Finite Fields and Their Applications 5, 254}265 (1999) Article ID !ta.1999.0243, available online at http://www.idealibrary.com on
Power Sums over Finite Subspaces of a Field Nigel P. Byott* and Robin J. Chapman School of Mathematical Sciences, University of Exeter, Exeter EX4 4QE, United Kingdom E-mail:
[email protected],
[email protected] Communicated by Peter Jau-Shyong Shiue Received June 10, 1998; revised October 6, 1998
Let < be a "nite additive subgroup of a "eld K of characteristic p'0. We consider sums of the form S (< : a)" (v#a)F for h50 and a3 K. In particular, we give F TZ4 for the vanishing of S (<; a), in terms of the digit necessary and su$cient conditions F sum in the base-p expansion of h, in the case that < has index p in K. The proof involves the polynomial f (x)" (x!v). We de"ne
1.
INTRODUCTION AND STATEMENT OF RESULTS
Let K be a "eld of characteristic p'0, and let < be a "nite additive subgroup of K. We will consider a class of sums running over a coset of <. In the case that K is "nite, we will also de"ne a kind of generalised trace Tr : KP<, which coincides with the usual trace whenever < is a sub"eld of 4 K. There is also an associated cotrace map Cotr : KP
0, if hI0 (mod q!1) or h"0, vF" !1, if h,0 (mod q!1) and h'0. TZ %O
* Corresponding author. 254 1071-5797/99 $30.00 Copyright 1999 by Academic Press All rights of reproduction in any form reserved.
(1.1)
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(The exceptional behaviour for h"0 is because of the contribution 03"1 for v"0.) We will consider more general sums of a similar form. More precisely, for any a3 K and any integer h50, we de"ne S (< ; a)" (v#a)F3 K. F TZ4
(1.2)
We then have the following partial analogue of (1.1). THEOREM 1. ¸et < and a be as above, and let < have order pK. ¹hen S (<; a)"0, if 04h4pK!2, F S (<; a)O0, if h"pK!1. F For any power q of p, we write ds (h) for the base-q digit sum of h: O L L ds (h)" h , where h" h qG with 04h 4q!1. O G G G G G
(1.3)
When K is "nite and < has index p in the additive group of K, we can give necessary and su$cient conditions, in terms of ds (h), for S (<; a) to vanish. N F THEOREM 2. ¸et K"%pK> , let < be an additive subgroup of K of order pK, and let 04h4pK>!1. (¹hus 04ds (h)4(p!1) (m#1)). N (i) Suppose that a3 K!<. ¹hen S (<; a)O0 if and only if (p!1)m4 F ds (h)((p!1) (m#1). N (ii) Suppose that a3 <. ¹hen S (<; a)O0 if and only if ds (h)"(p!1)m or F N h"pK>!1. Remark. There is no loss of generality in taking h to lie in the range indicated, since for h'0 it is clear that S (<; a) depends only on h mod F pK>!1. The need to consider the sums S (<; a) arose in an investigation [1] of the F integral Galois module structure of certain extensions of p-adic "elds. In a brief appendix to the present paper, we describe this problem and the relevance for it from our results here. For the purposes of [1] it is useful to consider % -subspaces < of K, where O % is a "nite sub"eld of K. If q"p, this reduces to the situation discussed O above. In Theorem 3 below, we gather various results on the vanishing or otherwise of the sums S (<; a). Theorem 2 will follow from Theorem 3, F although in the general case our results are less compete than Theorem 3. It is Theorem 3(iv) that is needed in [1].
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Before stating Theorem 3, we introduce the key ingredient in its proof. To the % -subspace < of K we attach the polynomial O f (x)" (x!v)3K [x]. 4 TZ4 By [4, Theorem 3.52], f (x) is a q-polynomial; i.e., it has the form 4 K f (x)" a xqG, a 3K. G 4 G G For the proof of Theorem 2 we will need to know that none of the coe$cients a vanish in the case that < has codimension 1. This is proved in G Lemma 2.10 below. Some further comments on the vanishing or otherwise of the a are given in the remark after the proof of Theorem 2. G THEOREM 3. ¸et K be a ,nite ,eld containing % , and let < be an % O O subspace of K of dimension m. (i) ¸et a 3 <. If hI0 (mod q!1) then S (<; a)"0. F (ii) ¸et a3 K!< and let 04h4qK>!2. ¹hen (a) S (<; a)"0 if ds (h)((q!1) m. F O (b) Suppose that h(pqK and that a O0 for 04i4m. ¹hen S (<; a)O0 G F if ds (h)5(q!1)m. O (iii) ¸et a3 < and let 04h4qK>!2. ¹hen (a) S (V; a)"0 if ds (h)O(q!1) m. F O (b) Suppose that h(pqK and that a O0 for 04i4m. ¹hen S (<; a)O0 G F if ds (h)"(q!1)m. O (iv) ¸et 04k4q!2. (a) If 04h4(k#1)qK!2 and h,k (mod q!1) then S (<; a)"0 for F any a 3K. (b) S (<; a)"0 if and only if kO0 and a3<. I>qK\ Now let K be "nite and consider the function f : KPK induced by 4 evaluation of the polynomial f (x). As f (x) is a q-polynomial, this function is 4 4 an % -linear endomorphism of K, whose kernel is obviously <. It seemed of O interest to investigate the properties of f and, particuarly, of its image. It is 4 this which gives our generalised trace and cotrace maps. We would like to thank the referee for suggesting this formulation of the ideas which were implicit in the original version of our paper. Let
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We now de"ne our trace and cotrace functions: Tr "f
(2.1)
Here a O0, since f (x) has only simple zeros. 4 LEMMA 2.2. As a formal power series in the indeterminate y, we have
K\ H S (<; a) yF>"a yqK f (a) yqK! a yqK!qG . F 4 G h50 j50 G
(2.3)
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Proof. We evaluate the logarithmic derivative of f (x!a) in two ways. 4 First, d log f (x!a)" (x!a!v)\ 4 dx TZ4 " x\ (1!(a#v)x\)\ TZ4 " x\ ((a#v)Fx\F) h50 TZ4 " S (<; a)x\F\. F F5
(2.4)
On the other hand, as f (x)"a and f (x!a)"f (x)!f (a), we have 4 4 4 4 d f (x!a) log f (x!a)" 4 4 dx f (x!a) 4 "a ( f (x)!f (a))\ 4 4 \ K\ "a xqK# a xqG!f (a)) G 4 G K\ \ "a x\qK 1! f (a) x\qK! a xqG!qK 4 G G K\ H "a x\qK f (a) x\qK! a xqG!qK . 4 G 5 H G
Equating (2.4) and (2.5) and setting x\"y, we obtain (2.3).
(2.5)
䊏
Proof of ¹heorem 1. Put q"p in (2.3). Equating coe$cients of yF>, we "nd that S (<; a)"0 for 04h4pK!2 and S (<; a)"a O0 for F F h"pK!1. 䊏 To prove Theorems 2 and 3, we must examine (2.3) more carefully. Expanding the right-hand side and equating coe$cients of yF>, we obtain the following result. COROLLARY 2.6. S (<; a) is a sum of terms of the form F r#r #2#r K\ K\ f (a)P arG , $a 4 G r, r , 2 , r K\ G
(2.7)
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where r, r , 2 , r 50 and K\ K\ qK (1#r)# (qK!qG) r "h#1. G G
(2.8)
Here
r#r #2#r K\ r, r , 2 , r K\
denotes (r#r #2#r )! K\ , r! r !2r ! K\ the multinomial coe$cient associated with the sum r#r #2#r . K\ PROPOSITION 2.9. ¸et 04h4qK>!2 and write h"h #qh #2# qK h as in (1.3). ¹hen S (<; a) contains at most one term (2.7), namely that with K F (i) r "q!1!h for 04i4m!1; G G (ii) r"h !r !r !2!r . K K\ ¹his term occurs precisely when the values of r, r , 2 , r given by (i) and (ii) K\ satisfy r50. Moreover, we then have (iii) r,h (mod q!1). Proof. As h#1(qK>, it follows from (2.8) that r, r , 2 , r (q for K\ any term (2.7) occurring in S (<; a). Reading (2.8) mod qK, we have F K K\ K\ h qG"h,!1! qG r , qG (q!1!r ) (mod qK). G G G G G G Hence h "q!1!r for 04i4m!1. This proves (i), and (ii) then follows G G from (2.8) by an easy calculation. Thus, for the given value of h, the only possible term (2.7) occurring in S (<; a) is that with r, r , 2 , r as F K\ determined by (i) and (ii). These values do give a term precisely when r50. Finally, (iii) is immediate from (2.8). 䊏 Proof of ¹heorem 3. (i) As a3 <, we have S (<; a)"S (<; 0). Let m be F F a primitive element of % . Then O S (<; 0)"S (m<; 0)"mF S (<; 0). F F F If hI0 (mod q!1) then mFO1; so S (<; 0)"0. F
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(ii) Let 04h4qK>!2 and consider the values of r, r , 2 , r deter K\ mined by h in Proposition 2.9. We calculate ds (h)"h #2#h O K K\ " (q!1!r )#(r #2#r #r) G K\ G "(q!1) m#r. For any term (2.7) occurring in S (<; a) we have r50. Thus, if S (<; a)O0, F F then ds (h)5(q!1) m. This proves (a). Conversely, if ds (h)5(q!1) m then O O a unique term (2.7) occurs. We have to show, under the hypotheses of (b), that this term does not vanish. Now f (a)O0, since a,<, and each a O0 by 4 G hypothesis, so the term in question vanishes only if the multinomial coe$cient is divisible by p. This happens precisely when there are carries in the base-p addition of r , 2 , r and r (see, for instance, [5, p. 24] for the case of K\ binomial coe$cients). But r#r #2#r "h by Proposition 2.9(ii), K\ K and h (p, since h(pqK, so no carry is possible. Hence, S (<; a)O0, K F proving (b). (iii) As above, there is only one term (2.7) to consider and for this we have ds (h)"(q!1) m#r. As a3 <, however, we now have f (a)"0, O 4 so this term vanishes unless r"0. This proves (a), and (b) follows as in (ii). (iv) Suppose that 04h4(k#1)qK!1 and h,k (mod q!1). Then h 4k(q!1. If S (<; a)O0, then S (<; a) again consists of a single term K F F (2.7). For this term we have 04r4h and r,k (mod q!1) by Proposition K 2.9(ii), (iii). Hence, r"k, and from (2.8) we have h5(k#1)qK!1, which shows (a). Moreover, if h"(k#1)qK!1, then r"k and r "2"r K\ "0. Thus, (2.7) reduces to $a f (a)I. This vanishes if and only if k'0 and 4 a3<, as asserted in (b). 䊏 Before proving Theorem 2, we verify that none of the coe$cients a in f (x) G 4 vanish in the case that < has codimension 1. LEMMA 2.10. ¸et < be an % -subspace of dimension m in the ,nite ,eld O K"%qK>. ¹hen, K f (x)" a xqG with a O0 for 04i4m. 4 G G G
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Proof. Write K"< % b as % -vector spaces, where b3K!<. Then O O f (b)O0. Let j"f (b)\. We calculate 4 4 f (x)" (x!v!gb) ) g3% O TZ4 " f (x!gb) 4 g3% O
" ( f (x)!g f (b)) 4 4 g3% O
"f (x)O! f (b)O\ f (x) 4 4 4 K K " aO xqG>! j\Oa xqG . G G G G But f (x)"xqK>!x, since K"%qK>. Equating coe$cients, we obtain ) j\O a "1 and aO "j\Oa for 14i4m. Hence, a "jqG>!1O0 for G\ G G 04i4m. 䊏 Proof of ¹heorem 2. We have ds (h)"(p!1) (m#1) if and only if N h"pK>!1, and in this case (v#a)F"1 for all v3<, with the exception of v"!a in case (ii). Thus, SpK>!1 (<; a)"0 in case (i) but not in case (ii). We may therefore suppose that 04h4pK>!2. Then, in view of Lemma 2.10, parts (i) and (ii) of Theorem 2 follow immediately from Theorem 3(ii), (iii), respectively. 䊏 Remark. There are two other cases where we can guarantee the nonvanishing of the coe$cients a : G (i) if < has dimension 1 then f (x)"xO#a x with a O0; 4 (ii) If < has dimension 2 then f (x)"xq#a xO#a x with a O0. Here 4 a "0 if and only if < is a vector space over % (since this occurs precisely O when f (x) is a q-polynomial). In particular, Theorem 3(ii) (b) and (iii) (b) 4 apply when < has dimension 2 and K has "nite odd dimension over % . N More generally, if d divides the dimension of < over % , then < is a vector O space over %qB if and only if f (x) is a qB-polynomial, i.e. if and only if a "0 for 4 G all i not divisible by d. One can easily "nd other cases in which some of the a G vanish; for example, the polynomial x#x#x splits in %3, and its zeros form a subspace < of dimension 3 over % for which a "0. 3. THE FUNCTIONS <>
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We "rst record some easy facts. Let N denote the kernel of the usual trace Tr"Tr % , let
(a<)R" (g!av) " g3K TZ4
" (ag!av) " g3K TZ4
" aqK (g!v) " g3K TZ4 "aqK
Remark. Proposition 3.1 (i) shows that we do not always have <5L f 3 g(x)" a bqG xqG>H" c xqI , say. G H I GH I Moreover, if a O0Ob then c "a bqKO0, so R contains no zeroK L K>L K L divisors, except 0 itself. Also, f (x)"xqB!x is central in R, since aqB"a for ) all a3K. We will use these facts about R to prove the "rst part of Theorem 4. Proof of ¹heorem 4. (i) For any subspace <, the polynomial f
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Since f (x)O0 is central in R, we may cancel to obtain f
B\ Tr(x)" xqG . G
Hence, Tr "Tr. 4 (iii) Let d42. Let <> denote the orthogonal complement of < with respect to the trace pairing. By dimension considerations, it is enough to show that
(3.2)
for all <. This is clear if <"+0, or <"K, so we need only consider the case d"2, dim<"1. Let <"% a. Then f (x)"xO!aO\x, and it will su$ce to O 4 verify that Tr(af (b))"0 for all b3K. We calculate 4 Tr(af (b))"Tr (a (bO!aO\ b)) 4 "Tr (abO!aOb) "Tr (abO!(abO)O) "0. (iv) If the map < C
(3.3)
It therefore su$ces to show that (3.3) implies d42. By dimension considerations, it follows from (3.3) that %R 5(% a)R" O O (% #% a)R for any a3K!% . Since % #% a"% #% (1#a), we then have O O O O O O O %R 5 (% a)R"%R 5(% (1#a))R. O O O O Using Proposition 3.1, we may rewrite this as N5aON"N5(1#a)ON.
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Setting b"aO and taking orthogonal complements with respect to the trace pairing, we obtain % #% b\"% #% (1#b)\. O O O O
(3.4)
Thus, b\"j#k(1#b)\ for some j, k3% , so b has degree 2 over % . O O Since a, and, hence b, may be chosen arbitrarily in K!% , this shows that O d42 as required.
APPENDIX We outline here how the results of this paper are applied in [1] to study the integral Galois module structure of p-adic "elds. Let K be a "nite extension of 0 . Its valuation ring O has a unique N ) maximal ideal, nO say, and the residue "eld O /nO is isomorphic to % for ) ) ) O some power q of p. For a "nite normal extension ¸ of K with Galois group Gal (¸/K)"G, we consider O as a module over its associated order * A
"+a3K[G] " a ) O -O , *) * *
in the group algebra K[G]. If ¸ is abelian over 0 then O is free over N * A 0 [2] and, in fact, free over A whenever 0 -K-¸ [3]. It is, * . *) N therefore, of interest to consider extensions ¸/K, where ¸ is abelian over K, but not necessarily over 0 , and to seek criteria which will guarantee that O N * is, or is not, free over A . In [1] we give some partial results in this direction *) for certain "elds obtained using Lubin}Tate formal groups. (For background on these, see, e.g., [6].) Let f (X) be a Lubin}Tate series for K, corresponding to the uniformising parameter n. There is a formal group over O for which f (X) is an endomor) phism. We can regard f (X) as multiplication by n. For n51, let KL be the "eld obtained by adjoining to K the nL-torsion points of the formal group. Then KL is a totally rami"ed, abelian extension of K whose Galois group is naturally isomorphic to (O /nLO );. Every "nite abelian extension of K is ) ) contained in the compositum of some KL and some unrami"ed extension of K. For the rest of this appendix, we assume that K is rami"ed over 0 and N that pO2. In [1] we show that OK is not free over AK and then consider ) some of the intermediate "elds ¸. Writing C to denote a cyclic group of K order m, we have G"Gal (K/K)"C ;! with ! isomorphic to % . O\ O Thus ¸ is the "xed "eld of Gal(K/¸)"C ;&, for some d dividing q!1 B and some subgroup & of !. We then have the following result, which is a slight specialisation of Theorem 6 of [1].
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THEOREM A.1. =ith the above hypotheses and notation, suppose that dOq!1 and that & corresponds to a subspace < of % of codimension 1 over O some sub,eld %q . ¹hen O is not free over A . In particular (taking q "p), * *) if K-¸-K and [¸ : K]"pr, where p / r and r'1, then O is not free * over A . *) To prove Theorem A.1, we start with an explicit description of O as an ) O [G]-module, and examine its "xed points under & to determine the ) module structure of O . When < has %q -codimension 1, we are able to * describe O explicitly. In doing so, we encounter certain sums in O which * ) reduce modulo n to the sums (1.2). The required description of O is obtained * by applying Theorem 3(iv): Theorem 2 will su$ce in the case q "p. We can then deduce Theorem A.1. Note that if we considered, say, only those subspaces stable under the Frobenius automorphism, rather than allowing arbitrary subspaces < in (1.2), then Theorem A.1 would be considerably weakened. When < is not of codimension 1 over some sub"eld, we are not able to analyse the module of "xed points under &. Thus the results of [1] are far from complete. For an arbitrary intermediate "eld ¸ of K/K, we suspect, however, that O is free over A if and only if [¸ : K] is either prime to p or * *) a power of p.
REFERENCES 1. N. P. Byott, Integral Galois module structure of some Lubin}Tate extensions, to appear. 2. H.-W. Leopoldt, Uber die Hauptordnung der ganzen Elemente eines abelschen ZahlkoK rpers, J. Reine Angew. Math. 201 (1959), 119}149. 3. G. Lettl, Relative Galois module structure of integers of local abelian "elds, Acta Arith. 85 (1998), 235}248. 4. R. Lidl and H. Niederreiter, &&Introduction to Finite Fields and their Applications,'' rev. ed., Cambridge Univ. Press, Cambridge, 1994. 5. P. Ribenboim, &&The Book of Prime Number Records,'' 2nd ed., Springer-Verlag, New York, 1989. 6. J.-P. Serre, Local class "eld theory, in &&Algebraic Number Theory'' (J. W. S. Cassels and A. FroK hlich, Eds.), Academic Press, London, 1967.