Power sums of polynomials in a Galois field

POWER SUMS OF POLYNOMIALS IN A GALOIS FIELD

BY HERBERT LEONARD LEE

1. Introduction. Let GF(pn) denote the Galois field [4; 1-70] of order pn. Let

M CoX clx"-1

c,_ix C,n

denote a polynomial in an indeterminate x, with coefficients in GF(pn). If Co 0,we write deg M m, and if Co 1, we call M primary. In this paper we evaluatecertain power sums, viz.

deg ,l=m deg M<mM primary all MO

In 3 we discuss some of the conditions under which the power sums vanish.4-6 are devoted to the evaluation of certain power sums in which the exponent kis of a special form. In 7 we evaluate R*m and S*m where k a.p am p.(m-i)- ap ao, for in this case R and S* are given in one term only. 8gives the development of recursion formulas involving power sums. Finally, in9, we show the connection between Rkm and pm and the complete symmetricpolynomial.

2. Definitions and notation. Let [1; 141-143]

(2.1) /m(t) 1--[ (t- M),deg M<

b0(t)- t,

where is another indeterminate, and the product extends over all polynomials(including 0) of degree less than m; we then have

(2.2) bm(t) (-- 1)m-; "n’,iffi0

where

(2.3) j FiL’_ 0

Received November 15, 1942. The writer is indebted to Professor Leonard Carlitz forencouragement and assistance in writing this paper, which was offered as a dissertation forthe Ph.D. degree at Duke University.

277

278 HERBERT LEONARD LEE

and

Fm(2.4)

Lm[m] [m- 1]".-- [11’’m-’’, Fo 1,

[m][m- 1] [11, L0 1, [m] x

We also write for brevity,

F,(2.5)

FF,_

and

(2.6)

X.

We put

(2.7) k a,p + + alp - ao

and denote by G,(t) the product

(0_ a, _,pn_ 1),

and

)(t)’(t).-.

tt(k) as -]- a._l -- -[- ao.

3. Conditions for vanishing sums. In this section we state as lemmas certainconditions for vanishing and non-vanishing sums. We do not prove them sincethe proof is very simple, but we do prove theorems related to such sums.

LEMMA 3.1. For k pm 1,

L,’ R ).-1.L,,.S OandR O.For O k pn,, 1,

LEMMA 3.2. Unless (pn 1) k, R 0 and p O.

LEMMA3.3. $2 0if(pn-- 1) Xk, -S,if(p" 1) tk, where

E ideg M-

all M

THEOREM 3.1. For k O, am O.

Note that am-L is integral, since M]Lm for all M. Take any irreduciblepolynomial P of degree m. Then am" L,, =-- L/P 0 (mod P), since

k kandP’Lm. Thus am’Lm 0, and thereforea 0for] 0. We mayM/L where M is integral and different from 0.write am

In precisely the same way, we may prove

SUMS OF POLYNOMIALS 279

COROLLARY 3.1. a, O, where

deg M <_

In the remainder of this section we take n 1.

THEOREM 3.2. If k ap + -t- ao where 0

_a

_p 1, then S

0 if and only if t(k) < m(p 1).

The proof is by induction. We write

(3.1)

s} E (x + c)

since ai

_p- 1,

i;

for every j. Assume now (k) < 1 (p 1).their sum is(k). Butc- 0, ifg < p- lorifg r(p- 1).,(k) < p- 1,c O. Thus if,(/) < l(p- 1),S O.Assume now S O. The sum cannot vanish because of any

for

Then i < p 1 since at mostSince g

_

E

(3.2)

S E E’ (xM -- c)deg Mfm-1

for every j; nor can the sum vanish due to addition of different terms in x,since for each different i, x is different. S can vanish only if c O, i.e.,if g r(p 1), where g i. Since i takes on all values from 0 toit is necessary that (]c) be such that i r(p 1) for all r > 0 includingr 1. Therefore (k) < l(p 1), and the theorem is proved for m 1.We now suppose the theorem is true for deg M m 1. Then we write


Suppose t(k) < m(p 1). Then we have

(a,- i,) + + (ao- io) <_ (m- e)(p- 1),(3.3)

i + + io

_e(p- 1) (e 1,2,...,m- 1),

where the inequality holds for one of the expressions (3.3) in view of our assump-tion that0; hence

Suppose nowS 0. Then(k) < re(p- 1). For ift(k) _> re(p- 1)itwould be possible to choose(m- 1)(p- 1). Thenc 0and

S(.-.)’+...+(o-io) O.

S could not then vanish. Our theorem is therefore proved.By a similar argument we may prove

THEOREM 3.3. Ifthen R 0 if and only if (k) < m(p 1).

4. Evaluation of S-1 for q[2; 497]

LEMMA 4.1.

_,, G(M)_deg Mffim M

We shall now prove

THEOREM 4.1.

We put [3; 943]

We state without proof

--1) F,/L,,forlc ip and O <_ i < p

lO otherwise.

For q pl p,, and s < pn, we have

Sq_l (- 1)m[/,]m [/C.]mL

for s values of k and form their product, obtaining

(4.1) t= -’o.= .,(t).,-o

We now put M, a primary polynomial of degree m, in both sides of (4.1),divide by M, and sum. Then

(4.2) Sq-- o k " E’ Gh(M)


where h pn-i + pn;.. By Lemma 4.1,

_,, Gh(M)deg M--m M

0 if h ip’m; i.e., j m,

(-1)"F:/Lm if h sp=; i.e., all j m.

This yields, if we substitute in (4.2) the values of_,, Gh(M)deg Mm M

the term

(4.3) (-- 1)m[ll]m [ks]mL

and proves the theorem.In Lemma 4.1, if i p, we then may write

"(t) .+(t) -t-- F’2-a.(t).With this revision we modify our lemma to read

’0if j> mor < m- 1,

1) Fm/Lm if j m,.,M- M degM-m M (- "

F ’ 1 (-)F/L i j - 1.

The last remark enables us to prove

THEOREM 4.2. For s p,

(--])m[]m V]cs]m..._ (--1)mill]m--1S.- L. L._

Also, by a slight elaboration of the proof, we get

THEOREM 4.3. For s p + 1,

(--1)m[kl]m [ks] + E (--1)m[kl]m--1 [kPn]m--l[kpn+l]mL.. L._

We also state the following theorems"

THEOREM 4.4. For r < pn, s p + r,

Lm Lm-


THEOREM 4.5. For s 2p",

--1)"[kl] ..-[k.].,Lm

Lm-1THEOREM 4.6. For s- 2p + 1,

(--1)[kl] [/c,],,Lm

_{_(--1)’[k],,_-.. [k.],,_l[k,.+].,

Lm--1

THEOREM 4.7. For s ap, a < p",

Lm

(-- 1)’[,]._1 [].],_,

THEOREM 4.8. For s aop + al where a, <_ p" 1,

E E (--l)’[m]"’[]l]’ [k"’n+a’]m[l"’1"+a’+1]m-I [k’]m-IL,n

THEOREM 4.9. For s p2,,

-, (- )-[,].... [.]. + (- )-{ [m]"-’ + }[,].-,L,_ L,_I

(-- 1)’n[m 1]"-l[]Q],n-Lm--2

pn_ 1+,=i

{(--1)"(m)"-1(/- i)P-i[kl]m-1


5. Evaluation of R-’, where q p"’ W p"’. We state without proofthe following lemma, which may be proved by induction"

LEMMA 5.1. For all s >_ O,+ 1 pn’ (-- 1)i+l[k]i (-- 1)’ ’[/c ].(5.1)

i=o L; L,

Note that for k <_ s, the sum is 0.

LEMMA 5.2.

Edeg M<all M#0

For k p,i, + + p,O,

O for lc >_ p’,

otherwise.

We shall now prove the following"

THEOREM 5.1. For s < p,

R-" (- 1)s--1 * sm[k lipton--1 [kL_,

As in Theorem 4.1,

to=(5.3) jl ,(t)"’",o

.(t).

We now put M, a polynomial of degree less than m, divide by Ms, and sum.This yields

(5.4) Re-S= o k, k,

_Gh(M)

=0 deg M<m M"all M#0

where h p"’ + + p’. We now apply Lemma 5.2 to obtain the lastsummation above and substitute these values in (5.4). This yields on simplifica-tion

which by Lemma 5.1 gives the theorem.We could extend the results to s >_ pn, but, since the formulas so derived are

rather complicated, we omit them.

nka6. Evaluation of S-; for q pk, + + P wherej _< s.require some preliminary theorems which we state without proof.

We shall

LEMMA 6.1.


LEMMA 6.2. If Gk(t) is written in the form Aot() + Alt() /.-1 + wherek bo blp T T bpn’, then

LEMMA 6.3.

(6.2) T ’ .G(M)deg M--m M

"(a) 0 for k < p" and k

_p.(.,+ j

_(k)

but arbitrary,

(b) (- mi1) F,/L,, if k ip’",

(c) 0 if j <_ (ko) for k ip" T ko,

(d) (-1)’i-"’’)’FAfor k ip" W ko,

j > (ko).

Proof. (a) Suppose k < pn. Then

--,, G(M)_ .,, /’=0deg M=m M deg M=m

since r < pnm__ 1, j __< (k). Suppose k >_ pn(m/). Then

-,, Gk(M). , b,/,(M)Gh(M) 0deg M=m M M=m M

since /,(M) 0 for s > 0, a condition which is satisfied if k >_ p*l).(b) Suppose k ip". Then, G(M)_ , (M)_ F: ’ 1 (_l)f__(c) Suppose k ip + ko, where 0 _< i _< p 1 and 0 < ko < p*’. Then

b,,(M)G,o(M) , G,o(M)Z’ G(M)_ Z’dou M d. U- M; Fm

deg M-,,, Mi0

for j _< (ko) by (a).(d) Suppose jl > (ko). Thenj

_(k) i + (ko) _< p 1 + (ko);

0 < j- (ko) <_ p’- 1., Gk(M)= Fdeg M=m M

, G,o(M)deg Mffim M

Aodeg M-- M

(-- 1)[-"(*)FAo


by (a) since j > (ko) or j (ko) > 0. This completes the proof of the lemma.We prove

THEOREM 6.1. For j

_s

Sqji (-])mi[ki]mL

1]._1+ (-1F’[] [._][_+ " ]_k LL:_

We write for s values of k and form the product. Then we put M, apmary polynomial of degree m, divide by M, and sum. This gives

(6.3) ’J= o k, k. E’ G(M)".= deg Mfm M

where h p + + p’. From Lemm 6.2,

0 for j <

F ’ (--1)Fif ll j m,

do= M LLG(M)E" F’-r E’ Gh,(M)

deg M,=m M deg Mfm M

(- 1)_I-__)F (-1)’, (-,)’,

where h (s r)p + h’.

If we substitute these values in (6.3), we get the result as stated in the theorem.In precisely the same way, we obtain for s p, if we remember we get

additional terms, when j m 1,

EOnEM6.2. Forj g s p’,q p’" + + p’,

L2

t-1 L- Lm-1

+ (- )"[]._i--1L L._

Extension of this type to s > p" can be made with more complicated results.

7. Evaluation of R and S for k < p.(m/l). The value of k in this sectionis chosen because for every such k, R and S are expressible in one term only.


We omit k p.(m/l) 1, which is a special case of a more general theoremalready known. We state

LEMMA 7.1. For tt(k) < m(p 1), R 0 and S O.

The proof of this lemma is similar to the proof of Theorem 3.2. We shalltherefore omit the proof.

THEOREM 7.1. If k amp + am_lP’(’-) + "3t- ao where 0 <_ ai <__

p"-- 1, am # 0 and tt(k) m(p 1),

R (_l)m+ (ao + a + + am-,)!o!! am-! (o + 1, C a.,_)

in which ao

(ao + 1, a

Write

then

am,a 1-- aifor i O,l, ,m-- land

(7.1) M’""’- ([ m ]M’"-" m-- ]M,,.,r.-,) +’’’)amin whichdegM < m. We putk amp"m+ ko,andobtain

M m M,.(m-,) m M,,-(m Mom-1 m-2 +""

(7.2) am’ lm,...E (__I)ao!al!-.. L]m1X M(a+a)+(at+a)pn+ +(tm--t+am-t)pn(m--i)I

ThereSa a. Now(aoand ko (ao + ao) + (a + a)p + + (am- + am-i)pn(m-i) p..m 1.Ifko < pm 1, then g(ko) < m(p" 1) and we would not get terms. Ifko > p=m, then for at least one j, a. + a. >_ p" and again g(ko) < m(p 1).Thus ko p=m 1, and a,. + ai 1,p for i O, 1 m- 1 Thisgivesa p"- 1 a.fori O, 1, ,m- 1. We then have

[Rkm (-- 1) .... ....

(_l)m+ (O0 + O/1 + + m--1) (ao + 1, a,


since (-- 1) "’ (-- 1) a’ and a am. We should note that we no longer needa in (7.2) since the a are now fixed quantities. This proves the theorem.Example 1. ]c pnm_ (p,__ 2)p,(m-a)

__(p_ 1)p(m_2) .... -F (p-- 1)p

+ (pn-- 1). Hereao 0, am- 1, am-- 1.

(-1) 0 m-- 1

Example 2. ]c- 2pnm- (p’-- 2)p(m-1) - (p-- 2)pn(m-2) -!- (p-- 1)p(-a

-I-’’"-t-(pn__ 1)p_t_ (p__ 1). We have ao--0, czm-1-- 1, am-2 1, am 2

0!1!1! m-- 1 m-- 2

THEOREM 7.2. If k ampnm+ am_lp(m-) + -t- alP -t ao where 0 <_ at <_

p’ 1, am O, and m(p 1) _< u(k) < (m - 1)(p- 1), then

..(am-- (ao 4-"’+ am-)S (-1)m+ + /

\ ]/"m "(ao + 1, a,,’’"

in which a p 1- a, for i O, 1, m 1, and p(k) m(p" 1).

Write

We put M a primary polynomial of degree m and obtain

If we substitute (7.3) in M and expand, we have

(7.4)

whereIn precisely the same way as in Theorem 7.1 we show that we obtain terms


only whena p"- 1 afori 0, 1, ,m- 1. This fixes alla, andtherefore f. Thus we have

S (l).... + +...(am-.-. (ao’+al-- --am-l)!Fml’k / LJF"y--.ao. a_ 0

Xm S_

m- 1

in which a, are constant. On simplifying we obtain the theorem.Example. 2p+ (p- 2)p(-) + (p- 1)p(-) + + (p- 1).

Here ao 0, a_ 1, a 2, and 1,

S= (-1)(21)F[][m :8. Recursion formulas for R and S. We develop here recursion formulas

that enable us to evaluate new power sums from sums already known. Supposewe write

o(t) + ,(t)" + + ,_""-"(t) +(S.1)

_‘+’," t"".

Since (M) 0 for deg M < m, the left side of (8.1) vanishes when we putM, a polynomial of degree less than m. Then we obtain the equations

m--1

(8.2) M""‘+’’ M""’ 0.i=0

If we replace M by x where i 0, 1, m 1, we write

X?

Xxi’ni[i xx" x(i_l)pn(i-,) (m+,)

(s.3)x(m-1)(-) [.

By Cramer’s rule we determine fl

(8.4) fl (_1)_,_ F+, 1

We now multiply both sides of (8.2) by M and sum. This gives

(s.5) R+,+.>.= Fro+ (_l)m-l-.[m- lJ Rk+"iF,_--i =o J [m + s -j]";’

which is the recursion formula desired.We develop a similar relation for S. Write

(8.6)


Since hm/(M) 0 for deg M <_ m, the left side of (8.6) vanishes when we replaceby such a value of M. We then get the system of equations

(8.7) _,BiMi=0

Then, by puUng M z for i O, 1, m,

The solution by Cramer’s rule gives

(8.9) B; (- 1)m-’F/,

If we substitute this value of B; in (8.7), multiply both sides of the equation byMk, and sum over primary polynomials of degree m, we have

(8 .10) p .,,,, (-- 1),,. ,_ [m + s

In precisely the same way similar recursion formulas may be obtained forp and a

Example. lc 3pn’- pn(,,-l) 1, s 2.

Sk/,,( F. SFF,(+.) (- 1)- [m + 2

We obtain terms when j mandj m- 1. Forj m, 3p- p(-)1 + p 4p"’- p("-) 1 2p + (p 1) + (p p(-)). Forj m 1, 3p pn(,n-) 1 p 3p 1 2p (p 1). Apply-ing Theorem 7.2, we have

,q+.,+) F+ 1)+ m-- 1 .+ 0 m-- 1

9. The complete symmetric polynomial. We now develop R and S in adifferent form. To do so we require the following.DEFINITION 9.1. If f(xx X,) is a homogeneous polynomial of degree k

in the indeterminates x, and in which all terms of degree k appear, f(xx, x,,)is called the complete symmetric polynomial of total degree k. We denote it byW(x Xn).We need the preliminary

LEMM& 9.1.

(9.1)m--1 k+raxx x,_x,, W(Xl, ..., =),

X XXl 2" m--lXm


where xlxl x:_-llx/m is a determinant of m rows and m columns, the termsof the main diagonal appearing.

We may reduce (9.1) to

1 m--2

(9.2)1 X,-X.,-1

This becomes [5; 329] (x x)(+-)-(-) W(Xl x).We prove the following

THEOREM 9.1. For any k such that p 1 k,

FR=’’+-’ (- 1) W(M, deg M < m).

By proper choices of the a, we may write

(9.3) G+(t) + aG+_(t) + + a_,G+l(t) + Bt.If we put s p’ 1, (t) G.++(t) for all j 0. Then on replacing by

M, a polynomial of degree less than m, the left side of the equation (9.3) vanishesfor all such M. There are s such polynomials. We have

(9.4)i=1

where deg M < m.If we solve by Cramer’s rule,

s--1 k+s

(9.5) B.= s--1

where x is used in place of M. In particular

(9.6) B, W(M, deg M < m).

If we sum over M in (9.4) we obtain

(0ifj < p -1,

" BR- (--1)(F/L)B, if j p 1.

If we substitute in (9.7) the value of B obtained in (9.6) and write s p 1,we have

F(9.8) R2"+- (- 1) W(M, deg i < m),

result which is the theorem.We state without proof the following


LEMMA 9.2.

then

T"(xl x.)U:_ (x,

(x, x)

where U-I (xl x,n) is a symmetric polynomial in xi every term of which is

of degree (m 1)r and no x is of degree greater than r.

THEOREM 9.2. If is any number for which p 1 k,

(9.9) L U (M, degM < m)-- m--1

PAs in Theorem 9.1, we may write

+i(9.10) aoG,/ (t) + aiGs+2(t) -t- + a_Gs/(t) B

where s pn 1. G(t) vanishes for M, a polynomial of degree less thanm, provided i > s. On substituting M in (9.10), we get the equations

(9.11) B+M+ M,

where

(9.12) Bk+l k+i-1 k+i+l k+8

Xl Xi-1 XiXi+I X8k+l k+s

We evaluate B. by substituting xi 1/y. Then (9.12) reduces to

(9.13)

Replacing the y by 1/x we have

(9.14) B. (x x)

since s is even. Then from Lemma 9.2 we have

(9.15) B8k-1U,-I(M, deg M < m) 1)./1 L

_rI M)

(- FW[ U,_,(M, deg M < m),deg M<m


since

I-I M=(-1)"F--vdeg M<M0

If we divide (9.11) by M and sum, we have

(9.16) p-’ BiM (- 1)roB. -.We substitute in (9.16) the value of B, in (9.15) obtaining

(9.17) p-’ (-1)’+(k/’)m L-I U-1 (M, deg M < m)Y--7 -1

FWe know p" 1 ] 1 since ]c 1 is the exponent of p.

even andThen /c - 1 is

(9.18) F_. U.,_(M,’- deg M < m).

Since k is any number, we may replace/ by/c - 1, obtaining the result statedin the theorem.

BIBLIOGRAPHY

1. L. CARLITZ, On Certain functions connected with polynomials in a Galois field, this Journal,vol. 1(1935), pp. 137-168.

2. L. CARLTZ, A set of polynomials, this Journal, vol. 6(1940), pp. 486-504.3. L. CARLTZ, Some sums involving polynomials in a Galois field, this Journal, vol. 5(1939),

pp. 941-947.4. L. E. DICKSON, Linear Groups, Leipzig, 1901.5. THOMAS MUIR AND WILLIAM H. METZLER, Theory of Determinants, Albany, New York, 1930.

UNIVERSITY OF TENNESSEE.

Power sums of polynomials in a Galois field

Documents

Transcript of Power sums of polynomials in a Galois field