ha v (l 1 ) v (l 2 ) v (l q ) for C. Then by roposiion 3.3, v (l n+1 ) = = v (l q ) = min{v (x i )} = e. Le I = {1,..., n + 1}. Then (3.3) and (3.4) imply (q n 1) min{v (x i )} = v (G) + v (de( Dɛi (l ij e p ) n i=0 l )) i j e By roposiion 3.2, we have he following wo inequaliies: n i=0 ɛ i v ( d d ). (3.5) v (de( Dɛi (l ij e ) n i=0 l )) min i j {ɛ i, v e (l ij ) + e }. (3.6) 0 i n v (de( Dɛ i (l ij e ) n i=0 l )) i j e n ɛ i. (3.7) Le S be a finie se of poins of C. Then (3.5), (3.6), (3.7) and he Riemann-Roch heorem (cf. [Ma]) imply (q n 1)h(x 0,..., x n ) q min{ɛ n, v (l j ) + e } / S j=1 + (ɛ ɛ n ) max{0, 2g 2 + S }. (3.8) i=0 This complees he proof. References [B-M] Brownawell, D and Masser, D., Vanishing Sums in Funcion Fields, Mah. roc. Cambridge hilos. Soc. 100 (1986), [G-V] Garcia, A and Voloch, J. F., Wronskians and Linear Independence in Fields of rime Characereisic, Manuscripa Mah. 59 (1987), [La] Lang, S., Inroducion o Complex Hyperbolic Spaces, Springer-Verlag, [Ma] Mason, R. C., Diophanine Equaions over Funcion Fields LMS. Lecure Noes 96, Cambridge Univ. ress, [No] Noguchi, J., Nevanlinna-Caran Theory and a Diophanine Equaion over Funcion Fields, J. rein angew. Mah. 487 (1997), [S-V] Söhr, K-O. and Voloch, J. F., Weiersrass oins and Curve over Funcion Fields, roc. London Mah. Soc.(3) 52 (1986), [Si] Silverman, J. H., The Arihmeic of Ellipic Curves, Springer-Verlag, [Vol] Voloch, J. F., Diagonal Equaions over Funcion Fields, Bol. Soc. Brazil Mah. 16 (1985), [Wa1] Wang, J., T.-Y., The Truncaed Second Main Theorem of Funcion Fields, J. of Number Theory 58 (1996),

11 WRONSKIANS AND ABC THEOREM 11 [Wa2] Wang, J., T.-Y., S-inegral poins of n {2n + 1hyperplanesingeneralposiion} over number fields and funcion fields, Trans. Amer. Mah. Soc. 348 (1996), [Wa3] Wang, J., T.-Y., Inegral poins of projecive spaces omiing hyperplanes over funcion fields of posiive characerisic, preprin [Wa4] Wang, J., T.-Y., abc esimae, inegral poins, and geomery of n minus hyperplanes, preprin Insiue of Mahemaics, Academia Sinica, Nankang, Taipei Taiwan, R.O.C. address: jwang@ mah.sinica.edu.w