Đa Thức Hoán Vị Được

I HC THI NGUYN

TRNG I HC KHOA HC

VNG TH YN

A THC HON V C

LUN VN THC S TON HC

Chuyn ngnh : PHNG PHP TON S CP

M s : 60 46 40

Gio vin hng dn:

PGS.TS. L TH THANH NHN

THI NGUYN, 2012

S ha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn

2Mc lc

Mc lc . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

Li cm n 3

Li ni u 4

1 Kin thc chun b 6

1.1 Kin thc chun b v nhm . . . . . . . . . . . . . . . 6

1.2 Kin thc chun b v vnh . . . . . . . . . . . . . . . . 10

1.3 Kin thc chun b v trng . . . . . . . . . . . . . . . 14

1.4 Kin thc chun b v a thc . . . . . . . . . . . . . . 17

2 a thc hon v c 20

2.1 Khi nim a thc hon v c . . . . . . . . . . . . . 20

2.2 Mt s lp a thc hon v c trn mt trng . . . . 26

2.3 a thc hon v c modulo 2k . . . . . . . . . . . . 30

Kt lun . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

Ti liu tham kho . . . . . . . . . . . . . . . . . . . . . 40


3Li cm n

ti c thc hin ti trng i hc Khoa hc - i hc Thi

Nguyn di s hng dn ca PGS.TS L Th Thanh Nhn. Ti xin

by t lng bit n su sc, chn thnh nht i vi C. Bi s gip ,

ch bo, khuyn khch n cn ca C gp phn rt ln cho s thnh

cng ca lun vn ny.

Ti cng xin c by t lng cm n chn thnh nht ti Ban lnh

o, Phng o to - Khoa hc v Quan h quc t, Khoa Ton - Tin

Trng i hc khoa hc - i hc Thi Nguyn to iu kin thun

li ti v cc bn hc vin cao hc Kha 4 (2010 - 2012) c hc

tp, nghin cu.

Ti cng xin cm n cc Thy, C l GS.TSKH H Huy Khoi,

GS.TSKH Nguyn Vn Mu,... l nhng nh ton hc hng u Vit

Nam ging dy cc chuyn cho lp chng ti.

Cui cng, ti xin c gi li cm n ti gia nh, bn b, nhng

ngi thn lun bn, ng vin, gip ti c th hon thnh

lun vn.


4Li ni u

Ta bit rng mt a thc fpxq trn mt vnh hu hn R c

gi l hon v c nu a thc hon v c cc phn t ca vnh

R, tc l nh x : R R cho bi paq fpaq phi l mt song nh.

Trong cun "Finite fields" xut bn ln u tin nm 1983, Lidl v

Niedereiter [LN] nghin cu cc tiu chun ca a thc hon v c,

cc dng c bit ca a thc hon v c, nhm cc a thc hon v

c, trng hp ngoi l ca a thc hon v c v a thc hon v

c mt s dng bt nh. Lidl v Mullen [LM1,2] cng nghin

a thc hon v c trn trng hu hn. Nm 1986, R. A. Mollin v

C. Small [MS] a ra tiu chun a thc hon v c dng xn. Nm

1999, R. Rivest [Riv] a ra tiu chun a thc hon v c modulo

2k.

Trong ti ny chng ti trnh by li cc kt qu trong hai bi

bo ca R.A.Mollin v C.Small [MS] v ca R.Rivest [Riv] v c trng

tnh hon v c ca a thc dng xn v a thc dng xk bxj c vi

pk j 1q trn mt trng hu hn, ng thi xt tnh hon v c

ca a thc dng P pxq a0 a1x ... anxn vi n 2k trn vnh

Z2k .Lun vn gm 2 chng. Chng 1 trnh by kin thc chun b v

nhm, vnh, trng v a thc nhm phc v cho vic chng minh cc

kt qu chng sau. Trong phn u ca Chng 2 trnh by khi

nim a thc hon v c v mt s v d n gin. Phn th 2 ca

Chng 2 ginh chng minh tiu chun hon v c trn mt trng

hu hn ca mt s lp a thc dng xn (nh l 2.1.7) v a thc dng

xk bxj c vi k j 1 (nh l 2.2.1). Phn cui ca Chng 2

nhm trnh by mt iu kin cn v mt a thc vi h s nguyn


5hon v c theo modulo 2k, tc l hon v c trn vnh Z2k (nhl 2.3.10).


6Chng 1

Kin thc chun b

Chng ny trnh by khi nim v nhng kt qu chun b v nhm,

vnh, trng v a thc phc v cho chng minh cc kt qu ca chng

sau.

1.1 Kin thc chun b v nhm

1.1.1 nh ngha. Nhm l mt tp G cng vi mt php ton (k

hiu theo li nhn) tho mn cc iu kin

(i) Php ton c tnh kt hp: apbcq pabqc, @a, b, c P G.

(ii) G c n v: De P G sao cho ex xe x, @x P G.

(iii) Mi phn t ca G u kh nghch: Vi mi x P G, tn ti x1 P G

sao cho xx1 x1x e.

Mt nhm G c gi l nhm giao hon (hay nhm Abel) nu php

ton l giao hon. Nu G c hu hn phn t th s phn t ca G c

gi l cp ca G. Nu G c v hn phn t th ta ni G c cp v hn.

Sau y l mt s v d v nhm: Z,Q,R,C l cc nhm giao honcp v hn vi php cng thng thng. Vi mi s nguyn m 1, tp

Zm ta | a P Z, a b nu v ch nu a b chia ht cho mu

cc s nguyn modulo m vi php cng a b a b l mt nhm giao


7hon cp m. Tp

Zm ta P Zm | pa,mq 1u

cc s nguyn modulo m nguyn t cng nhau vi m vi php nhn

a b ab l mt nhm giao hon cp pmq, trong l hm Euler,

tc l p1q 1 v khi m 1 th pmq l s cc s t nhin nh hn m

v nguyn t cng nhau vi m.

1.1.2 nh ngha. Mt nhm G c gi l xyclic nu tn ti a P G

sao cho mi phn t ca G u l mt lu tha ca a. Trong trng

hp ny ta vit G paq v ta gi G l nhm xyclic sinh bi a. Phn t

a c gi l mt phn t sinh ca G.

1.1.3 B . Nhm con ca nhm xyclic l xyclic.

Chng minh. Gi s G paq l nhm xyclic. Cho H l nhm con ca

G. Nu H teu th H l nhm xyclic sinh bi e. Gi s H teu.

Chn e x P H. Vit x ak. Do x e nn k 0. V H l nhm con

nn ak P H. Trong hai s k v k t phi c mt s nguyn dng.

V th H cha nhng ly tha nguyn dng ca a. Gi r l s nguyn

dng b nht sao cho ar P H. R rng H parq. Cho y P H. Vit

y at vi t rq s, trong 0 s r. Ta c y at parqqas.

Do as yparqq P H. T cch chn ca r ta suy ra s 0. Do

y at parqq P parq. Vy H parq l nhm xyclic.

1.1.4 nh ngha. Tp con H ca mt nhm G c gi l nhm con

ca G nu e P H, a1 P H v ab P H vi mi a, b P H.

Cho G l mt nhm. Khi teu l nhm con b nht ca G v G l

nhm con ln nht ca G. Cho a P G. t paq tan | n P Zu. Khi paq l nhm con ca G, c gi l nhm con xyclic sinh bi a. Cp

ca nhm con paq c gi l cp ca phn t a.


81.1.5 B . Cho G l mt nhm v a l mt phn t ca G. Cc pht

biu sau l tng ng

(i) a c cp n.

(ii) n l s nguyn dng b nht sao cho an e.

(iii) an e v nu ak e th k l bi ca n vi mi k P Z.

Chng minh. (i)(ii). Trc ht ta khng nh tn ti mt s nguyn

dng k sao cho ak e. Gi s ngc li, vi mi cp s t nhin k k1

ta c ak1

k e. Suy ra ak ak1

. iu ny chng t paq c cp v hn,

v l vi gi thit (i). Do , tn ti nhng s nguyn dng k sao cho

ak e. Gi r l s nguyn dng b nht c tnh cht ar e. Ta thy

rng cc phn t e, a, a2, . . . , ar1 l i mt khc nhau. Tht vy, nu

ai aj vi 0 i j r th aji e v 0 j i r, do theo cch

chn ca r ta c i j. By gi ta chng minh G te, a, a2, . . . , ar1u.

R rng G te, a, a2, . . . , ar1u. Cho b P G. Khi b ak vi k P Z.Vit k rq s trong q, s P Z v 0 s r 1. Ta c

b ak arqs parqqas as P te, a, a2, . . . , ar1u.

V th G te, a, a2, . . . , ar1u l nhm cp r. Suy ra r n v (ii) c

chng minh.

(ii)(iii). Gi s ak e. Vit k nq r vi 0 r n. V an e nn

e ak anqar ar. Theo cch chn n ta phi c r 0, suy ra k chia

ht cho n.

(iii)(i). Gi r l s nguyn dng b nht sao cho ar e. Theo (iii),

r l bi ca n. Do n l s nguyn dng b nht tha mn an e.

Tng t nh chng minh (i)(ii) ta suy ra cp ca a l n.

1.1.6 H qu. Cho G paq l nhm xyclic cp n. Khi phn t

b ak l phn t sinh ca G nu v ch nu pk, nq 1.


9Chng minh. Gi s b ak l phn t sinh ca G. Khi b c cp n.

t d pk, nq. Ta c bn{d panqk{d e. Theo B 1.1.5, n{d l bi

ca n. V th d 1.

Ngc li, gi s pk, nq 1. Ta c bn panqk e. Gi s bt e.

Khi akt e. Theo B 1.1.5, kt l bi ca n. Do pk, nq 1 nn t

l bi ca n. Theo B 1.1.5, b c cp n. Vy G pbq.

1.1.7 nh ngha. Cho G l nhm v H l nhm con ca G. Vi mi

a P G, k hiu Ha tha | h P Hu. Ta gi Ha l mt lp ghp tri hay

lp k tri ca H trong G ng vi phn t a. Tp cc lp ghp tri ca

H trong G c k hiu l G{H. Khi H ch c hu hn lp ghp tri th

s cc lp ghp tri ca H c gi l ch s ca H trong G v c k

hiu l pG : Hq. Trong trng hp ny, ch s ca H chnh l s phn

t ca G{H. c bit, cp ca G chnh l pG : eq, ch s ca nhm con

tm thng teu.

Vi H l nhm con ca nhm G v a, b P G, ta d dng kim tra

c Ha Hb nu v ch nu ab1 P H.

1.1.8 nh l. (Lagrange). Trong mt nhm hu hn, cp v ch s

ca mt nhm con l c ca cp ca ton nhm.

Chng minh. Gi s G l nhm c cp n v H l nhm con ca G c

cp m. Vi mi a P G ta c a ea P Ha. V th, mi phn t ca

G u thuc mt lp ghp tri ca H. Gi s Ha X Hb H. Khi

tn ti h, h1 P H sao cho ha h1b. Suy ra a h1h1b. Cho xa P Ha,

trong x P H. Khi xa pxh1h1qb P Hb. Suy ra Ha Hb. Tng

t, Hb Ha v do Ha Hb. Vy hai lp ghp tri bt k ca

H nu khc nhau th phi ri nhau. Vi mi a P G, r rng nh x

f : H Ha xc nh bi fphq ha l mt song nh. V th mi lp

ghp tri ca H u c ng m phn t. Gi ch s ca H l s. T cc

lp lun trn ta suy ra n sm. V th s v m u l c ca n.


10

1.1.9 H qu. Cho G l nhm cp n v a P G. Khi cp ca a l

c ca n. Hn na, an e.

Chng minh. Gi cp ca a l r. Khi nhm con xyclic paq c cp r.

Theo nh l Lagrange, r l c ca n. Theo B 1.1.5 ta c ar e.

Suy ra an e.

1.1.10 H qu. Mi nhm cp nguyn t u l nhm xyclic.

Chng minh. Gi s G l nhm cp p nguyn t. Ly a P G, a e.

Theo nh l Lagrange, a c cp l c ca p. V p nguyn t nn cp

ca a l 1 hoc l p. Do a e nn cp ca a ln hn 1. Vy cp ca a

l p, tc G l nhm xyclic sinh bi a.

1.2 Kin thc chun b v vnh

1.2.1 nh ngha. Vnh l mt tp V c trang b hai php ton

cng v nhn tha mn cc iu kin sau y:

(i) V l mt nhm giao hon vi php cng;

(ii) V l mt v nhm vi php nhn: Php nhn c tnh cht kt hp

v tn ti phn t 1 P V (gi l phn t n v) sao cho 1x x1 x

vi mi x P V ;

(iii)Php nhn phn phi i vi php cng.

Nu php nhn l giao hon th V c gi l vnh giao hon. Sau

y l mt s v d thng gp v vnh:

1.2.2 V d. a) R rng Z,Q,R,C l nhng vnh giao hon vi phpcng v nhn thng thng;

b) Vi mi s t nhin n 0, tp Zn cc s nguyn modulo nlm thnh mt vnh giao hon vi php cng v php nhn cho bi:

a b a b v a b ab vi mi a, b P Zn.


11

c) Cho R l mt vnh. K hiu

Rrxs ta0 a1x . . . anxn| n P N, ai P R, @iu

l tp cc a thc mt bin x vi h s trong R. Khi Rrxs l mt

vnh giao hon vi php cng v nhn cc a thc:

aixi

bixi

pai biqxi v

aixi

bixi

ckxk vi ck

ijk aibj. Ta gi

Rrxs l vnh a thc mt bin x trn R. R rng R giao hon nu v

ch nu Rrxs l giao hon.

1.2.3 nh ngha. Cho V l mt vnh. Mt tp con S ca V c gi

l vnh con ca V nu 1 P S v x y, xy P S vi mi x, y P S.

D thy rng tp con S ca vnh V l vnh con ca V nu v ch nu

php cng v php nhn ng kn trong S v S lm thnh mt vnh

cng vi hai php ton ny.

1.2.4 nh ngha. Cho V l vnh v I l tp con ca V. Ta ni rng

I l ian ca V nu I l nhm con ca nhm cng V v xa, ax P I vi

mi a P I, x P V.

Cho V l mt vnh. D thy rng t0u l ian b nht ca V v V

l ian ln nht ca V. Idan t0u c k hiu l 0. Cc ian ca V

khc V c gi l cc ian thc s.

1.2.5 nh ngha. Cho V l vnh v I l ian ca V. Xt tp V {I

tx I | x P V u - tp cc lp ghp ca V theo nhm con I. R rng hai

phn t x I, y I P V {I l bng nhau nu v ch nu xy P I. Trong

tp V {I, ta nh ngha quy tc cng pxIqpyIq pxyqI v quy

tc nhn px Iqpy Iq xy I. Ta c th kim tra rng quy tc cng

v nhn trn khng ph thuc vo vic chn i din ca cc phn t,

tc l nu x1 I x I v y1 I y I th x y I x1 y1 I

v xy I x1y1 I. V th cc quy tc cng v nhn ny l hai php


12

ton trn V {I. Hn na, tp V {I cng vi php cng v nhn xc nh

nh trn l mt vnh giao hon, c n v l 1 I v phn t khng l

0 I. Vnh V {I va xy dng trn c gi l vnh thng ca V

theo ian I.

Ch rng vnh thng ca vnh Z theo ian mZ chnh l vnhZm cc s nguyn modulo m.

1.2.6 nh ngha. Mt ng cu vnh l mt nh x f t vnh V n

vnh S sao cho

(i) fpa a1q fpaq fpa1q vi mi a, a1 P V.

(ii) fpaa1q fpaqfpa1q vi mi a, a1 P V.

(iii) fp1q 1.

ng cu f c gi l n cu (ton cu, ng cu) nu f l n

nh (ton nh, song nh). Vnh V c gi l ng cu vi vnh S nu

tn ti mt ng cu gia chng. Mt ng cu (n cu, ton cu,

ng cu) t vnh S n S c gi l mt t ng cu (t n cu, t

ton cu, t ng cu).

Mnh sau y cho ta tnh cht ca vnh con v ian khi tc ng

qua mt ng cu vnh.

1.2.7 Mnh . Cho f : V S l ng cu vnh, B l vnh con ca

V v J l ian ca S. Cc pht biu sau l ng.

(i) fpBq l vnh con ca S.

(ii) f1pJq l ian ca V.

Chng minh. piq. Cho s, r P fpBq. Khi s fpbq v r fpcq vi

b, c P B. V b c, bc P B nn s r fpbq fpcq fpb cq P fpBq v

sr fpbqfpcq fpbcq P fpBq. V 1 P B nn

1 fp1q fp1q P fpBq.


13

Vy fpBq l vnh con ca S.

piiq. Do fp0q 0 P J nn 0 P f1pJq. Cho a, b P f1pJq. Khi

fpaq, fpbq P J. Suy ra fpa bq fpaq fpbq P J. Do ta c a b P

f1pJq. V th f1pJq l nhm con ca nhm cng V. Cui cng, cho

a P f1pJq v x P V. Khi fpaq P J. Suy ra fpaxq fpaqfpxq P J,

tc l ax P f1pJq. Tng t, xa P f1pJq. Vy f1pJq l ian ca

V.

1.2.8 nh ngha. Cho f : V S l mt ng cu vnh. V V l

vnh con ca V nn fpV q l vnh con ca S. Vnh con fpV q c gi

l nh ca f v c k hiu bi Im f . t Ker f ta P V | fpaq 0u.

Khi Ker f f1p0q. V 0 l ian ca S nn theo Mnh 1.2.7,

Ker f l ian ca V. Ta gi Ker f l ht nhn ca f .

1.2.9 Mnh . Cho f : V S l ng cu vnh. Khi f l n

cu nu v ch nu Ker f 0. Trong trng hp ny, V ng cu vi

vnh con Im f ca S.

Chng minh. Gi s f l n cu. R rng 0 P Ker f. Cho a P Ker f.

Khi fpaq 0 fp0q. Suy ra a 0. V th Ker f 0. Gi s

Ker f 0. Cho a, b P V sao cho fpaq fpbq. Khi fpa bq 0. Suy

ra a b P Ker f 0. V th a b 0 hay a b. Vy f l n cu.

1.2.10 nh l. (nh l ng cu vnh). Cho f : V S l ng cu

vnh. Khi V {Ker f Im f.

1.2.11 nh ngha. Cho V l vnh. Gi s tn ti s nguyn n 0

sao cho n1 0. Khi pnq1 0. Trong hai s n v n t phi c

mt s nguyn dng. Trong trng hp ny, ta gi c s ca V l s

nguyn dng n nh nht sao cho n1 0. Nu n1 0 ch xy ra khi

n 0 th ta ni V c c s 0.


14

D thy rng vnh Z cc s nguyn, vnh Q cc s hu t, vnh Rcc s thc, vnh C cc s phc u c c s 0. Vnh Zm cc s nguynmodulo m c c s m.

1.2.12 Mnh . Cho V l mt vnh. Cc pht biu sau l ng.

piq Nu V c c s 0 th V cha mt vnh con ng cu vi vnh Z.piiq Nu V c c s m th V cha mt vnh con ng cu vi Zm.

Chng minh. Xt nh x f : Z V xc nh bi fpnq n1 vi min P Z. D thy rng f l ng cu vnh. Gi s V c c s 0. Khi fpnq 0 khi v ch khi n 0. V th f l n cu. Do Z Im f.V th Im f l vnh con ca V ng cu vi Z. Gi s V c c s m.Khi Ker f mZ. Theo nh l 1.2.10, Z{mZ Im f. V th Im fl vnh con ca V ng cu vi Zm.

1.3 Kin thc chun b v trng

1.3.1 nh ngha. Mt phn t a ca vnh giao hon R c gi l

kh nghch nu tn ti b P R sao cho ab 1. Trng l mt vnh giao

hon, khc 0 v mi phn t khc 0 u kh nghch.

Ch rng vnh Z6 khng l trng v 2 P Z6 khng kh nghch.Vnh Z khng l trng v 2 P Z khng kh nghch. Cc vnh Q, R vC u l trng.

1.3.2 B . c s ca trng l 0 hoc l s nguyn t.

Chng minh. Gi s T l trng c c s n 0. V 1 0 nn n 1.

Nu n khng nguyn t th n ab vi 1 a, b n. V n l s nguyn

dng b nht tho mn n1 0 nn a1, b1 0. Do tn ti cc phn

t x, y P T sao cho xpa1q 1 ypb1q. V th ta c

0 pn1qxy xpa1qypb1q 1.1 1.


15

iu ny l v l.

1.3.3 B . Vnh Zn l trng khi v ch khi n l s nguyn t.

Chng minh. Cho Zn l trng. V Zn c c s n nn theo B 1.3.2,n l s nguyn t. Cho n l s nguyn t. Khi n 1. V th Zn 0.Cho 0 a P Zn. Khi a khng l bi ca n. V n nguyn t nn av n nguyn t cng nhau, tc l tn ti x, y P Z sao cho 1 ax ny.Suy ra 1 a x, tc l a kh nghch. Vy Zn l trng.

1.3.4 nh ngha. Mt tp con A ca trng T c gi l mt trng

con nu php cng v nhn l ng kn trong A v A lm thnh mt

trng cng vi hai php ton ny.

Gi s T l mt trng c c s m 0. Theo B 1.3.2, m phi

l s nguyn t. Theo Mnh 1.2.12, T cha mt trng con ng cu

vi trng Zm.Trong phn cui ca mc ny, chng ta nghin cu s phn t ca

mt trng hu hn. Trc ht ta cn nhc li mt s khi nim v tnh

cht ca khng gian vc t.

1.3.5 nh ngha. Cho K l mt trng. Mt tp V c trang b mt

php cng v mt nh x KV V (gi l php nhn vi v hng)

c gi l mt khng gian vc t trn trng K hay mt K-khng gian

vec t nu pV,q l mt nhm giao hon v tch v hng tho mn

cc tnh cht sau y: vi mi x, y P K v mi , P V ta c

(i) Phn phi: px yq. x. y. v x.p q x. x.;

(ii) Kt hp: xpyq px.yq.;

(iii) Unita: 1 .

1.3.6 nh ngha. Gi s V l mt K-khng gian vc t.

(i) Mt h vc t tviuiPI trong V c gi l mt h sinh ca V nu

mi phn t x P V u c th biu th tuyn tnh theo h , tc l


16

tn ti hu hn phn t vi1, . . . , vik ca h tviuiPI v hu hn phn t

ai1, . . . , aik ca K sao cho x k

j1aijvij. Nu V c mt h sinh gm

hu hn phn t th V c gi l K-khng gian hu hn sinh.

(ii) Mt h vc t tviuiPI trong V c gi l mt h c lp tuyn

tnh nu t mi rng buc tuyn tnh ca hk

j1aijvij 0 ta u c

aij 0 vi mi j 1, . . . , k.

(iii) Mt h vc t trong V c gi l mt c s ca V nu n l

mt h sinh v c lp tuyn tnh.

Ch rng mt h vc t ca V l mt c s ca V nu v ch nu

mi vc t ca V u c th biu th tuyn tnh mt cch duy nht qua

h . Ta c th ch ra rng mi K-khng gian vc t V 0 u c t

nht mt c s v cc c s ca V u c cng lc lng. Lc lng

chung ny c gi l s chiu ca V v k hiu l dimK V. c bit,

nu V c mt c s gm n phn t th cc c s khc ca V cng c n

phn t v ta c dimK V n.

1.3.7 Mnh . Cho T l trng hu hn c n phn t. Khi T c

c s p nguyn t v n l ly tha no ca p.

Chng minh. Theo B 1.3.2, c s ca T l 0 hoc l s nguyn t.

Nu T c c s 0 th theo Mnh 1.2.12, T cha mt vnh con ng

cu vi Z, do T c v hn phn t, v l. V th T c c s p 0.Theo B 1.3.2, p l s nguyn t. Theo Mnh 1.2.12, T cha mt

vnh con ng cu vi Zp. Ch rng Zp l trng theo B 1.3.3.V th ta d dng kim tra rng T c cu trc t nhin l mt khng

gian vc t trn trng Zp. Do T c hu hn phn t nn khng gianny c chiu hu hn. Gi s dimZp T k. Khi s phn t ca T l

n pk.


17

1.4 Kin thc chun b v a thc

Trong mc ny ta gi thit K l mt trng. Nhc li rng mt

a thc mt bin x vi h s trong K l mt biu thc dng fpxq

anxn an1x

n1 ... a0 trong ai P K. Nu an 0 th an c gi

l h s cao nht ca fpxq v s t nhin n c gi l bc ca fpxq.

Ta k hiu bc ca fpxq l deg fpxq. K hiu Krxs l tp cc a thc

mt bin x vi h s trong K. Gi s fpxq

aixi v gpxq

bixi,

ta nh ngha fpxq gpxq

pai biqxi v fpxqgpxq

ckxk, trong

ck

ijk aibj. Khi Krxs l mt vnh, gi l vnh a thc mt

bin x vi h s trong K.

1.4.1 Ch . Vi fpxq, gpxq P Krxs ta lun c

degpfpxq gpxqq maxtdeg fpxq, deg gpxqu

degpfpxq.gpxqq deg fpxq deg gpxq.

Tip theo l nh l php chia vi d cho a thc mt bin.

1.4.2 nh l. Cho fpxq, gpxq P Krxs vi gpxq 0. Khi tn ti duy

nht mt cp a thc qpxq, rpxq P Krxs sao cho

fpxq gpxqqpxq rpxq, vi rpxq 0 hoc deg rpxq deg gpxq.

Chng minh. Chng minh tnh duy nht. Gi s

fpxq gpxqqpxq rpxq gpxqq1pxq r1pxq,

trong rpxq, r1pxq bng 0 hoc c bc nh hn bc ca gpxq. Khi

gpxqpqpxq q1pxqq r1pxq rpxq.

Nu r1pxq rpxq th gpxqpqpxq q1pxqq 0. V gpxq 0 v K l

trng nn qpxq q1pxq 0, tc l qpxq q1pxq. Nu rpxq r1pxq th

degpr r1q deg

gpq q1q

deg g degpq q1q.


18

Ch rng

degpr r1q maxtdeg r, deg r1u deg g deg g degpq q1q.

iu ny mu thun vi ng thc trn.

Ta chng minh s tn ti ca qpxq v rpxq. Nu deg fpxq deg gpxq

th ta chn qpxq 0 v rpxq fpxq. Gi s deg fpxq deg gpxq. Cho

fpxq amxm . . . a0 v gpxq bnx

n . . . b0 vi am, bn 0 v

n m. Chn hpxq ambnxmn. t f1pxq fpxq gpxqhpxq. Khi

f1pxq 0 hoc f1pxq c bc thc s b hn bc ca fpxq. Trong trng

hp f1pxq 0, ta tm c d ca php chia fpxq cho gpxq l rpxq 0

v thng l qpxq hpxq. Nu f1pxq 0 th ta tip tc lm tng t

vi f1pxq v ta c a thc f2pxq. C tip tc qu trnh trn ta c

dy a thc f1pxq, f2pxq, . . ., nu chng u khc 0 th chng c bc

gim dn. V th sau hu hn bc ta c mt a thc c bc b hn

bc ca gpxq v chnh l a thc d rpxq. Nu mt a thc ca dy

bng 0 th d rpxq 0. C th, ta c

f1pxq fpxq gpxqhpxq

f2pxq f1pxq gpxqh1pxq

. . . . . . . . .

fkpxq fk1pxq gpxqhk1pxq

vi fkpxq 0 hoc deg fkpxq deg gpxq. Cng v vi v cc ng thc

li, ta c

fpxq gpxqphpxq h1pxq ... hk1pxqq fkpxq.

T ta c qpxq hpxq h1pxq . . . hk1pxq v rpxq fkpxq.

Trong nh l trn, nu rpxq 0 th qpxq c gi l thng ht v

rpxq c gi l d ca php chia fpxq cho gpxq. Nu rpxq 0 th ta

ni rng fpxq chia ht cho gpxq hay gpxq l c ca fpxq.


19

1.4.3 H qu. Cho K l mt trng v a P K. Khi d ca php

chia fpxq P Krxs cho x a l fpaq.

Chng minh. Chia fpxq cho x a, d hoc bng 0 hoc l mt a thc

bc 0 v bc ca pxaq bng 1. V vy, d l mt phn t r P K. Ta c

fpxq pxaqqpxqr. Thay x a vo ng thc ta c r fpaq.

Mt phn t a P K c gi l nghim ca fpxq P Krxs nu fpaq 0.

T H qu 1.4.3 ta c ngay kt qu sau.

1.4.4 H qu. Cho K l mt trng v a P K. Khi a l nghim ca

a thc fpxq P Krxs nu v ch nu tn ti a thc gpxq P Krxs sao

cho fpxq px aqgpxq.

T H qu 1.4.4 ta c ngay kt qu sau.

1.4.5 H qu. Cho fpxq P Krxs l a thc khc 0 v a1, a2, . . . , ar P K

l cc nghim phn bit ca fpxq. Khi deg fpxq r.


20

Chng 2

a thc hon v c

2.1 Khi nim a thc hon v c

2.1.1 nh ngha. (i) Cho R l mt vnh giao hon c hu hn phn

t. Cho fpxq P Rrxs. Ta ni rng fpxq l a thc hon v c trn R

nu nh x : R R cho bi paq fpaq l mt song nh.

(ii) Gi s fpxq l a thc vi h s nguyn. Vi n l mt s nguyn

dng cho trc, xt fpxq nh a thc trong Znrxs. Ta ni fpxq l honv c modulo n nu n l a thc hon v c trn Zn.

Di y l mt s v d v a thc hon v c.

2.1.2 V d. Xt R Z2 - trng cc s nguyn modulo 2. Cho fpxq 1 5x 2x2 3x3 v gpxq 1 x 4x2. Trong Z2rxs, a thc fpxqc dng fpxq 1 x x3 v a thc gpxq c dng gpxq 1 x. Ta c

fp0q 1 v fp1q 1. V th nh x : Z2 Z2 cho bi paq fpaqkhng phi l song nh. Ta c gp0q 1 v gp1q 0. V th nh x

: Z2 Z2 cho bi paq gpaq l mt song nh. Do fpxq l athc khng hon v c modulo 2 v gpxq l a thc hon v c

modulo 2.

2.1.3 V d. Xt R Z4 - vnh cc s nguyn modulo 4. Cho fpxq 2 3x 2x2 v gpxq 3 2x x2. Ta c fp0q 2, fp1q 3, fp2q 0

v fp3q 1. V th nh x : Z4 Z4 cho bi paq fpaq l song


21

nh. Ta c gp0q 3, gp1q 2, gp2q 3 v gp3q 2. V th nh x

: Z4 Z4 cho bi paq gpaq khng l mt song nh. Do fpxq la thc hon v c modulo 4 v gpxq l a thc khng hon v c

modulo 4.

Mt bi ton rt t nhin t ra l: Cho trc a thc fpxq

a0 a1x . . . anxn. Vi iu kin no ca cc h s a0, . . . , an, a

thc fpxq l hon v c. V nhng ng dng ca a thc hon v

c, c rt nhiu ngi quan tm n bi ton ny. Tuy nhin, cho n

nay, bi ton ny mi ch c gii quyt cho mt s trng hp c

bit. Bi ton tng qut vn ang l mt vn m thch thc cc nh

ton hc trn th gii.

Phn cn li ca mc ny l trnh by tnh hon v c cho cc a

thc bc khng qu 2 v a thc dng xn. Trc ht chng ta a ra

tiu chun hon v c ca cc a thc bc khng qu 1.

2.1.4 Mnh . Cc pht biu sau l ng.

(i) Trn mt vnh nhiu hn 1 phn t, mi a thc bc khng u

khng hon v c.

(ii) Trn mt trng T hu hn, mi a thc bc nht l hon v

c.

Chng minh. (i) Gi s fpxq a l a thc bc 0 trn vnh R c

nhiu hn 1 phn t. Khi fpbq a vi mi b P R. Do nh x

: R R xc nh bi pbq fpbq khng l ton nh. V th fpxq

khng hon v c.

(ii) Gi s fpxq abx P T rxs l a thc bc nht. Khi b 0. V

T l trng nn tn ti phn t b1 P T l nghch o ca b. Do vi

mi c P T , phng trnh abx c lun c nghim l x b1pcaq P T.

iu ny chng t nh x : T T cho bi pq fpq a b l


22

mt ton nh. Do T l hu hn nn l song nh. Vy fpxq l hon v

c trn T .

Phn tip theo, chng ta s a ra tiu chun hon v c cho nhng

a thc dng fpxq xn. Trc ht, ta cn hai b h tr sau. Nhc

li rng hm Euler c nh ngha trn tp cc s nguyn dng nh

sau: p1q 1 v pnq l s cc s nguyn dng nh hn n v nguyn

t cng nhau vi n.

2.1.5 B . Cho G l mt nhm c cp n. Khi G l nhm xyclic

nu v ch nu vi mi c d ca n tn ti nhiu nht mt nhm con

xyclic cp d.

Chng minh. Gi s G paq l xyclic. Cho d l c ca n. t b an{d.

Ta c bd an e. Nu bk e th ank{d e. Do nk{d l bi ca n.

Suy ra k l bi ca d. Theo B 1.1.5, b c cp d. t H pbq. Khi

H l nhm xyclic cp d. Gi s H 1 cng l mt nhm con xyclic cp

d ca G. Vit H 1 pcq v biu din c at. Do H 1 c cp d nn cd e.

Suy ra atd e. V th td l bi ca n v do t l bi ca n{d. V th

c at P H. Suy ra H 1 H. Do H v H 1 cng c s phn t l d nn

H H 1. Vy G c duy nht mt nhm con xyclic cp d.

Ngc li, gi s mi c d ca n c nhiu nht mt nhm con xyclic

cp d ca G. Ta xt quan h trn G nh sau: x y nu v ch nu

pxq pyq. D thy l quan h tng ng trn G. Vi mi x P G,

k hiu clpxq l lp tng ng ca x. Khi ta c

clpxq ty P G | pxq pyqu ty P G | y l phn t sinh ca pxqu.

Gi s x c cp dx. Theo nh l Lagrange, dx l c ca n. Theo H qu

1.1.6, phn t y xk l phn t sinh ca nhm xyclic pxq nu v ch

nu pk, dxq 1. V th clpxq c ng pdxq phn t vi l hm Euler.


23

Gi clpx1q, . . . , clpxtq l tt c lp tng ng. Khi G

i1,...,t

clpxiq

v clpxiq X clpxjq H vi mi i j. Do ta c

n pdx1q . . . pdxtq.

Ch rng dx1, . . . , dxt u l c ca n. Theo gi thit, mi c d ca

n ch c ng mt nhm con xyclic cp d nn cc nhm px1q, . . . , pxtq

phi c cp i mt khc nhau. Do dx1, . . . , dxt l cc s t nhin

i mt khc nhau. By gi ta xt nhm Zn. Khi vi mi c d can, tn ti duy nht mt nhm con xyclic cp d ca Zn. Do ta cn

d|n pdq. V th n pdx1q . . . pdxtq

d|n pdq n.

V th dx1, . . . , dxt l tt c cc l c ca n. c bit, trong cc c

dxi, i 1, . . . , t, t c mt c bng n. Do tn ti phn t xi P G c

cp dxi n. V vy G pxiq l xyclic.

Nhc li rng nu T l mt trng th mi a thc mt bin bc n

vi h s trong T c nhiu nht n nghim trong T .

2.1.6 B . Cho T l trng hu hn. t T T zt0u. Khi T l

nhm xyclic vi php nhn.

Chng minh. Gi s T c q phn t. Khi T c q 1 phn t.

Trc ht ta chng minh T l nhm vi php nhn. Cho a, b P T .

Khi a, b 0. Do tn ti c, d P T sao cho ac 1 bd. Suy ra

pabqpsdq 1 0. V th ab 0, tc l ab P T . Do php nhn

ng kn trong T . R rng php nhn trong T c tnh cht kt hp.

Do 1 0 nn 1 P T . Cho a P T . Khi a 0. V th a c nghch o

a1 P T. R rng a1 0. Suy ra a1 P T . V th mi phn t trong

T u kh nghch. Vy, T l mt nhm nhn vi cp l q 1.

Tip theo ta chng minh T l nhm xyclic. Gi s d l mt c ca

q 1. Theo B 2.1.5, chng minh T l xyclic, ta ch cn chng


24

minh T c nhiu nht mt nhm con xyclic cp d. Gi s H paq v

H 1 pbq l hai nhm con xyclic cp d khc nhau ca G. Khi ad 1

v bd 1. V th cc phn t ca H v H 1 u l nghim ca a thc

xd 1 P T rxs. V H v H 1 khc nhau nn H YH 1 c t nht d 1 phn

t. iu ny chng t a thc bc d c t nht d 1 nghim. iu ny

l v l.

By gi chng ta trnh by tiu chun hon v c ca a thc xn.

2.1.7 nh l. Trn mt trng c q phn t, a thc fpxq xn l

hon v c nu v ch nu q 1 v n l nguyn t cng nhau.

Chng minh. Gi s q 1 v n l nguyn t cng nhau. Khi tn ti

cc s nguyn r, s sao cho 1 rpq 1q sn. Gi s T l trng c

q phn t. Vi mi 0 c P T, ta c c c1 crpq1qpcsqn. Ta khng

nh crpq1q 1. Tht vy, t T T zt0u. Khi T l mt nhm

nhn vi cp l q 1 v nhm nhn ny l xyclic. Theo H qu 1.1.9 ta

c cq1 1. Suy ra crpq1q 1. Khng nh c chng minh. Do

c pcsqn fpcsq. Khi c 0 th ta lun c c pcqn fpcq. Suy ra nh

x : T T cho bi paq an l ton nh. V T l tp hu hn nn

l song nh. Do a thc fpxq xn l hon v c.

Gi s xn l hon v c trn trng T c q phn t. Ta chng minh

q 1 v n nguyn t cng nhau. Gi d l c chng ln nht ca q 1

v n. Theo B 2.1.6, T l nhm xyclic cp q 1 vi php nhn. Gi

s T paq. V T c cp q 1 nn aq1 1 theo H qu 1.1.9. Do

paq1qn{d 1. V th papq1q{dqn 1 1n. t fpxq xn. Khi

ta c fpapq1q{dq fp1q. Do fpxq hon v c theo gi thit nn ta c

apq1q{d 1. Do a c cp q 1 nn theo B 1.1.5, pq 1q{d l bi

ca q 1. V th d 1.

S dng nh l 2.1.7, ta c th c trng tnh hon v c ca cc


25

a thc bc 2 nh sau.

2.1.8 Mnh . a thc fpxq ax2 bx c vi a 0 l hon v c

trn mt trng hu hn T nu v ch nu b 0 v T c c s 2.

Chng minh. Gi s b 0 v T c c s 2. Khi fpxq ax2 c. Gi

s T c q phn t. Do T c c s 2 nn theo Mnh 1.3.7, tn ti

s t nhin k sao cho q 2k. Do q 1 l s l. V th q 1 nguyn

t cng nhau vi 2. Theo nh l 2.1.7, a thc x2 l hon v c trn

T . Gi s fprq fpsq vi r, s P T. Khi ar2 c as2 c. Suy ra

ar2 as2. Do a 0 v T l trng nn a kh nghch. Nhn hai v vi

nghch o ca a ta c r2 s2. V a thc x2 l hon v c trn T

nn r s. Vy fpxq l hon v c trn T .

Ngc li, gi s fpxq l hon v c trn T . Trc ht ta chng

minh b 0. Gi s b 0 v ta cn tm mu thun. Do a 0 nn tn

ti phn t a1 P T sao cho aa1 1. V th ta c

fpa1bq apa1bq2 bpa1bq c a1b2 a1b2 c c.

R rng fp0q c. Do fpxq l hon v c trn T nn 0 a1b. V

b 0 nn tn ti b1 P T sao cho bb1 1. Nhn c hai v ca ng

thc 0 a1b vi ab1 ta c 0 1, iu ny l v l. Vy b 0.

Tip theo ta chng minh c s ca T l 2. Do T l trng nn T c

c s 0 hoc c s p nguyn t. Nu T c c s 0 th vi mi s t

nhin n m ta c pm nq1 0, tc l n.1 m.1. V th T cha mt

tp con tn.1 | n P Zu gm v hn phn t, iu ny l v l. V th Tc c s p nguyn t. Ta cn chng minh p 2. Nu p 2 th p l s

nguyn t l. Do s phn t ca T l mt ly tha ca p nn q l s l.

V th q 1 l s chn. V th q 1 v 2 khng nguyn t cng nhau.

Theo nh l 2.1.7, a thc x2 khng hon v c trn T . V th tn

ti hai phn t r s trong T sao cho r2 s2. Suy ra ar2 c as2 c,


26

tc l fprq fpsq, trong khi r s. Do fpxq khng hon v c

trn T , v l.

2.2 Mt s lp a thc hon v c trn mt

trng

Mc ny dnh trnh by tiu chun hon v c trn mt trng

ca mt s lp a thc. Trc ht, chng ta trnh by mt m rng ca

Mnh 2.1.8.

2.2.1 nh l. Cho T l trng c q phn t. Gi s k, j l cc s

nguyn dng sao cho k j 1 v gcdpk j, q 1q 1. Khi a

thc fpxq axk bxj c vi a 0 l hon v c trn T nu v ch

nu b 0 v gcdpk, q 1q 1.

Chng minh. Gi s fpxq hon v c trn T . Trc ht ta chng minh

b 0. Gi s b 0 v ta cn tm mu thun. t gpxq xka1bxj. Ta

khng nh gpxq hon v c trn T . Cho r, s P T sao cho gprq gpsq.

Khi agprqc agpsqc. Suy ra fprq fpsq. Do fpxq hon v c

trn T nn r s.V th gpxq hon v c. Ta c gpxq xjpxkja1bq.

V j 1 nn gp0q 0. Theo gi thit, gcdpk j, q 1q 1. Do tn

ti cc s nguyn m,n sao cho 1 mpk jqnpq1q. t d a1b.

Do b 0 v a 0 nn d P T . Ta c

d d1 pdmqkjpdq1qn.

Do T l nhm cp q 1 vi php nhn v d P T nn dq1 1. V th

d pdmqkj. Suy ra

gpdmq pdmqjppdmqkj dq pdmqjppdmqkj pdmqkjq 0.

Do d 0 nn dm 0. Nh vy, gp0q gpdmq, trong khi dm 0.

iu ny chng t gpxq khng hon v c trn T , v l. Vy b 0.


27

Tip theo, ta chng minh gcdpk, q 1q 1. Do b 0 nn fpxq

axk c. t hpxq xk. Cho hprq hpsq vi r, s P T. Khi rk sk.

Suy ra ark c ask c, tc l fprq fpsq. Do fpxq hon v c trn

T nn r s. V th hpxq l hon v c trn T . Theo nh l 2.1.7,

gcdpk, q 1q 1.

Ngc li, gi s b 0 v gcdpk, q1q 1. Ta cn chng minh fpxq

hon v c trn T . V b 0 nn fpxq axk c. Do gcdpk, q 1q 1

nn theo nh l 2.1.7, a thc xk l hon v c trn T . Gi s fprq

fpsq. Khi ark c ask c. Suy ra rk sk. Do xk hon v c trn

T nn r s. Vy fpxq l hon v c trn T .

H qu sau y cho ta mt lp cc a thc hon v c.

2.2.2 H qu. Cho T l trng c q phn t. Gi s q 1 khng l bi

ca 3, 5, 7. Khi a thc x8 bxt, vi t l mt s l nh hn 8, l a

thc hon v c trn T nu v ch nu b 0 v T c c s 2.

Chng minh. Xt a thc fpxq xkaxt vi k 8. V t l v t 8 nn

k t 8 t P t7, 5, 3, 1u. Theo gi thit, q 1 khng l bi ca 3, 5, 7.

Do gcdpk t, q 1q 1. Theo nh l 2.2.1, a thc fpxq l hon v

c trn T nu v ch nu a 0 v gcdpk, q 1q gcdp8, q 1q 1.

Ch rng gcdp8, q 1q 1 nu v ch nu q l s chn. Gi p l c

s ca T . Khi p l s nguyn t v q l mt ly tha ca p. V th q

l s chn nu v ch nu T c c s nguyn t chn, tc l c s ca

T bng 2. Vy, fpxq l hon v c trn T nu v ch nu a 0 v T

c c s 2.

2.2.3 H qu. Cho T l trng c q phn t. Cho fpxq axk bxj c

l mt a thc trn T , trong a 0, k j 1 v ba1 l mt ly

tha bc k j ca mt phn t trong T . Khi fpxq l hon v c

trn T nu v ch nu b 0 v gcdpk, q 1q 1.


28

Chng minh. Gi s fpxq hon v c trn T . Theo gi thit, tn ti

phn t r P T sao cho ba1 rkj. t gpxq xkba1xj. Ta khng

nh gpxq l hon v c trn T . Tht vy, gi s gpsq gps1q vi

s, s1 P T. Khi agpsq c agps1q c. Do fpsq fps1q. V fpxq

hon v c trn T nn s s1. Do gpxq hon v c trn T .

Ta c

gpxq xjpxkj pa1bqq xjpxkj rkjq.

Do j 1 nn gp0q 0. R rng gprq 0. V gpxq hon v c nn

r 0. Suy ra a1b 0. V a 0 nn b 0. Do gpxq xk. V gpxq

hon v c nn theo nh l 2.1.7 ta c gcdpk, q 1q 1.

Ngc li, gi s b 0 v gcdpk, q 1q 1. Khi fpxq axk c.

Do gcdpk, q 1q 1 nn theo nh l 2.1.7 ta suy ra xk l hon v c

trn T . Gi s fprq fpsq vi r, s P T. Khi ark c ask c. Suy

ra rk sk. Do gpxq hon v c trn T nn r s. V th fpxq l hon

v c trn T .

Trong phn cui ca mc ny, chng ta trnh by mt s v d v a

thc hon v c.

2.2.4 V d. Cc pht biu sau l ng:

(i) a thc x8 2x3 hon v c trn trng Z11.(ii) a thc x8 4x hon v c trn trng Z29.

Chng minh. Trong H qu 2.2.2, vi gi thit T l trng c q phn

t vi q 1 khng l bi ca 3, 5, 7 th a thc fpxq x8 bxt, trong

t l nh hn 8, l hon v c trn T nu v ch nu b 0 v T c

c s 2. V d 2.2.4 ch ra rng gi thit q 1 khng l bi ca 3, 5, 7

l khng th b i c. Mc d a thc x8 2x3 trong mnh (i) c

h s b 2 0 v c s ca trng T Z11 l 11 2, nhng a


29

thc ny vn hon v c trn T v q 1 10 l mt bi ca 5. Tht

vy, xt a thc fpxq x8 2x3 ta c fp0q 0, fp1q 10, fp2q

9, fp3q 6, fp4q 2, fp5q 7, fp6q 1, fp7q 5, fp8q 4, fp9q 8

v fp10q 3 . V th nh x : Z11 Z11 cho bi paq fpaq l songnh. Do fpxq l a thc hon v c modulo 11. Tng t nh vy,

mc d a thc x8 4x trong mnh (ii) c h s b 4 0 v c

s ca trng T Z29 l 29 2, nhng n vn hon v c trn Z29.S d nh vy v q 1 28 l mt bi ca 7. Tht vy, xt a thc

fpxq x8 4x ta c fp0q 0, fp1q 5, fp2q 3, fp3q 19, fp4q

12, fp5q 15, fp6q 18, fp7q 6, fp8q 23, fp9q 27, fp10q



8, fp23q 28, fp24q 4, fp25q 9, fp26q 24, fp27q 16 v fp28q

26 . V th nh x : Z29 Z29 cho bi paq fpaq l song nh. Do fpxq l a thc hon v c modulo 29

2.2.5 V d. Xt a thc fpxq x8 3x. Khi

(i) fpxq khng hon v c trn trng Z5;(ii) fpxq khng hon v c trn trng Z11;(iii) fpxq khng hon v c trn trng Z13.

Chng minh. v d (i) a thc fpxq x8 3x c q 1 4 khng l

bi ca 3, 5, 7 nhng b 3 0 v c s ca trng T Z5 l 5 2.ng thi a thc fpxq c fp0q 0, fp1q 3, fp2q 0, fp3q 2 v

fp4q 4. V th nh x : Z5 Z5 cho bi paq fpaq khng l songnh. Do fpxq l a thc khng hon v c modulo 5. v d (ii)

a thc fpxq x8 3x c q 1 10 l bi ca 5 v c b 3 0 v

c s ca trng T Z11 l 11 2. ng thi a thc fpxq c fp0q 0, fp1q 9, fp2q 8, fp3q 7,fp4q 8, fp5vq 0, fp6q 8, fp7q

10, fp8q 3, fp9q 9 v fp10q 4. V th nh x : Z11 Z11 cho


30

bi paq fpaq khng l song nh pfp2q fp4q fp6q 8q. Do

fpxq l a thc khng hon v c modulo 11. Tng t nh vy, v d

(iii) a thc fpxq x83x c q1 12 l bi ca 3 v c b 3 0

v c s ca trng T Z13 l 13 2. ng thi a thc fpxq cfp0q fp3q 0, fp1q fp6q 11, fp2q fp8q 3, fp4q fp12q 4

v fp9q fp11q 2 V th nh x : Z13 Z13 cho bi paq fpaqkhng l song nh. Do fpxq l a thc khng hon v c modulo

13

2.3 a thc hon v c modulo 2k

Mc ch ca mc ny l trnh by tiu chun hon v c ca a

thc vi h s nguyn theo modulo 2k trong bi bo ca Ronald Rivest

[R]. T nay n ht lun vn, ta lun gi thit fpxq l a thc vi h

s nguyn. Nhc li rng fpxq c gi l hon v c modulo n nu

nh x : Zn Zn cho bi paq fpaq l mt song nh.

2.3.1 Ch . Gi s X l mt tp hu hn. Khi mt nh x :

X X l song nh nu v ch nu n l n nh, nu v ch nu n

l ton nh. Do , a thc fpxq hon v c modulo n nu v ch nu

nh x : Zn Zn cho bi paq fpaq l n nh, nu v ch nufpaq fpbq pmod nq ko theo a b pmod nq vi mi a, b P Z.

2.3.2 V d. Cho fpxq 37x8x23x34x4 v gpxq 22x2x2.

Ta xt tnh hon v c ca fpxq v gpxq theo modulo 23. Trong Z8rxs,a thc fpxq c dng fpxq 37x3x34x4. Ta c fp0q 3, fp1q

1, fp2q 4, fp3q 7, fp4q 6, fp5q 2, fp6q 0 v fp7q 5. V

th nh x f : Z8 Z8 cho bi fpaq fpaq l song nh. Ta cgp0q 2, gp1q 6, gp2q 6, gp3q 2, gp4q 2, gp5q 6, gp6q 6

v gp7q 2. V th nh x g : Z8 Z8 cho bi gpaq gpaq khng


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l song nh. Vy, fpxq hon v c modulo 23 v gpxq khng hon v

c modulo 23.

Trc ht chng ta trnh by mt tiu chun a thc fpxq l hon

v c modulo 2.

2.3.3 B . a thc fpxq a0 a1x . . . adxd vi h s nguyn

l hon v c modulo 2 nu v ch nu a1 . . . ad l l.

Chng minh. Ta lun c fp0q a0 P Z2. Ta xt hai trng hp.a) Gi s a0 chn. Khi fp0q 0 P Z2. V th, fpxq l hon v c

modulo 2 nu v ch nu fp1q 1 P Z2. Ta c fp1q a0a1 . . .ad.V a0 chn nn fp1q 1 P Z2 nu v ch nu a1 . . . ad l l. Do fpxq l hon v c modulo 2 nu v ch nu a1 . . . ad l l.

b) Gi s a0 l. Khi fp0q 1 P Z2. V th, fpxq l hon v cmodulo 2 nu v ch nu fp1q 0 P Z2. Ta c fp1q a0a1 . . .ad.V a0 l nn fp1q 0 P Z2 nu v ch nu a1 . . .ad l l. Do fpxql hon v c modulo 2 nu v ch nu a1 . . . ad l l.

2.3.4 B . Cho fpxq a0 a1x . . . adxd l a thc vi h s

nguyn. Cho n 2m vi m l mt s nguyn chn. Gi s fpxq l hon

v c modulo n. Khi a1 l s l.

Chng minh. Ta c 0 m P Zn. V fpxq hon v c modulo n nnfp0q fpmq P Zn. Ta c

fpmq a0 a1m

a2m2 . . . adm

d

.

Do m l s chn nn m2 chia ht cho n. Do

fpmq a0 a1m pmod nq.

Gi s a1 l chn. V m chn nn a1m 0 pmod nq. Do fpmq

a0 pmod nq. R rng fp0q 0 pmod nq. V th fpmq fp0q P Zn, trong


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khi 0 m P Zn. V th fpxq khng hon v c modulo n, v l.Vy, a1 l s l.

2.3.5 V d. Cho fpxq 2x 4x2 6x3. Ta xt tnh hon v c ca

fpxq theo modulo n 8. R rng n 2.m, vi m 4 l s chn. Ta

c a1 2 l s chn. V th, theo B 2.3.4, a thc fpxq khng hon

v c modulo 8. iu ny cng c th thy d dng t nh ngha a

thc hon v c. Trong Z8 ta c fp0q 0, fp1q 4, fp2q 4, fp3q 4, fp4q 0, fp5q 4, fp6q 4 v fp7q 4. V th nh x g : Z8 Z8cho bi fpaq fpaq khng l song nh.


nguyn. Cho n 2k vi k 0 v cho m 2k1 n{2. Nu fpxq hon

v c modun n th n hon v c modulo m.

Chng minh. Gi s fpxq hon v c modulo n. Vi mi a P Z ta c

fpamq a0 a1pxmq . . . adpxmqd

a0 a1a . . . adadmt, vi t l mt s nguyn no

a0 a1a . . . adadpmod mq

fpaq pmod mq.

Gi s fpxq khng hon v c modulo m. Khi c hai gi tr r

s P Zm sao cho fprq fpsq P Zm. Suy ra r s pmod mq v fprq fpsq pmod mq. V vy, theo tnh cht va chng minh trn ta c

fprq fpr mq fpsq fpsmq pmod mq.

Do , tn ti cc s nguyn t1, t2 sao cho fprq fpr mq mt1 v

fprq fpsq mt2. V pr mq r khng l bi ca n 2m nn ta

c r m r P Zn. Do fpxq hon v c modulo n nn fpr mq fprq pmod nq. Do t1 l. Theo gi thit, r s pmod mq. V th


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r s pmod nq. V fpxq hon v c modulo n nn fprq fpsq pmod nq.

V th t2 l l. Suy ra t2 t1 chn. Do

fpr mq fpsq

fprq fpsq

fprq fpr mq

mpt2 t1q 0 pmod nq.

Do fpxq hon v c modulo n nn r m s 0 pmod nq. Do m l

c ca n nn rm s 0 pmod mq. V th r s 0 pmod mq. iu

ny l v l vi cch chn ca r v s.

Nhn chung, bi ton kim tra (theo nh ngha) tnh khng hon

v c ca fpxq modulo 2k khi k ln l rt kh. V th, trong nhiu

trng hp ta c th s dng B 2.3.6 gii quyt bi ton ny.

Di y l mt v d minh ha.

2.3.7 V d. Cho a thc fpxq 12xx2. Khi fpxq khng hon

v c modulo 128.

Chng minh. Trong vnh Z8 ta c

fp0q 1, fp1q 4, fp2q 1, fp3q 0, fp4q 1, fp5q 4, fp6q 1.

c bit, ta c 0 2 P Z8, nhng fp0q fp2q 1 P Z8. V th nhx f : Z8 Z8 cho bi fpaq fpaq khng l song nh. V th fpxqkhng hon v c modulo 8. Theo B 2.3.6, fpxq khng hon v

c modulo 16. Tip tc p dng B 2.3.6, ta suy ra fpxq khng

hon v c modulo 2k vi mi k 3. c bit, v 128 l ly tha ca

2 nn fpxq khng hon v c modulo 128.

2.3.8 B . Cho fpxq a0 a1x . . . adxd l a thc vi h

s nguyn. Cho n 2m. Gi s fpxq hon v c modulo n. Khi

fpamq fpaq m pmod mq vi mi s nguyn a.


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Chng minh. Cho a P Z. Theo chng minh B 2.3.6, tn ti s nguynt sao cho fpa mq fpaq mt. V n 2m nn pa mq a khng

l bi ca n, tc l am a P Zn. Do fpxq hon v c modulo nnn fpamq fpaq P Zn. Do fpaq fpa mq pmod nq. V thmt khng l bi ca n. Suy ra t l s l. V th

fpamq fpaq mt fpaq m pmod 2mq

fpaq m pmod nq.


nguyn. Cho n 2m vi m l mt s chn. Gi s fpxq hon v c

modulo m. Khi fpxq hon v c modulo n nu v ch nu a1 l s

l v a3 a5 a7 . . . l s chn.

Chng minh. Do m l s chn nn m2 chia ht cho n 2m. Do vi

mi i 2, . . . , d v vi mi s nguyn a ta c

pamqi ai imai1 tm2 vi s nguyn t no

ai imai1 pmod nq.

V i 2 nn i 1 1. Do nu c t nht mt trong 3 s i, a, ai l

chn th aiimai1 l bi ca n. Trong trng ny ta c

aipamqi aia

ipmod nq.

Cn nu c 3 s i, a, ai u l s l th aipimai1q m pmod nq v do

ta c

aipamqi aia

im pmod nq.

By gi ta xt hai trng hp.


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Trng hp 1: a l s chn. Theo trn ta c

fpamq a0 a1pamq pa2a2 . . . ada

dq

fpaq a1m pmod nq.

Trng hp 2: a l s l. Theo trn ta c

fpamq a0 a1pamq

a2a2 a4a

4 . . .

pa3a3mq pa5a

5mq . . .

fpaq pa1 a3 a5 . . .qm pmod nq.

By gi ta chng minh b . Gi s fpxq hon v c modulo n.

Theo B 2.3.4, h s a1 l l. Ch rng am a pmod nq vi

mi s nguyn a. V th fpa mq fpaq pmod nq vi mi s nguyn

a. Chn a l mt s l (chng hn a 1). Theo Trng hp 2 ta c

fpamq fpaq pa1 a3 a5 . . .qm pmod nq.

Do ta c pa1a3a5 . . .qm 0 pmod nq. Suy ra a1a3a5 . . .

phi l s l. V a1 l nn a3 a5 a7 . . . l s chn.

Ngc li, gi s a1 l s l v a3a5a7. . . l s chn. Gi s phn

chng rng fpxq khng hon v c modulo n. Khi tn ti a b P Znsao cho fpaq fpbq P Zn. Do c hai s nguyn a b pmod nq saocho fpaq fpbq pmod nq. V m l c ca n nn fpaq fpbq pmod mq.

Theo gi thit, fpxq hon v c modulo m, v th a b pmod mq. Do

a b pmod nq v n 2m nn a b mt vi t l mt s l. V th c

mt s nguyn t1 sao cho b amt am 2t1m am pmod nq.

Suy ra b am P Zn. V fpaq fpbq pmod mq nn ta c

fpaq fpamq pmod nq.

Nu a l s chn th theo Trng hp 1 ta c

fpamq fpaq a1m pmod nq.


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Do a1m l bi ca n. Suy ra a1 chn, iu ny mu thun vi gi

thit a1 l. Do a l s l. Theo Trng hp 2 ta c

fpamq fpaq pa1 a3 a5 . . .qm pmod nq.

Do pa1 a3 a5 . . .qm l bi ca n. Suy ra a1 a3 a5 . . . l

chn. Theo gi thit a1 l s l. V th a3 a5 . . . l s l, iu ny

l mu thun vi gi thit a3 a5 . . . chn. Vy fpxq hon v c

modulo n.

nh l sau y l kt qu chnh ca mc ny, v l mt trong 3 kt

qu chnh ca lun vn.

2.3.10 nh l. Cho fpxq a0a1x . . .adxd l mt a thc vi h

s nguyn. Cho n 2k vi k 2. Khi fpxq hon v c modulo n

nu v ch nu a1 l s l, a2a4a6. . . l s chn v a3a5a7. . .

l s chn.

Chng minh. Gi s fpxq hon v c modulo n. t m1 2k1. Khi

n 2m1. Do k 2 nn m1 l s chn. Do theo B 2.3.4, h s

a1 l s l. Do fpxq hon v c modulo n 2m1 nn theo B 2.3.6,

a thc fpxq hon v c modulo m1. V th, p dng B 2.3.9, ta

c a3 a5 a7 . . . l s chn.

Tip theo, ta khng nh bng quy np theo i 1, . . . , k 1 rng

fpxq l hon v c modulo 2ki. Vi i 1, ta chng minh phn

trn rng fpxq hon v c modulo m1 2k1, do khng nh ng

vi i 1. Gi s kt qu ng cho i 1 vi i k 1, tc l fpxq

hon v c modulo 2ki1. t mi 2ki v mi1 2

ki1. Khi

mi1 2mi. Do fpxq hon v c modulo mi1 nn theo B 2.3.6,

fpxq hon v c modulo mi 2ki. Khng nh c chng minh.

Vi i k 1, t khng nh trn ta suy ra fpxq hon v c modulo


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2. Do theo B 2.3.3 ta c a1 a2 a3 . . . ad l s l. V a1 l

nn

pa2 a4 a6 . . .q pa3 a5 a7 . . .q

l s chn. Do a3 a5 a7 . . . l s chn nn a2 a4 a6 . . . l s

chn.

Ngc li, gi s a1 l s l, a2 a4 a6 . . . l s chn v a3

a5 a7 . . . l s chn. Ta cn chng minh fpxq hon v c modulo

n 2k. Ta chng minh iu ny bng quy np theo k. Cho k 1. V

a1 l s l, a2 a4 a6 . . . l s chn v a3 a5 a7 . . . l s chn

nn a1 a2 a3 . . . ad l s l. Do , theo B 2.3.3 ta suy ra

fpxq hon v c modulo 2 21. Vy, kt qu ng cho trng hp

k 1.

Cho k 1 v gi thit kt qu ng cho trng hp k 1,

tc l fpxq hon v c modulo 2k1 trong trng hp a1 l s l,

a2 a4 a6 . . . l s chn v a3 a5 a7 . . . l s chn. t

m 2k1. Khi n 2m. Do k 1 nn m l s chn. V a1 l s l,

a3 a5 a7 . . . l s chn v fpxq hon v c modulo m nn theo

B 2.3.9 ta suy ra fpxq hon v c modulo n.

Phn cui ca mc ny trnh by mt s v d v a thc hon v

c modulo 2k.

2.3.11 V d. Cc a thc sau l hon v c modulo 2k

(i) ax bx2 vi mi s l a v mi s chn b.

(ii) x x2 x4.

(iii) 1 x x2 . . . xd, trong d 1 pmod 4q.

Chng minh. (i) Ta c a1 a l, a2 b chn. Theo nh l 2.3.10,

ax bx2 hon v c modulo 2k.


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(ii) Ta c a1 1 l, a2a4 2 chn. Theo nh l 2.3.10, xx2x4

hon v c modulo 2k.

(iii) T gi thit ta suy ra d l s l. Ta c a1 1 l, a2 a4 . . .

ad1 pd 1q{2 v a3 a5 . . . ad pd 1q{2. Do d 1 pmod 4q

nn d 1 l bi ca 4. V th pd 1q{2 l s chn. Theo Theo nh l

2.3.10, 1 x x2 . . . xd hon v c modulo 2k.


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Kt lun

Lun vn trnh by mt s kt qu v a thc hon v c trong

hai bi bo:

1. R. Rivest, Permutation polynomials modulo 2w, Finite fields and

their applications, 7 (2001), 287-292.

2. R. A. Mollin and C. Small, On permutation polynomials over

finite fields, Inter. J. Math. and Math. Sciences, 10 (1987), 535-544.

Ni dung chnh ca lun vn l:

Trnh by mt s kin thc chun b v nhm, vnh, trng, a

thc phc v chng minh cc kt qu chnh Chng 2.

a ra tiu chun a thc dng xn l hon v c trn mt

trng hu hn (nh l 2.1.7). T suy ra tiu chun hon v c

ca a thc bc khng qu hai.

Chng minh mt iu kin cn v cho cc a thc dng

xk bxj c, trong k j 1 l hon v c trn mt trng hu

hn (nh l 2.2.1). T thu c tnh cht hon v c ca mt

s a thc c bit.

Gii quyt trn vn bi ton v tnh hon v c modulo n cho

cc a thc vi h s nguyn trong trng hp n l ly tha ca 2

(nh l 2.3.10).


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Ti liu tham kho

[C] Nguyn T Cng, i s hin i, tp 1, NXB HQGHN, 2001.

[HT] Bi Huy Hin v Phan Don Thoi, Bi tp i s v s hc,

Tp 2, NXB Gio dc, 1986.

[La] Ng Thc Lanh, i s v s hc, Tp 2, NXB Gio dc, 1986.

[LN] R. Lidl and H. Niedereiter, Finite Fields, Addison - Wesley, 1983

[LM1] R. Lidl and G. L. Mullen, When does a polynomial over a

finite field permute the elements of the field? The American Math,

Monthly, 3 (1988), 243-246

[LM2] R. Lidl and G. L. Mullen, When does a polynomial over a finite

field permute the elements of the field? II. The American Math,

Monthly, 1 (1993), 71-74

[MS] R. A. Mollin and C. Small, On permutation polynomials over

finite fields, Inter. J. Math. and Math. Sciences, 10 (1986), 535-

544.

[Riv] R. Rivest, Permutation polynomials modulo 2w, Finite fields and

their applications, 7 (1999), 287-292.


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