LV Da Thuc Bat Kha Quy

I HC THI NGUYN TRNG I HC KHOA HC

NGUYN H LINH

A THC BT KH QUY

LUN VN THC S TON HC

Thi Nguyn 2012

1S ha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn

Mc lc

Mc lc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

Li ni u . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3

1 a thc bt kh quy 5

1.1 Khi nim a thc . . . . . . . . . . . . . . . . . . . . . . 5

1.2 a thc bt kh quy . . . . . . . . . . . . . . . . . . . . . 9

1.3 Trng phn r ca a thc . . . . . . . . . . . . . . . . . 13

2 Mt s phng php xt tnh bt kh quy trn Q 202.1 Nghim hu t v tnh bt kh quy trn Q . . . . . . . . . 212.2 Phng php dng B Gauss . . . . . . . . . . . . . . . 24

2.3 Phng php dng tiu chun Eisenstein . . . . . . . . . . 28

2.4 Rt gn theo mun mt s nguyn t . . . . . . . . . . . 30

3 Tnh bt kh quy trn trng Zp 343.1 Kin thc chun b v nhm nhn Zp . . . . . . . . . . . . 343.2 Tnh bt kh quy trn trng Zp . . . . . . . . . . . . . . . 37Kt lun . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44

Ti liu tham kho . . . . . . . . . . . . . . . . . . . . . . . . 45

1


2Li cm n

Ti xin gi li bit n chn thnh nht n PGS.TS L Th Thanh Nhn.

C dnh rt nhiu thi gian v tm huyt trong vic hng dn ti. Cho

n hm nay, lun vn thc s ca ti c hon thnh cng chnh l

nh s nhc nh, n c, s gip nhit tnh ca C.

Ti xin trn trng cm n Ban Gim hiu, Khoa Ton - Tin v Phng

o to - Khoa hc v Quan h quc t ca trng i hc Khoa hc -

i hc Thi Nguyn. Ti xin trn trng cm n cc Thy C tn tnh

truyn t nhng kin thc qu bu cng nh to mi iu kin thun li

nht ti hon thnh lun vn ny.

Ti xin chn thnh by t lng bit n n gia nh, bn b, nhng

ngi khng ngng ng vin, h tr v to mi iu kin tt nht cho

ti trong sut thi gian hc tp v thc hin lun vn.


3Li ni u

Trong l thuyt a thc, a thc bt kh quy ng mt vai tr quan

trng ging nh vai tr ca s nguyn t trong tp cc s nguyn. Nu

nh l c bn ca S hc cho php coi cc s nguyn t nh l nhng

vin gch xy nn tp cc s nguyn, th cc a thc bt kh quy chnh l

nhng vin gch xy nn tp tt c a thc. Bi v mi a thc bc dng

dng chun (tc l h s cao nht bng 1) vi h s trn mt trng u

vit c thnh tch ca hu hn a thc bt kh quy dng chun v s

phn tch l duy nht nu khng k n th t cc nhn t.

Bi ton xt tnh bt kh quy ca cc a thc trn trng phc C vtrn trng thc R c gii quyt t u th k 19, khi ngi ta chngminh c nh l c bn ca i s. C th, cc a thc bt kh quy

trn C l v ch l cc a thc bc nht; cc a thc bt kh quy trn Rl v ch l cc a thc bc nht hoc bc hai vi bit thc m. Tuy nhin

bi ton xt tnh bt kh quy ca a thc trn trng hu t Q hoc trntrng thng d Zp (vi p l s nguyn t) vn ang th thch cc nhton hc trn th gii.

Mc ch ca lun vn l trnh by mt s kt qu v a thc bt kh

quy trn mt trng, c bit l trn trng Q v trng Zp. Ni dung calun vn c vit da theo cun sch ``L thuyt Galois" ca J. Rotman

[Rot], cun sch ``a thc v tnh bt kh quy" ca A. Schinzel [Sc], bi

bo ``Tnh bt kh quy ca a thc" ng trn Tp ch i s ca I. Seres

[S] v bi bo ``Tiu chun bt kh quy ca a thc" ng trn tp ch ni

ting Ann. Math ca H. L. Dorwart - O. Ore [DO].

Lun vn gm 3 chng. Chng 1 trnh by mt s kin thc c s v

a thc bt kh quy v s dng a thc bt kh quy chng minh nh


4l Kronecker v s tn ti ca trng phn r ca a thc (nh l 1.3.2)

v nh l ca Galois v s tn ti mt trng c hu hn phn t (nh

l 1.3.5). Chng 2 trnh by mt s phng php xt tnh bt kh quy

ca a thc trn trng Q nh phng php tm nghim hu t, phngphp dng B Gauss, tiu chun Eisenstein v phng php rt gn

theo mun mt s nguyn t. Bng cch s dng nh l Kronecker v

s tn ti trng phn r v nh l Lagrange v cp ca nhm hu hn

(nh l 3.1.7), tnh bt kh quy ca mt s a thc trn trng Zp (vi pl mt s nguyn t) c trnh by trong Chng 3.


Chng 1

a thc bt kh quy

Trc khi trnh by khi nim v mt s kt qu v a thc bt kh quy,

chng ta trnh by kin thc c s v a thc.

1.1 Khi nim a thc

1.1.1 nh ngha. Mt tp F cng vi hai php ton, k hiu l php cng

v php nhn, c gi l trng nu cc tnh cht sau tha mn

(i) Kt hp: a + (b + c) = (a + b) + c v (ab)c = a(bc) vi mi

a, b, c F.(ii) Giao hon: a+ b = b+ a v ab = ba vi mi a, b F.(iii) Lut phn phi: a(b+ c) = ab+ ac vi mi a, b, c F.(iv) Tn ti phn t n v 1 F sao cho a1 = 1a = a vi mi a F.(v) Tn ti phn t 0 F sao cho a+ 0 = 0 + a = a vi mi a F.(vi) Mi a F , tn ti phn t i a F sao cho a+ (a) = 0.(vii) Mi 0 6= a F , tn ti phn t nghch o a1 F sao cho

aa1 = 1.

1.1.2 nh ngha. Cho F l mt trng v a0, a1, . . . , am F . Mt biuthc c dng f(x) = amx

m+am1xm1+. . .+a1x+a0 c gi l mt a

thcmt bin x. Tp cc a thc vi h s trn F c k hiu l F [x]. Nu

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6am 6= 0 th ta ni bc ca f(x) lm v k hiu l deg f(x) = m. H s amc gi l h s cao nht ca f . Nu am = 1 th f(x) c gi l a thc

dng chun (monic polynomial). Hai a thc l bng nhau nu n c cng

bc v cc h s tng ng l bng nhau. Vi hai a thc f(x) =aix

i

v g(x) =bix

i, ta nh ngha tng f(x) + g(x) =

(ai + bi)x

iv tch

f(x)g(x) =ckx

k, trong ck =

i+j=k aibj.

T nh ngha trn ta c ngay cc tnh cht sau y.

1.1.3 B . Cho f(x), g(x), h(x) F [x]. Khi (i) deg(f(x) + g(x)) 6 max{deg f(x), deg g(x)}.(ii) Nu f(x) 6= 0 v g(x) 6= 0 th f(x)g(x) 6= 0 v

deg(f(x)g(x)) = deg f(x) + deg g(x).

(iii) Nu f(x) 6= 0 v f(x)g(x) = f(x)h(x) th g(x) = h(x).

1.1.4 nh ngha. Cho f(x), g(x) F [x]. Nu f(x) = q(x)g(x) viq(x) F [x] th ta ni rng g(x) l c ca f(x) hay f(x) l bi ca g(x)v ta vit g(x)|f(x). Tp cc bi ca g(x) c k hiu l (g).

Ta c ngay cc tnh cht n gin sau y.

1.1.5 B . Cc pht biu sau l ng.

(i) Vi c F v k l s t nhin ta c (x c)|(xk ck).(ii) Nu f(x) F [x] v c F th tn ti q(x) F [x] sao cho

f(x) = q(x)(x c) + f(c).

1.1.6 nh ngha. Cho f(x) = amxm + . . . + a0 F [x]. Gi s K lmt trng cha F . Mt phn t c K c gi l nghim ca f(x)nu f(c) = amc

m + . . . + a0 = 0. Trong trng hp ny ta cng ni c l

nghim ca phng trnh f(x) = 0.


71.1.7 B . Cho f(x) F [x] v c F. Khi (i) c l nghim ca f(x) nu v ch nu f(x) l bi ca x c.(ii) S nghim ca f(x) khng vt qu deg f(x).

1.1.8 Mnh . (Thut ton chia vi d). Cho f(x), g(x) F [x] vig(x) 6= 0. Khi tn ti duy nht cp a thc q(x), r(x) F [x] sao cho

f(x) = q(x)g(x) + r(x)

trong r(x) = 0 hoc deg r(x) < deg g(x).

1.1.9 nh ngha. Mt tp con I 6= ca F [x] c gi l mt ian caF [x] nu n tha mn cc iu kin sau

(i) Nu f(x), g(x) I th f(x) + g(x) I;(ii) Nu f(x) I v q(x) F [x] th q(x)f(x) I .

Ch rng tp con I 6= ca F [x] l ian nu v ch nu f g Iv fh I vi mi f(x), g(x) I v h(x) F [x].

1.1.10 Mnh . Nu I 6= {0} l mt ian trong F [x] v d(x) 6= 0 l athc c bc b nht trong I th

I = (d) = {d(x)q(x) | q(x) F [x]}.

Chng minh. Cho a thc f(x) I. Vit f(x) = d(x)q(x) + r(x) trong r(x) = 0 hoc deg r(x) < deg d(x). V f(x), d(x) I nn ta cr(x) = f(x) d(x)q(x) I . Do r(x) = 0 theo cch chn d(x). Suyra f(x) = d(x)q(x). Ngc li, v d(x) I nn d(x)q(x) I vi miq(x) F [x].

1.1.11 nh ngha. Mt a thc dng chun d(x) F [x] c gi l cchung ln nht ca f(x), g(x) F [x] nu d(x)|f(x), d(x)|g(x) v nuh(x)|f(x) v h(x)|g(x) th h(x)|d(x). Ta k hiu c chung ln nht ca


8f(x) v g(x) l gcd(f(x), g(x)). Nu gcd(f(x), g(x)) = 1 th ta ni f(x)

v g(x) l nguyn t cng nhau.

T Mnh 1.1.10 ta c kt qu sau.

1.1.12 Mnh . Nu f(x), g(x) l hai a thc khng ng thi bng 0

th gcd(f(x), g(x)) lun tn ti v l t hp tuyn tnh ca f(x) v g(x),

tc l tn ti a(x), b(x) F [x] sao chogcd(f(x), g(x)) = a(x)f(x) + b(x)g(x).

1.1.13 H qu. Cho p(x), f(x), g(x) F [x]. Nu gcd(p(x), f(x)) = 1v p(x)|f(x)g(x) th p(x)|g(x).Chng minh. Theo gi thit, 1 = p(x)a(x) + f(x)b(x). Suy ra

g(x) = p(x)a(x)g(x) + f(x)b(x)g(x).

Do p(x) l c ca a thc v phi nn p(x)|g(x).Vi 0 6= g(x) F [x], k hiu g(x) = g(x)/an trong an l h scao nht ca g(x). Ch rng g(x) l a thc dng chun. tm c

chung ln nht ta c thut ton sau:

1.1.14 Mnh . (Thut ton Euclid tm c chung ln nht). Cho hai a

thc f(x), g(x) F [x] vi g(x) 6= 0. Nu g(x)|f(x) thgcd(f(x), g(x)) = g(x).

Nu ngc li, chia lin tip ta c

f(x) = q(x)g(x) + r(x), r(x) 6= 0, deg r(x) < deg g(x).g(x) = q1(x)r(x) + r1(x), r1(x) 6= 0, deg r1(x) < deg r(x).

. . . . . . . . .

rn2(x) = qn(x)rn1(x) + rn(x), rn(x) 6= 0, deg rn(x) < deg rn1(x).rn1(x) = qn+1(x)rn(x).


9Khi gcd(f(x), g(x)) = rn(x).

Chng minh. T ng thc cui ta c rn(x)|rn1(x). Thay vo ng thcth hai t di ln ta c rn(x)|rn2(x). C tip tc lp lun vi cc ngthc t di ln trn ta suy ra rn(x)|g(x) v rn(x)|f(x). Do rn(x)|f(x)v rn(x)|g(x). Gi s h(x)|f(x) v h(x)|g(x). T ng thc u tinta c h(x)|r(x). T ng thc th hai ta c h(x)|r1(x). C tip tc lplun trn vi cc ng thc t trn xung di ta c h(x)|rn(x). Do h(x)|rn(x).

1.2 a thc bt kh quy

1.2.1 nh ngha. Mt a thc f(x) F [x] c gi l bt kh quy nudeg f(x) > 0 v f(x) khng phn tch c thnh tch ca hai a thc c

bc b hn. Nu deg f(x) > 0 v f(x) l tch ca hai a thc c bc b

hn th ta ni f(x) l kh quy.

Sau y l mt s v d v a thc bt kh quy.

1.2.2 B . Cc pht biu sau l ng.

(i) a thc bc nht lun bt kh quy.

(ii) Nu f(x) bc ln hn 1 v c nghim trong F th f(x) kh quy.

(iii) a thc bc 2 v bc 3 l bt kh quy nu v ch nu n khng c

nghim trong F.

(iv) a thc f(x) c bc dng l bt kh quy nu v ch nu f(x+a)

l bt kh quy vi mi a F.Chng minh. (i) R rng a thc bc nht khng th l tch ca hai a

thc bc thp hn.

(ii) Nu deg f(x) > 1 v f(x) c nghim x = a F th f = (x a)gtrong deg g = deg f 1 1. V th f kh quy.


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(iii) Cho f(x) c bc 2 hoc 3. Nu f kh quy th n phn tch c

thnh tch ca hai a thc bc thp hn, mt trong hai a thc phi c

bc 1, do f(x) c nghim trong F . Nu f(x) c nghim trong F th

theo (ii), f(x) l kh quy.

(iv) Cho a thc f(x) F [x] c bc dng v a F. Vi mi h F ,t h1(x) = h(x a). Ch rng deg h1(x) = deg h(x) vi mi h F .V th f(x + a) = k(x)g(x) l phn tch ca f(x + a) thnh hai a thc

c bc thp hn khi v ch khi f(x) = k1(x)g1(x) l phn tch ca f(x)

thnh tch ca hai a thc c bc thp hn. V vy f(x) kh quy khi v

ch khi f(x+ a) kh quy.

Tip theo, chng ta nh ngha khi nim a thc bt kh quy ca mt

phn t cha F trong mt trng. Trc ht ta cn kt qu sau.

1.2.3 nh ngha. Cho K l mt trng cha F v a K. Ta ni a lphn t i s trn F nu tn ti mt a thc 0 6= f(x) F [x] nhn alm nghim. Nu a khng i s trn F th ta ni a l siu vit trn F .

1.2.4 Mnh . Cho K l mt trng cha F v a K l phn t is trn F . Khi tn ti duy nht mt a thc p(x) F [x] bt kh quydng chun nhn a lm nghim, v mi a thc g(x) F [x] nhn a lmnghim u l bi ca p(x).

Chng minh. V a l nghim ca mt a thc khc 0 vi h s trong F

nn tn ti a thc khc 0 vi h s trong F c bc b nht nhn a lm

nghim. Gi p(x) F [x] l dng chun ca a thc ny. Khi a lnghim ca p(x). Ta chng minh p(x) bt kh quy. Gi s p(x) khng

bt kh quy. Khi p(x) phn tch c thnh tch ca hai a thc trong

F [x] vi bc b hn, v do mt trong hai a thc ny phi nhn a lm

nghim, iu ny l mu thun vi cch chn p(x). Gi s g(x) F [x]


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nhn a lm nghim. Nu p(x) khng l c ca g(x) th v p(x) bt

kh quy nn gcd(g(x), p(x)) = 1, do 1 = p(x)q(x) + g(x)h(x) vi

q(x), h(x) F [x]. Thay x = a vo c hai v ta c 1 = 0, iu ny l vl. Vy g(x) chia ht cho p(x). Gi s q(x) F [x] cng l a thc bt khquy dng chun nhn a lm nghim. Theo chng minh trn, q(x) l bi

ca p(x). Vit q(x) = p(x)k(x). V q(x) bt kh quy nn k(x) = b F.Do q(x) = bp(x). ng nht h s cao nht ca hai v vi ch rng

q(x) v p(x) u c dng chun, ta suy ra b = 1. V th p(x) = q(x).

1.2.5 nh ngha. a thc p(x) F [x] bt kh quy dng chun xc nhnh trong mnh trn c gi l a thc bt kh quy ca a.

1.2.6 V d. a thc x3 2 Q[x] l a thc bt kh quy ca 32 R;a thc x2 + 1 R[x] l a thc bt kh quy ca i C.

a thc bt kh quy c tnh cht tng t nh tnh cht ca s nguyn

t. Trc ht, chng ta bit, B Euclid pht biu rng s t nhin

p > 1 l s nguyn t nu v ch nu p|ab ko theo p|a hoc p|b vi mis t nhin a, b. Mnh sau y l iu tng t cho a thc bt kh

quy.

1.2.7Mnh . Nu p(x) F [x] bt kh quy v p(x)|a(x)b(x) th p(x)|a(x)hoc p(x)|b(x) vi mi a(x), b(x) F [x]. c bit, mt a thc bt khquy l c ca mt tch hu hn a thc th n phi l c ca t nht

mt trong cc a thc .

Chng minh. Cho p(x)|a(x)b(x). Gi s p(x) khng l c ca a(x) vcng khng l c ca b(x). Khi gcd(p(x), a(x)) = 1. Do tn ti

s(x), r(x) F [x] sao cho 1 = s(x)p(x) + r(x)a(x). Tng t, tn tie(x), f(x) F [x] sao cho 1 = e(x)p(x) + f(x)b(x). Nhn v vi v ca


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hai ng thc ny ta c

1 = p(x)g(x) + r(x)f(x)a(x)b(x)

vi g(x) F [x]. a thc bn v phi ca ng thc trn l bi ca p(x),trong khi a thc bn v tri l 1 khng chia ht cho p(x). iu ny l

v l.

Tip theo, nh l c bn ca S hc ni rng mi s t nhin ln hn

1 u phn tch c thnh tch cc tha s nguyn t v s phn tch ny

l duy nht nu khng k n th t cc tha s. Kt qu sau y l mt

s tng t ca nh l ny i vi a thc.

1.2.8 nh l. Mi a thc dng chun bc dng c th phn tch c

thnh tch cc a thc bt kh quy dng chun v s phn tch ny l duy

nht nu khng k n th t cc nhn t.

Chng minh. Trc ht, chng ta chng minh s tn ti phn tch bng

quy np theo bc ca a thc. Gi s f(x) F [x] l a thc dng chunbc d > 0. Nu d = 1 th f(x) l bt kh quy nn s phn tch bt kh

quy ca f(x) l f(x) = f(x), kt qu ng cho trng hp d = 1. Cho

d > 1 v gi s kt qu ng cho cc a thc bc nh hn d. Nu

f(x) bt kh quy th f(x) c s phn tch bt kh quy l f(x) = f(x).

V th ta gi thit f(x) khng bt kh quy. Khi f(x) = g(x)h(x) vi

deg g(x), deg h(x) < deg f(x). t g(x) = g(x)/ak vi ak l h s cao

nht ca g(x). Khi ta c f(x) = g(x)(akh(x)). ng nht h s cao

nht hai v ta c 1 = akbt, trong bt l h s cao nht ca h(x).

t h(x) = akh(x). Khi f(x) = g(x)h(x) vi g(x), h(x) l cc

a thc dng chun c bc nh hn d. Theo gi thit quy np, g(x) v

h(x) phn tch c thnh tch cc a thc bt kh quy dng chun. V


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th f(x) phn tch c thnh tch ca hu hn a thc bt kh quy dng

chun.

By gi ta chng minh tnh duy nht ca phn tch. Gi s f(x) c hai

s phn tch thnh nhn t bt kh quy dng chun

f(x) = p1(x)p2(x) . . . pn(x) = q1(x)q2(x) . . . qm(x).

Ta chng minh bng quy np theo n rng n = m v sau mt php hon

v ta c pi(x) = qi(x) vi mi i = 1, . . . , n. Cho n = 1. Khi ta c

p1(x) = q1(x)q2(x) . . . qm(x). Suy ra p1(x)|q1(x)q2(x) . . . qm(x). Do p1(x)l bt kh quy nn p1(x) l c ca mt nhn t qi(x) no , khng mt

tnh tng qut ta c th gi thit p1(x)|q1(x). Biu din q1(x) = p1(x)t1(x).V q1(x) bt kh quy nn t1(x) = a F . ng nht h s cao nht cahai v ca ng thc q1(x) = ap1(x) vi ch rng p1(x) v q1(x) l dng

chun, ta c 1 = 1.a. Suy ra a = 1 v do p1(x) = q1(x). Num > 1 th

1 = q2(x) . . . qm(x), iu ny l v l. Vy, kt qu ng cho n = 1. Cho

n > 1. V p1(x)|q1(x)q2(x) . . . qm(x) v p1(x) l bt kh quy nn khngmt tnh tng qut ta c th gi thit p1(x)|q1(x). Li do q1(x) l bt khquy v p1(x), q1(x) u c dng chun nn tng t nh lp lun trn ta

c p1(x) = q1(x). Gin c c hai v cho p1(x) ta c

p2(x)p3(x) . . . pn(x) = q2(x)q3(x) . . . qm(x).

Theo gi thit quy np ta c n 1 = m 1 v bng vic nh s li ccnhn t qi(x) ta suy ra pi(x) = qi(x) vi mi i = 2, . . . , n.

1.3 Trng phn r ca a thc

Trong tit ny, da vo tnh cht bt kh quy, chng ta ch ra rng vi mi

a thc f(x) F [x], tn ti mt trng K ti thiu cha trng F v


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cha tt c cc nghim ca f(x). Trng K c tnh cht trn c gi l

trng phn r ca a thc f(x) trn F .

1.3.1 nh ngha. Cho F v F l hai trng.

(i) Mt tp con T ca F c gi l trng con ca F nu x1 Tvi mi 0 6= x T v x+y, xy,1 T vi mi x, y T. Ch rng tpcon T ca F l trng con ca F nu php cng v nhn ng kn trong

T v T l mt trng vi hai php ton ny.

(ii) Mt nh x : F F c gi l mt ng cu nu (1) = 1,(x+ y) = (x) + (y) v (xy) = (x)(y) vi mi x, y F .(iii) Mt ng cu : F F c gi l n cu nu l n nh.Trong trng hp ny (F ) l mt trng con ca F . V th ta ni F

nhng c vo F v ta cng c th coi F l mt trng cha F .

(iv) Mt ng cu : F F c gi l ton cu nu l ton nh.(v) Mt ng cu : F F c gi l ng cu nu l song nh.Trong trng hp ny ta ni F v F l ng cu vi nhau v ta c th

ng nht hai trng F v F vi nhau.

1.3.2 nh l. (Kronecker). Cho f(x) F [x] l mt a thc c bc dng.Khi tn ti mt trng ti thiu cha F v cha tt c cc nghim ca

f(x). c bit, mi a thc trn mt trng u c trng phn r.

Chng minh. K hiu f (x) l a thc dng chun ca f(x). V cc nghim

ca f(x) cng l cc nghim ca f (x) nn ta c th gi thit f(x) c dng

chun. Ta chng minh nh l bng quy np theo deg f(x) = n. Gi s

n = 1. Khi f(x) = xa vi a F. Do a l nghim duy nht ca f(x)nn ta ch vic chn K = F. Gi thit rng n > 1 v nh l ng cho

trng hp a thc bc nh hn n. Trc ht ta chng minh cho trng

hp f(x) bt kh quy. t

I = (f) = {g(x)f(x) | g(x) F [x]}.


15

D kim tra c I l mt ian ca F [x]. Vi mi g(x) F [x] ta t

g(x) + I = {g(x) + h(x) | h(x) I}.

Ta c th ch ra rng g(x) + I = h(x) + I nu v ch nu g(x)h(x) I.t K = {g(x) + I | g(x) F [x]}. Trc ht ta kim tra quy tc cng

(g(x) + I) + (h(x) + I) = (g(x) + h(x)) + I

l mt php ton trn K. Tht vy, nu g + I = g1 + I v h+ I = h1 + I

th g g1 I v h h1 I. Do g g1 v h h1 l bi ca f.Suy ra (g + h) (g1 + h1) = (g g1) + (h h1) l bi ca f . V th(g + h) (g1 + h1) I hay (g + h) + I = (g1 + h1) + I. Suy ra quy tccng trn l mt php ton trn K. Hon ton tng t, ta c th ch ra

rng quy tc nhn

(g + I)(h+ I) = gh+ I

l mt php ton trn K. D thy php cng v php nhn trn K c

tnh cht kt hp, giao hon; Php nhn phn phi vi php cng; Phn t

khng caK l 0+I; Phn t n v caK l 1+I; Phn t i xng ca

g + I K l g + I K. Ta chng minh mi phn t khc 0 + I Ku c nghch o. Ly g + I K vi g + I 6= 0 + I. Khi g / I . Do g khng l bi ca f . V f bt kh quy nn gcd(f, g) = 1. V th ta

c biu din 1 = f(x)p(x) + g(x)q(x) vi p(x), q(x) F [x]. Ch rngfp I . Do fp+ I = 0 + I . Suy ra

1 + I = (fp+ gq) + I = (fp+ I) + (gq+ I) = gq+ I = (g+ I)(q+ I).

Do g + I kh nghch trong K. Vy K lm thnh mt trng vi php

cng v nhn trn. Xt nh x : F K cho bi (a) = a + I. Rrng l mt ng cu. Nu (a) = (b) vi a, b F th a+ I = b+ I.V th a b I. Suy ra a b l bi ca f(x). Nu a b 6= 0 th a b


16

l a thc c bc 0 nn n khng th l bi ca a thc f(x) bc dng,

iu ny l v l. Do a b = 0. Suy ra a = b. V vy l n cu,do ta c th xem K l mt trng cha F . t = x + I K. Gis f(x) = xn + an1xn1 + . . .+ a1x+ a0. ng nht phn t a F viphn t a+ I K, khi trong trng K ta c

f() = (x+ I)n + (an1 + I)(x+ I)n1 + . . .+ (a0 + I)

= (xn + I) + (an1xn1 + I) + . . .+ (a0 + I)

= (xn + an1xn1 + . . .+ a0) + I

= f(x) + I = 0 + I.

V vy l mt nghim ca a thc f(x) trong trng K. Do tn ti

f1(x) K[x] sao cho f(x) = (x )f1(x), trong deg f1(x) = n 1.Theo gi thit quy np, tn ti mt trng K1 cha K v cha tt c cc

nghim ca f1(x). Do K1 cha F v cha tt c cc nghim ca f(x).

Gi K l giao ca tt c cc trng con ca K1 cha F v cha tt c

cc nghim ca f(x). Khi K l trng ti thiu cha F v cha cc

nghim ca f(x).

Tip theo, ta chng minh cho trng hp f(x) kh quy. Trong trng

hp ny, tn ti hai a thc dng chun bc dng g(x), h(x) F [x] saocho f(x) = g(x)h(x) v deg g, deg h < n = deg f. Theo gi thit quy

np, tn ti mt trng K cha F v cha tt c cc nghim ca g(x). Ta

coi h(x) l a thc vi h s trong K. Theo gi thit quy np, tn ti mt

trng cha K1 v cha tt c cc nghim ca h(x). V th K1 l trng

cha F v cha cc nghim ca f(x). Ly giao ca tt c cc trng con

ca K1 cha F v cha cc nghim ca f(x), ta c trng ti thiu

cha F v cc nghim ca f(x).

1.3.3 V d. K hiu Q(i

3, 3

2) l giao ca cc trng con ca C cha


17

Q v cha cc phn t i

3, 3

2. Khi trng phn r ca f(x) = x3 2trn Q l Q(i

3, 3

2). Tht vy, d thy 3 nghim ca x3 2 l

x1 =3

2, x2 =3

2(12

+i

3

2), x3 =

3

2(12 i

3

2).

Do x1, x2, x3 Q(i

3, 3

2). Ngc li, trng b nht cha Q v ccnghim x1, x2, x3 phi cha i

3 v 3

2.

S dng nh l 1.3.2 v s tn ti trng phn r ca a thc, chng

ta c th ch ra s phn t ca mt trng hu hn v s tn ti mt trng

c hu hn phn t. Ch n gin sau y l rt c ch. Nu T l mt

trng con ca trng F th F c cu trc l khng gian vc t trn T vi

php cng l php cng ca F v tch v hng l php nhn cc phn t

ca T vi cc phn t ca F.

1.3.4 Mnh . Nu F l mt trng hu hn th s phn t ca F l ly

tha ca mt s nguyn t.

Chng minh. Vi mi s t nhin n, k hiu n1 = 1+ . . .+1 l tng ca n

phn t 1 v (n)1 l tng ca n phn t 1. Quy c 01 = 0. Ta khngnh rng tn ti mt s nguyn dng k sao cho k1 = 0. Gi s ngc

li, khi tng ng : Q F cho bi (n/m) = (n1)(m1)1 l ncu trng. Do Q c v hn phn t nn F c v hn phn t, iu nyl v l. Vy, tn ti s nguyn dng k k1 = 0. Gi p l s nguyn

dng b nht c tnh cht p1 = 0. Ta chng minh p l s nguyn t. Tht

vy, v p1 = 1 6= 0 nn p > 1. Nu p khng nguyn t th p = nm trong 1 < n,m < p. Suy ra 0 = p1 = (n1)(m1). Do n,m < p nn n1 6= 0v m1 6= 0. Do n1 v m1 kh nghch. Gi a, b F ln lt l nghcho ca n1 v m1. Khi

0 = 0(ab) = (n1)a(m1)b = 1.1 = 1,


18

iu ny l v l. Vy p l s nguyn t. Tip theo ta khng nh m1 = 0

nu v ch nu m pZ. Tht vy, nu m pZ th r rng m1 = 0. Num1 = 0 th ta vit m = ps + r vi s, r Z v 0 6 r < p, v do 0 = m1 = ps1 + r1 = r1. Do r < p nn r = 0 theo cch chn p, khng

nh c chng minh. t

T = {(n1)(m1)1 | n Z,m / pZ} F.

D kim tra c T l trng con ca F . Ta chng minh T c ng p

phn t. Tht vy, vi mi m / pZ, do p nguyn t nn gcd(m, p) = 1.Suy ra 1 = ms+ pt vi s, t Z. Suy ra

1 = 1.1 = (m1)(s1) + pt1 = (m1)(s1).

Do (m1)1 = s1 vi s Z. Suy ra T = {n1 | n Z}. Vi min Z, vit n = pt+ r vi 0 6 r < p ta c n1 = pt1 + r1 = r1. Do T = {n1 | 0 6 n < p}. Nu 0 6 n, n < p v n1 = n1 th (n n)1 = 0v do n n l bi ca p, do n = n. Suy ra T l trng con caF c ng p phn t. Ch rng F l khng gian vc t trn T . Gi q l

s phn t ca F v d = dimT F l chiu ca Tkhng gian vc t F .Gi {e1, . . . , ed} l mt c s ca F. Khi mi phn t x F c biudin mt cch duy nht di dng x = a1e1 + . . .+ anen, trong ai Tvi mi i = 1, . . . , n. V th, s phn t ca F l q = pd.

1.3.5 nh l. (Galois). Vi mi s nguyn t p v mi s nguyn dng

d, tn ti mt trng c ng pd phn t.

Chng minh. t q = pd. Do p nguyn t nn Zp l mt trng. Theonh l Kronecker, tn ti mt trng K cha Zp v cha cc nghimca a thc xq x. t E = { K | g() = 0}, tc l E l tp ttc cc nghim ca g(x) trong K. K hiu g(x) = qxq1 1 l o hm


19

ca g(x). V 1 Zp nn ta c p1 = 0. Do q1 = pd1 = 0. Suy rag(x) = (q1)xq1 1 = 1. Do c chung ln nht ca g(x) v g(x)bng 1. Suy ra g(x) khng c nghim bi trong K. iu ny cng c

ngha l E c ng q phn t. Nh vy, nh l c chng minh nu ta

ch ra E l mt trng. Nu p l th q l v do g(1) = 1 + 1 = 0v v th 1 E. Nu p chn th v p nguyn t nn p = 2. Suy ra

g(1) = (1)q + 1 = 1 + 1 = 2.1 = p.1 = 0.

Do 1 E. Vy trong mi trng hp ta u c 1 E. Choa, b E. Khi g(a) = g(b) = 0. Suy ra aq = a v bq = b. Suy ra(ab)q = ab. Do g(ab) = 0 v v th ab E. Ch rng

(a+ b)q = aq + C1qaq1b+ . . .+ Cq1q ab

q1 + bq = a+ b,

trong Ckq l s t hp chp k ca q phn t. V q = pdv p nguyn

t nn bng quy np ta d dng kim tra c Ckq l bi ca p vi mi

k = 1, . . . , q 1. Do ta c Ckq aqkbk = (Ckq 1)aqkbk = 0 vi mik = 1, . . . , q 1. V th (a+ b)q = aq + bq. Theo trn ta c aq = a vbq = b. Do (a+ b)q (a+ b) = 0, v v th a+ b E. Cho 0 6= a EKhi aq = a. Do a 6= 0 nn nhn c hai v vi a1 ta c aq1 = 1.Suy ra aq2 l nghch o ca a trong E. Vy E l mt trng.

T nh l trn ta suy ra cc tnh cht sau.

1.3.6 V d. Tn ti trng c 81 phn t. Khng tn ti trng c 100

phn t.


Chng 2

Mt s phng php xt tnh bt kh

quy trn Q

nh l c bn ca i s pht biu rng mi a thc bc dng vi h

s phc lun c t nht mt nghim phc. Ch rng nu f(x) C[x] cnghim x = C th f(x) = (x )g(x) vi g(x) C[x]. V th nudeg f(x) 2 th f(x) c th phn tch c thnh tch ca hai a thc cbc thp hn. Do cc a thc bt kh quy trn C l v ch l cc athc bc nht. Gi s f(x) R[x]. Ch rng nu s phc = a + bil nghim ca f(x) vi / R th s phc lin hp = a bi cng lnghim ca f(x). V th f(x) chia ht cho (xabi)(xa+bi). R rng(xa bi)(xa+ bi) = x22ax+a2 + b2 R[x] v x22ax+a2 + b2khng c nghim thc. Do cc a thc bt kh quy trn R l v ch lcc a thc bc nht hoc a thc bc hai khng c nghim thc.

Mc d bi ton xt tnh bt kh quy ca cc a thc trn C v trn R c gii quyt t khi ngi ta chng minh c nh l c bn ca

i s (chng minh hon chnh u tin cho nh l c bn ca i s

c a ra bi Gauss nm 1816), nhng bi ton xt tnh bt kh quy ca

cc a thc trn Q cho n nay vn l bi ton m. Mc tiu ca chngny nhm trnh by mt s tiu chun mt a thc l bt kh quy trn

Q.

20


21

Gi s f(x) Q[x]. Ch rng f(x) l bt kh quy trn Q khi vch khi af(x) l bt kh quy vi mi 0 6= a Z. V th, bng vicnhn vi mu s chung ca cc h s ca f(x), ta c mt a thc vi

h s nguyn m tnh bt kh quy trn Q ca n tng ng vi tnhbt kh quy ca f(x). Do ta ch cn xt tnh bt kh quy trn Q chocc a thc vi h s nguyn. T nay n ht chng ny, lun gi thit

f(x) = anxn + . . .+ a1x+ a0 Z[x], trong an 6= 0 v n > 0.

2.1 Nghim hu t v tnh bt kh quy trn Q

Nh trnh by trong chng trc, iu kin mt a thc bc ln

hn 1 kh quy trn Q l n c nghim trong Q. V vy, xt tnh khquy ca a thc, trong nhiu trng hp ta c th dng phng php tm

nghim hu t. Trc ht chng ta a ra cch tm nghim hu t cho cc

a thc vi h s nguyn.

2.1.1 Mnh . Nu

r

sl phn s ti gin v l nghim hu t ca f(x)

th r l c ca a0 v s l c ca an.

Chng minh. Gi s

r

s Q trong r, s l cc s nguyn, s > 0 v

(r, s) = 1. Nur

sl nghim ca a thc f(x) th f(

r

s) = 0. Ta c

0 = f(r

s) = an(

r

s)n + an1(

r

s)n1 + . . .+ a1

r

s+ a0.

Suy ra 0 = anrn + an1rn1s+ . . .+ a1rsn1 + a0sn. V th ta c

anrn = (an1rn1s+ . . .+ a1rsn1 + a0sn).

V phi ca a thc ny l bi ca s. V (r, s) = 1 nn s l c ca an

Tng t ta c a0sn = (anrn + an1rn1s+ . . .+ a1rsn1). V phi caa thc ny l bi ca r. V (r, s) = 1 nn r l c ca a0.


22

2.1.2 H qu. Nghim hu t (nu c) ca f(x) = xn + . . .+ a1x+ a0 l

nghim nguyn v s nguyn ny l c ca s hng t do.

2.1.3 Mnh . Nu phn s ti gin

r

sl nghim ca f(x) th r msl c ca f(m) vi m l s nguyn bt k. c bit, (r + s) l c ca

f(1) v (r s) l c ca f(1).

Chng minh. Phn tch f(x) theo cc lu tha ca xm ta c

f(x) = an(xm)n + bn1(xm)n1 + . . .+ b1(xm) + b0.

Cc h s b0, b1, . . . , bn1 Z v m Z. Ta c f(m) = b0 v cho x = rsta c f(

r

s) = 0. Ch rng

0 = f(r

s) = an(

r

sm)n + bn1(r

sm)n1 + . . .+ b1(r

sm) + f(m).

T ta c

0 = an(rms)n + bn1(rms)n1s+ . . .+ b1(rms)sn1 + f(m)sn.

Suy ra

f(m)sn = {an(rms)n + bn1(rms)n1s+ . . .+ b1(rms)sn1}.

V phi ca a thc l bi ca r ms. Do f(m)sn l bi ca r mshay r ms l c ca f(m).Trng hp c bit, vi m = 1 th r s l c f(1), m = 1 th r + sl c f(1).

Sau y l mt s v d minh ha v vic xt tnh kh quy ca a thc

bng phng php tm nghim hu t.

2.1.4 V d. Xt tnh bt kh quy trn Q ca cc a thc sau.(i) 10x3 + 3x2 106x+ 21;


23

(ii) 9x3 + 6x2 8x+ 7;(iii) x3 x2 + x 6;(iv) 6x4 + 19x3 7x2 26x+ 12.

Gii. (i) Gi s a thc f(x) = 10x3 + 3x2 106x + 21 c nghim hut

r

s, vi

r

sl phn s ti gin. Theo Mnh 2.1.1, r|21 v s|10. T suy ra r = 1,3,7,21 v s = 1,2,5,10. Ta tnh cf(1) = 72, f(1) = 120. Vy cc s hu t tho mn Mnh 2.1.3l 1

2,1

5,3, 3

2,3

5,7,7

2,7

5. Th li ta thy ch c

1

5, 3,7

2l

nghim ca f(x). Vy f(x) c nghim hu t, do n kh quy trn Q.(ii) xt tnh bt kh quy ca a thc f(x) = 9x3+6x28x+7 ta nhnhai v ca f(x) vi 3 ta c f(3x) = 27x3+18x224x+21. t y = 3xta c f(y) = y3+2y28y+21. Gi s f(y) c nghim hu t. V an = 1nn theo H qu 2.1.2, nghim hu t ca f(y) phi l nghim nguyn

v r|21. Suy ra r = 1,3,7,21. Ta thy f(1) = 16, f(1) = 30.Vy ch c cc s 3,7 tho mn Mnh 2.1.3 nhng th li ta thyf(3) 6= 0, f(7) 6= 0. Vy a thc f(y) khng c nghim hu t. Suyra f(x) bt kh quy trn Q.(iii) Gi s a thc f(x) = x3x2+x6 c nghim hu t v nghim phi l nghim nguyn v an = 1. Cc c ca 6 l r = 1,2,3,6.Ta thy f(1) = 5 v f(1) = 9 nn ch c r = 2 tho mn Mnh 2.1.3. Th li ta thy rng f(2) = 0. Vy a thc f(x) c nghim hu t,

do n kh quy.

(iv) Gi s a thc f(x) = 6x4 +19x37x226x+12 c nghim hut

r

s, vi

r

sl phn s ti gin. Theo Mnh 2.1.1 th r|12 v s|6. Suyra r = 1,2,3,4,6,12 v s = 1,2,3,6. Ta tnh cf(1) = 4 v f(1) = 18. Do cc s hu t tho mn Mnh 2.1.3l

1

2,1

3, 2,3. Th li ta thy f(1

2) = 0 v f(3) = 0. Vy f(x) c


24

nghim hu t, do n kh quy trn Q.

2.1.5 Ch . i vi a thc bc 4, ta khng th suy ra tnh bt khquy trn Q t vic kim tra a thc khng c nghim hu t. Tht vy, athc (x2 + 1)(x2 + 1) khng c nghim hu t, nhng n khng bt kh

quy.

2.2 Phng php dng B Gauss

Trong tit ny, chng ta trnh by mt tiu chun bt kh quy trn Q thngqua tiu chun khng phn tch c trn Z[x].

2.2.1 nh l. (B Gauss). Cho p(x) Z[x]. Nu p(x) = g(x)f(x) ls phn tch p(x) thnh tch ca hai a thc g(x), f(x) vi h s trong Qth p(x) cng phn tch c thnh tch ca hai a thc g(x), f(x) vi

h s trong Z vi deg g(x) = deg g(x), deg f(x) = deg f(x). c bit,nu p(x) l kh quy trn Q th n phn tch c thnh tch ca hai athc vi h s nguyn c bc thp hn.

chng minh nh l trn, chng ta cn nhc li khi nim a thc

nguyn bn v mt tnh cht ca a thc nguyn bn.

2.2.2 nh ngha. a thc f(x) Z[x] c gi l nguyn bn nu cchung ln nht ca cc h s ca f(x) l 1.

2.2.3 B . Tch ca hai a thc nguyn bn l a thc nguyn bn.

Chng minh. Gi s f(x) = g(x)h(x), trong

g(x) = bnxn + . . .+ b1x+ b0

h(x) = ckxk + . . .+ c1x+ c0.


25

l cc a thc nguyn bn. Vit f(x) = amxm + . . . + a1x + a0 Z[x].Nu f(x) khng nguyn bn th tn ti mt s nguyn t p sao cho n l

c ca mi h s ca f . V h(x), g(x) l nguyn bn nn tn ti ch s

b nht s sao cho bs khng l bi ca p v tn ti ch s t b nht sao cho

ct khng l bi ca p. Ch rng

as+t = bs+tc0 + bs+t1c1 + . . .+ bsct + bs1ct+1 + . . .+ b0ct+s.

Theo cch chn s v t ta c bi, cj l bi ca p vi mi i > s v j > t. V

as+t l bi ca p nn ta suy ra bsct l bi ca p, iu ny l v l vi cch

chn bs v ct.

By gi ta c th chng minh nh l 2.2.1.

Chng minh nh l 2.2.1. Gi s p(x) Z[x] v p(x) = f(x)g(x) vif, g Q[x]. Ta c th vit f = af1 v g = bg1 trong a, b Q vf1, g1 Z[x] l cc a thc nguyn bn. Theo B 2.2.3, f1g1 Z[x]l a thc nguyn bn. R rng p = abf1g1 Z[x]. Ta chng minhab Z. Tht vy, gi s ab / Z. Khi ab = r

svi r, s Z, s > 1 v

gcd(r, s) = 1. Vit f1g1 = anxn + . . .+ a1x+ a0. V f1g1 l nguyn bn

nn gcd(an, an1, . . . , a1, a0) = 1. V p(x) Z[x] nnrans, . . . ,

ra1s,ra0s Z.

Suy ra s l c chung ca an, . . . , a1, a0, iu ny l v l. Vy, ab Z.Do p = (abf1)g1. t f = abf1 v g = g1. Khi p = fg l s

phn tch ca p thnh tch ca hai a thc vi h s nguyn f v g vi

deg f = deg f v deg g = deg g.

Di y l mt s v d minh ha vic xt tnh bt kh quy trn Q cacc a thc bng vic s dng B Gauss.


26

2.2.4 V d. Xt tnh bt kh quy ca a thc f(x) = x4 + 3x3 + x2 + 3

trn Q.

Gii. Ta chng minh a thc ny bt kh quy trn Q. Gi s ngc li, khi theo nh l 2.2.1, f(x) c s phn tch thnh tch g(x)h(x) trong

g(x) Z[x] c bc 1 hoc bc 2, h(x) Z[x] c bc 3 hoc bc 2. V hs cao nht ca f(x) l 1 nn ta c th gi thit h s cao nht ca g(x)

v h(x) cng l 1. Nu g(x) bc 1 th theo H qu 2.1.2 f(x) c nghim

nguyn l 3 hoc 3. Tuy nhin, thay vo ta c f(3) 6= 0 v f(3) 6= 0.Do g(x) c bc 2 v v th h(x) c bc 2. Vit g(x) = x2 + ax+ b v

h(x) = x2 + cx+ d vi a, b, c, d Z. ng nht h s hai v ca ngthc f = gh ta c

bd = 3, bc+ ad = 0, ac+ d+ b = 1, c+ a = 3.

V bd = 3 v vai tr ca b, d l nh nhau nn khng mt tnh tng qut ta

c th gi thit b = 1, d = 3 hoc b = 1, d = 3. Nu b = 1, d = 3 thc + 3a = 0, ac = 3, a + c = 3. Suy ra a = 3

2/ Z, v l. Nu b = 1v d = 3 th c 3a = 0, ac = 5, c+ a = 3. Suy ra a = 3

2/ Z, v l.Nh vy, f(x) khng phn tch c thnh tch ca hai a thc bc nh

hn 4 vi h s nguyn. Suy ra f(x) bt kh quy trn Q.

2.2.5 V d. Xt tnh bt kh quy ca a thc f(x) = x5 + x3 + x2 + 5

trn Q.

Gii. Ta s chng minh f(x) l bt kh quy trn Q. Tht vy, gi s f(x)kh quy trnQ. Khi , theo nh l 2.2.1 ta c f(x) = g(x)h(x) trong g(x) Z[x] c bc 1 hoc bc 2, h(x) Z[x] c bc 4 hoc bc 3. TheoH qu 2.1.2, cc nghim hu t ca f(x) ch c th l 5. Tuy nhinf(5) 6= 0 v f(5) 6= 0. Do g(x) khng th c bc 1. Vy g(x) c bc


27

2 v h(x) c bc 3. Vit g(x) = x2 + ax+ b v h(x) = x3 + cx2 + dx+ e

vi a, b, c, d, e Z. ng nht h s hai v ca ng thc f = gh tac

a+ c = 0, b+ d+ ac = 1, bc+ ad+ e = 1, ae+ bd = 0, be = 5.

V be = 5 nn cc kh nng sau xy ra:

Trng hp 1: b = 1, e = 5. Khi

a+ c = 0, d+ ac = 0, c+ ad = 4, 5a+ d = 0.

Suy ra hoc a = 5, c = 5, d = 95/ Z, v l; hoc a = 0, c = 0 thay vo

c+ ad = 4, v l.Trng hp 2: Nu b = 1, e = 5. Khi

a+ c = 0, d+ ac = 2,c+ ad = 6,5a d = 0.

Suy ra a =517

2/ Z, v l.Trng hp 3: Nu b = 5, e = 1. Khi

a+ c = 0, d+ ac = 4, 5c+ ad = 0, a+ 5d = 0.

Suy ra d =1401

50/ Z, v l.Trng hp 4: Nu b = 5, e = 1. Khi

a+ c = 0, d+ ac = 6,5c+ ad = 2, a+ 5d = 0.

Suy ra d =25585

10/ Z, v l.Vy khng c cc s nguyn a, c, d no tho mn cc trng hp trn. V

vy f(x) khng phn tch c thnh tch ca hai a thc bc nh hn 5

vi h s nguyn. Do n bt kh quy trn Q.


28

2.3 Phng php dng tiu chun Eisenstein

2.3.1 nh l. [Tiu chun Eisenstein]. Gi s

f = anxn + an1xn1 + . . .+ a1x+ a0

l a thc vi h s nguyn sao cho c mt s nguyn t p tho mn cc

tnh cht

(i) p khng l c ca h s cao nht an.

(ii) p l c ca cc h s cn li.

(iii) p2 khng l c ca h s t do a0.

Khi f l bt kh quy trn Q.

Chng minh. Gi s f khng bt kh quy trn Q. Theo nh l 2.2.1 tac th biu din

f = gh = (b0 + b1x+ . . .+ bmxm)(c0 + c1x+ . . .+ ckx

k),

trong g, h Z[x] vi deg g = m < n v deg h = k < n. Do p l cca a0 = b0c0 nn p l c ca b0 hoc c0. Li do p

2khng l c ca a0

nn trong hai s b0 v c0, c mt v ch mt s chia ht cho p. Gi thit c0

chia ht cho p. Khi b0 khng chia ht cho p. V an = bmck v an khng

chia ht cho p nn bm v ck u khng chia ht cho p. Gi r l ch s b

nht sao cho cr khng l bi ca p. Ch rng s r nh vy lun tn ti

v ck khng l bi ca p. Trc ht ta xt trng hp r < n. Khi p l

c ca ar. V b0cr = ar (b1cr1 + b2cr2 + . . . + brc0) vi ch rngcc s c0, . . . , cr1 u l bi ca p nn b0cr l bi ca p. iu ny l v

l v c hai s b0 v cr u khng l bi ca p. Xt trng hp r = n. Khi

n = r 6 k 6 n. Suy ra k = n, v l. Vy f l bt kh quy trn Q.

Chng ta cng dng tiu chun Eisenstein kim tra tnh cht bt kh

quy ca cc a thc sau y gi l a thc chia ng trn: Cho p l s


29

nguyn t. a thc chia ng trn th p l

p(x) = xp1 + . . .+ x+ 1.

2.3.2 H qu. Vi mi s nguyn t p, a thc chia ng trn th p l

bt kh quy trn Q.

Chng minh. Ch rng a thc chia ng trn p(x) l bt kh quy

khi v ch khi p(x+ 1) l bt kh quy. Ta c

p(x+ 1) =(x+ 1)p 1

x.

Suy ra

p(x+ 1) = xp1 + C1px

p2 + . . .+ Ckpxpk1 + . . .+ Cp2p x+ p,

trong Ckp l s t hp chp k ca p phn t. Do p nguyn t nn Ckp l

bi ca p vi mi k = 1, . . . , p 2. V th p(x+ 1) l bt kh quy theotiu chun Eisenstein.

H qu n gin sau y ch ra rng vi mi s t nhin n lun tn ti

cc a thc bt kh quy trn Q bc n.

2.3.3 H qu. Cho a = pn11 pn22 . . . p

nkk l s phn tch tiu chun ca s t

nhin a thnh tch cc tha s nguyn t. Nu nj = 1 vi mt s j no

, 1 6 j 6 k, th xn a l bt kh quy vi mi n.

Chng minh. Theo gi thit, a l bi ca s nguyn t pj nhng khng l

bi ca p2j . V th theo tiu chun Eisenstein ta c kt qu.

2.3.4 V d. Theo tiu chun Eisenstein vi p = 2 ta suy ra a thc

x54x+2 l bt kh quy trn Q. Ch rng tnh cht bt kh quy ca athc ny khng d kim tra bng B Gauss trong nh l 2.2.1. Ngc

li, tnh bt kh quy ca a thc trong V d 2.2.5 c kim tra bng


30

vic dng B Gauss, nhng khng d kim tra tnh bt kh quy ca a

thc ny bng tiu chun Eisenstein.

2.3.5 V d. Xt tnh bt kh quy trn Q ca cc a thc sau.(i) x10 + 50;

(ii) 5x11 9x4 + 12x3 + 36x+ 6;(iii) x4 8x3 + 10x2 12x+ 3.

Gii. (i) D thy vi p = 2 a thc x10 + 50 bt kh quy trn Q (theo tiuchun Eisenstein).

(ii) a thc 5x11 9x4 + 12x3 + 36x+ 6 bt kh quy trn Q theo tiuchun Eisenstein vi p = 3.

(iii) t x = y + 3 ta c

f(x) = x4 8x3 + 10x2 12x+ 3 = y4 + 4y3 8y2 60y 78 = f(y).

a thc f(y) bt kh quy trn Q theo tiu chun Eisenstein vi p = 2. Do suy ra a thc f(x) = x4 8x3 + 10x2 12x+ 3 bt kh quy trn Q.

2.4 Rt gn theo mun mt s nguyn t

Cho p l mt s nguyn t. K hiu Zp = {a|a Z}, trong a = b Zpkhi v ch khi a b chia ht cho p. Ta c th kim tra c Zp l mttrng vi php cng a+ b = a+ b v php nhn ab = ab. Phn t khng

l 0, phn t n v l 1. Trng Zp c p phn t.Trong tit ny, chng ta trnh by mt phng php xt tnh bt kh

quy trn Q ca a thc bng cch da vo B Gauss v xt n trntrng Zp vi p l mt s nguyn t ph hp. Gi s

f(x) = anxn + . . .+ a1x+ a0 Z[x].


31

Vi mi s nguyn t p, k hiu ai Zp l s nguyn ai mun p (haylp thng d ca ai theo mun p). K hiu

f(x) := anxn + . . .+ a1x+ a0 Zp[x].

2.4.1 nh l. Cho f(x) Z[x]. Nu tn ti mt s nguyn t p sao chodeg f(x) = deg f(x) v f(x) bt kh quy trn Zp th f(x) bt kh quytrn Q.

Chng minh. Gi s f(x) bt kh quy trong Zp[x] v deg f(x) = deg f(x).Khi deg f(x) > 0. Suy ra deg f(x) > 0. Ta cn chng minh f(x) bt

kh quy trn Q. Gi s f(x) khng bt kh quy trn Q. Theo B Gauss, f(x) c s phn tch f(x) = g(x)h(x) trong g(x), h(x) Z[x],deg f(x) > deg g(x) v deg f(x) > deg h(x). Ch rng vi mi s

nguyn a, b ta c ab = a b v a+ b = a + b. V th ta c th kim

tra c f(x) = g(x)h(x). Do deg f(x) = deg g(x) + deg h(x). V

deg f(x) = deg f(x), deg g(x) deg g(x) v deg h(x) deg h(x) nnta suy ra deg g(x) = deg g(x) v deg h(x) = deg h(x). V th f(x) phn

tch c thnh tch ca hai a thc g(x), h(x) c bc thp hn, mu thun

vi tnh bt kh quy trn Zp ca f(x).

2.4.2 Ch . Gi thit deg f(x) = deg f(x) trong nh l 2.4.1 l cn

thit. Chng hn, xt a thc f(x) = 5(x 1)9 + (x 1) Z[x]. athc ny khng bt kh quy trn Q v n c c thc s l x 1. Ta cf(x) = x 1 Z5[x]. V f(x) c bc 1 nn r rng n l bt kh quytrn Z5.

2.4.3 V d. Xt tnh bt kh quy trn Q ca cc a thc sau.(i) 5x2 + 10x+ 4;

(ii) 3x3 + 7x2 + 10x 5;(iii) 11x4 5x3 + 21x2 9x+ 6.


32

Gii. (i) a thc f(x) = 5x2 + 10x + 4 Z[x]. Chn p = 3, ta cf(x) = 2x2 + x + 1 Z3[x]. a thc f(x) khng c nghim trong Z3nn l bt kh quy. V deg f(x) = 2 = deg f(x) nn theo nh l 2.4.1

a thc f(x) bt kh quy trn Q.(ii) Rt gn trong Z2[x], a thc f(x) = 3x3 + 7x2 + 10x 5 Z[x]tr thnh f(x) = x3 + x2 1 Z2[x]. a thc f(x) bt kh quy trn Z2v n khng c nghim trong Z2. V deg f(x) = 3 = deg f(x) nn theonh l 2.4.1 a thc f(x) bt kh quy trn Q.(iii) Rt gn trn Z5, a thc f(x) = 11x45x3+21x29x+6 Z[x]tr thnh f(x) = x4 +x2 +x+ 1 Z5[x]. D thy f(x) khng c nghimtrn Z5 nn n khng c nhn t bc mt. Gi s nu f(x) kh quy trongZ5[x] th f(x) = (x2 + ax+ b)(x2 + cx+ d) vi a, b, c, d Z5. ng nhth s hai v ca ng thc ny ta c

a+ c = 0, b+ ac+ d = 1, ad+ bc = 1, bd = 1.

V bd = 1 v vai tr ca b, d l nh nhau nn khng mt tnh tng qut ta c

th gi thit (b, d) = (1, 1) hoc (b, d) = (2, 3) hoc (b, d) = (4, 4). Nu

(b, d) = (1, 1) th a+ c = 0, ac = 1, a+ c = 1, v l. Nu (b, d) = (2, 3)th a + c = 0, ac = 4, 3a + 2c = 1. Suy ra a = 1, c = 1 nhng thayvo phng trnh ac = 4 ta c 1 = 4, v l. Nu (b, d) = (4, 4)th a + c = 0, ac = 7, a + c = 1

4, v l. Vy f(x) bt kh quy trn Z5.V deg f(x) = 4 = deg f(x) nn theo nh l 2.4.1 a thc f(x) bt kh

quy trn Q.

2.4.4 Ch . Peter Cameron s dng B Gauss v dng phng

php rt gn theo mun mt s nguyn t a ra mt chng minh

khc cho tiu chun Eisenstein nh sau: Gi s f = gh, trong g, h l

cc a thc vi h s nguyn v deg g = m < n, deg h = k < n. Gi

at, t = 0, 1, . . . , n; bi, i = 0, 1, . . . ,m v cj, j = 0, 1, . . . , k tng ng l


33

cc h s ca f , g v h. V cc h s a0, . . . , an1 ca f u chia ht cho

p nn ta c f(x) = anxn. Ch rng n = m + k v an = bmck. Do an

khng l bi ca p theo gi thit nn bm v ck u khng l bi ca p. V

th

g(x) = bmxm + a thc bc thp hn Zp[x],

h(x) = ckxk + a thc bc thp hn Zp[x].

Ta c th kim tra c f(x) = g(x)h(x). Do anxn = g(x)h(x). Ch

rng anxnch c duy nht mt c bt kh quy l x. V th g(x) v

h(x) cng ch c ng mt c bt kh quy l x. Do g(x) = bmxmv

h(x) = ckxk. Do bi = 0 v cj = 0 vi mi i < m v j < k. c bit,

c0 v b0 u chia ht cho p. Do a0 chia ht cho p2, v l.


Chng 3

Tnh bt kh quy trn trng Zp

Mc ch ca chng ny l xt tnh bt kh quy ca cc a thc trn

trng Zp vi p l mt s nguyn t. chng minh cc kt qu trongchng ny, ngoi vic s dng nh l Kronecker v s tn ti trng

phn r ca mt a thc, chng ta cn chun b thm mt s kt qu v l

thuyt nhm, c bit l nhm nhn Zp.

3.1 Kin thc chun b v nhm nhn Zp

3.1.1 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: a(bc) = (ab)c vi mi a, b, c G.(ii) G c n v: e G sao cho ex = xe = x vi mi x G.(iii) Mi phn t caG u kh nghch: Vi mi x G, tn ti x1 Gsao cho xx1 = x1x = e.

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

ton l giao hon, tc l ab = ba vi mi a, b G. Nu G c hu hnphn 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.

3.1.2 V d. Cho m > 1 l mt s nguyn. Vi a, b Z, ta nh ngha34


35

a = b nu v ch nu a b chia ht cho m. K hiu Zm = {a | a Z} ltp cc s nguyn munm (hay tp cc lp thng d theo munm). K

hiu Zm = {a Zm | (a,m) = 1} l tp cc s nguyn mun m nguynt cng nhau vi m. Khi Zm vi php nhn a b = ab l mt nhm giaohon cp (m), trong l hm Euler, tc l (1) = 1 v khi m > 1

th (m) l s cc s t nhin nh hn m v nguyn t cng nhau vi m.

Phn t n v ca Zm l 1. R rng nu m l s nguyn t th cp caZm l m 1.

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

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

Cho G l mt nhm vi php ton k hiu theo li nhn. Cho a G.t (a) = {an | n Z}. Khi (a) l nhm con ca G. Ta gi (a) lnhm con xyclic sinh bi a. Cp ca nhm con (a) c gi l cp ca

phn t a.

3.1.4 B . Cho G l mt nhm vi n v e. Vi mi a G, cc phtbiu 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 Z.

Chng minh. (i)(ii). Trc ht ta khng nh tn ti mt s nguyndng k sao cho ak = e. Gi s ngc li, vi mi cp s t nhin k < k

ta c akk 6= e. Suy ra ak 6= ak. iu ny chng t (a) 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 6 i 6 j < r th aji = e v 0 6 j i < r, do theo cch


36

chn ca r ta c i = j. By gi ta chng minh G = {e, a, a2, . . . , ar1}.R rng G {e, a, a2, . . . , ar1}. Cho b G. Khi b = ak vi k Z.Vit k = rq + s trong q, s Z v 0 6 s 6 r 1. Ta c

b = ak = arq+s = (ar)qas = as {e, a, a2, . . . , ar1}.

V th G = {e, a, a2, . . . , ar1} l nhm cp r. Suy ra r = n v (ii) cchng minh.

(ii)(iii). Gi s ak = e. Vit k = nq + r vi 0 6 r < n. V an = enn 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.

3.1.5 V d. Xt nhm nhn Z7. Cp ca Z7 l (7) = 6. Nhm conxyclic sinh bi 2 l (2) = {1, 2, 4}. Do cp ca 2 trong nhm nhn Z7l 3. Ch rng 3 l s nguyn dng b nht tha mn (2)3 = 1 Z7.V th theo b trn ta cng suy ra cp ca 2 l 3.

3.1.6 nh ngha. Cho H l nhm con ca G vi php ton k hiu theo

li nhn. Vi a G, k hiu Ha = {ha | h H}. Ch rng Ha = Hbnu v ch nu ab1 H. Ta gi Ha l lp ghp tri ca H trong G ngvi phn t a. 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.

3.1.7 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 G ta c a = ea Ha. V th, mi phn t ca G


37

u thuc mt lp ghp tri ca H. Gi s Ha Hb 6= . Khi tn tih, h H sao cho ha = hb. Suy ra a = h1hb. Cho xa Ha, trong x H. Khi xa = (xh1h)b Hb. Suy ra Ha Hb. Tng t,Hb Ha v do Ha = Hb. Vy hai lp ghp tri bt k ca H nukhc nhau th phi ri nhau. Vi mi a G, r rng nh x f : H Haxc nh bi f(h) = 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.

3.1.8 V d. Xt nhm nhn Z7. Cp ca Z7 l 6. Cp ca nhm con(2) = {1, 2, 4} l 3. Theo chng minh nh l Lagrange, ch s ca nhmcon {1, 2, 4} l 6 : 3 = 2.

3.2 Tnh bt kh quy trn trng Zp

Trong sut tit ny, lun gi thit p l mt s nguyn t. Khi Zp lmt trng vi php cng v php nhn cc s nguyn mun p. thun

tin, cc phn t ca Zp vn c k hiu nh cc s nguyn, trong tahiu hai phn t a, b Zp l bng nhau nu v ch nu a b l bi ca p.Gi s f(x) l a thc vi h s nguyn. Coi f(x) nh a thc trong

Zp[x]. Trong trng hp khi bc ca f(x) khng ln v p l s nguyn tnh th ta c th kim tra tnh bt kh quy ca f(x) trn Zp trc tip tnh ngha a thc bt kh quy. Sau y l mt v d minh ha.

3.2.1 V d. a thc x4 + x+ 1 l bt kh quy trn Z2.

Chng minh. Gi s f(x) = x4 + x + 1 khng bt kh quy trn Z2. Vf(x) khng c nghim trong Z2 nn n khng c nhn t bc nht. V thn phn tch c thnh tch ca hai a thc bc hai: f(x) = g(x)h(x)

vi g(x), h(x) Z2[x] v deg g(x) = deg h(x) = 2. Cc a thc bc hai


38

trong Z2[x] l x2, x2 + x, x2 + 1, x2 + x+ 1. D thy a thc bc hai duynht trong Z2[x] khng c nghim trong Z2 l x2 + x+ 1. V f(x) khngc nghim trong Z2 nn g(x) v h(x) cng khng c nghim trong Z2.Do g(x) = h(x) = x2 + x+ 1. R rng x4 + x+ 1 khng chia ht cho

x2 + x + 1, tc l f(x) khng chia ht cho g(x), iu ny l v l. Vy,

x4 + x+ 1 l bt kh quy trn Z2.

Trong trng hp tng qut, khi p l s nguyn t bt k, vic kim tra

tnh bt kh quy trn trng Zp nhn chung khng thc hin c. Mcch ca tit ny l s dng nh l Kronecker v trng phn r (nh

l 1.3.2) v nh l Lagrange trong l thuyt nhm (nh l 3.1.7) xt

tnh bt kh quy ca a thc trn Zp trong mt s trng hp c bit.

3.2.2 Mnh . a thc x2 + 1 l bt kh quy trn Zp vi mi s nguynt p tha mn p 3(mod 4).Chng minh. Cho p l s nguyn t tha mn p 3(mod 4). Gi s x2+1khng bt kh quy trn Zp. Khi x2 + 1 phn tch c thnh tch cahai a thc bc 1. Do x2 + 1 c nghim Zp. V th 2 = 1.Suy ra 4 = 1. Do p 3(mod 4) nn p 6= 2. V th 1 6= 1 Zp. Suyra 6= 1 v 2 6= 1. Ta khng nh 3 6= 1. Tht vy, nu 3 = 1 th2 = 1. Do = 1 hay = 1. iu ny l v l. Vy n 6= 1 vimi n = 1, 2, 3 v 4 = 1. Ch rng 6= 0 v 0 khng l nghim cax2 + 1. Do Zp. Theo B 3.1.4, cp ca phn t trong nhmnhn Zp l 4. V p l s nguyn t nn cp ca nhm nhn Zp l p 1.Theo nh l Lagrange, cp ca l c ca cp ca nhm nhn Zp. Do 4 l c ca p1. T gi thit ta ta c p1 ng d vi 2 theo mun4. iu ny l v l. Vy x2 + 1 l bt kh quy trn Zp.

3.2.3 Mnh . a thc x2 + x + 1 l bt kh quy trn Zp vi mi snguyn t p tha mn p 2(mod 3).


39

Chng minh. Cho p l s nguyn t tha mn p 2(mod 3). Gi sf(x) = x2 + x + 1 khng bt kh quy trn Zp. Khi f(x) l tch cahai a thc bc 1. Do f(x) c nghim Zp. Ch rng 0 khng lnghim ca f(x). V th 6= 0 v do d Zp. V

x3 1 = (x 1)(x2 + x+ 1) = (x 1)f(x),

v v l nghim ca f(x) nn l nghim ca x3 1. Suy ra 3 = 1.Ta khng nh 1 khng l nghim ca f(x). Tht vy, nu 1 l nghim

ca f(x) th 1 + 1 + 1 = 0 Zp v do 3 l bi ca p. Tuy nhin,p 2(mod 3). iu ny l v l. Vy 1 khng l nghim ca f(x) v do 6= 1. Nu 2 = 1 th 1 = 3 = 2 = , iu ny l v l. Nhvy, n 6= 1 vi n = 1, 2 v 3 = 1. Theo B 3.1.4, cp ca trongnhm nhn Zp l 3. Ch rng Zp c cp l p 1 ng d vi 1 theomun 3. Theo nh l Lagrange, 3 l c ca p 1. iu ny v l. Do, x2 + x+ 1 l a thc bt kh quy trn Zp.

3.2.4 Mnh . a thc f(x) = x4 + x3 + x2 + x+ 1 l bt kh quy trn

Zp vi mi s nguyn t p tha mn p 6= 5 v p 6 1(mod 5).

Chng minh. Cho p l s nguyn t tha mn p 6= 5 v p 6 1(mod 5).Theo nh l 1.3.2 (Kronecker), tn ti mt trng K cha Zp v chacc nghim ca f(x). Gi l mt nghim ca f(x). R rng 0 khng l

nghim ca f(x), v th 6= 0. Khng nh 1: n 6= 1 vi mi n = 1, 2, 3, 4 v 5 = 1. Tht vy, rrng x5 1 = (x 1)(x4 + x3 + x2 + x+ 1) = (x 1)f(x). Do lnghim ca x5 1. Suy ra 5 = 1. Nu = 1 th 5 = 0 Zp v do 5 l bi ca p, iu ny l mu thun vi gi thit p 6= 5. Suy ra 6= 1.Nu 2 = 1 th 1 = 5 = (2)2 = , iu ny l v l. Nu 3 = 1

th 1 = 5 = 32 = 2, v l. Nu 4 = 1 th 1 = 5 = 4 = , v


40

l. Nh vy, n 6= 1 vi mi n = 1, 2, 3, 4 v 5 = 1. Khng nh cchng minh.

Khng nh 2: f(x) khng c nhn t bc nht trong Zp[x]. Tht vy,gi s f(x) c nhn t bc nht. Khi f(x) c nghim Zp. V 6= 0nn Zp. Theo Khng nh 1, n 6= 1 vi mi n = 1, 2, 3, 4 v 5 = 1.V th, theo B 3.1.4, cp ca trong nhm nhn Zp l 5. Theo nhl Lagrange, 5 l c ca cp ca Zp. T gi thit, Zp c cp p 1 khngchia ht cho 5. iu ny l v l.

Khng nh 3: f(x) khng c nhn t bc hai trong Zp[x]. Thtvy, gi s f(x) c nhn t bc hai. Khi f(x) = q(x)r(x), trong

q(x), r(x) Zp[x] v deg q(x) = deg r(x) = 2. Ch rng q(x), r(x) lbt kh quy trn Zp v nu ngc li th q(x) hoc r(x) c nhn t bcnht v do f(x) c nhn t bc nht, iu ny mu thun vi Khng

nh 2. Do l nghim ca f(x) nn n l nghim ca q(x) hoc r(x).

Khng mt tnh tng qut, ta gi thit l nghim ca q(x). t F = Zpv T = {a + b | a, b Zp}. R rng php cng ng kn trong T .Cho u = a + b, v = c + d T. Vit q(x) = a0 + a1x + a2x2 via0, a1, a2 Zp. Do l nghim ca q(x) nn a0 + a1 + a22 = 0. Lido q(x) c bc hai nn a2 6= 0. V th 2 = a12 (a0 + a1). Do

uv = ac+ bd2 + (ad+ bc)

= ac bda0a12 + (ad+ bc bda1a12 ) T.

V th T ng kn vi php cng v php nhn. D kim tra c T l

mt trng con ca trng K. Xt T nh Fkhng gian vc t. R rng{1, } l mt h sinh ca Fkhng gian vc t T . Gi s a.1 + b = 0vi a, b F . Nu b = 0 th a = 0. Nu b 6= 0 th = ab1 F = Zp,v l vi Khng nh 2. Do {1, } l mt c s ca Fkhng gianvc t T , v v th chiu ca khng gian ny l 2. Do F c p phn t v


41

dimF T = 2 nn T c p2phn t. Suy ra nhm nhn T = T \ {0} c cpl p2 1 = (p 1)(p + 1). V 6= 0 nn T . Theo Khng nh 1ta c n 6= 1 vi mi n = 1, 2, 3, 4 v 5 = 1. V th, theo B 3.1.4,cp ca trong nhm nhn T l 5. Theo nh l Lagrange, 5 l c ca

(p 1)(p+ 1). V p 6 1(mod 5) theo gi thit nn 5 khng l c ca(p 1)(p+ 1), v l. Vy f(x) bt kh quy trn Zp.

3.2.5 Mnh . a thc f(x) = x6 + x5 + x4 + x3 + x2 + x + 1 l bt

kh quy trn Zp vi mi s nguyn t p 3(mod 7) hoc p 5(mod 7).

Chng minh. Cho p 3(mod 7) hoc p 5(mod 7) vi p l s nguynt. Theo nh l 1.3.2 (Kronecker), tn ti mt trng K cha Zp v chatt c cc nghim ca f(x). Gi K l mt nghim ca f(x). R rng0 khng l nghim ca f(x), v th 6= 0. Khng nh 1: n 6= 1 vi mi n = 1, 2, 3, 4, 5, 6 v 7 = 1. Thtvy, r rng x7 1 = (x 1)f(x). Do l nghim ca x7 1. Suy ra7 = 1. Nu = 1 th 7 = 0 Zp v do 7 l bi ca p, iu ny lmu thun vi gi thit v p. Nu 2 = 1 th 1 = 7 = (2)3 = , iu

ny l v l. Nu 3 = 1 th 1 = 7 = (3)2 = , v l. Nu 4 = 1 th

1 = 7 = 43 = 3, v l. Nu 5 = 1 th 1 = 7 = 52 = 2, v

l. Nu 6 = 1 th 1 = 7 = 6 = , v l. Nh vy, n 6= 1 vi min = 1, 2, 3, 4, 5, 6 v 7 = 1.

Khng nh 2: f(x) khng c nhn t bc nht trong Zp[x]. Tht vy,nu f(x) c nhn t bc nht th f(x) c nghim trong Zp. TheoKhng nh 1, n 6= 1 vi mi n = 1, 2, 3, 4, 5, 6 v 7 = 1. V 6= 0 nn Zp. Do cp ca trong nhm nhn Zp l 7. Theo gi thit, Zp ccp p 1, trong p 1 ng d vi 2 theo mun 7 hoc ng d vi4 theo mun 7. Theo nh l Lagrange, 7 l c ca p 1, iu ny lv l.


42

Khng nh 3: f(x) khng c nhn t bc hai trong Zp[x]. Tht vy,gi s f(x) c nhn t q(x) Zp[x] bc hai. Theo Khng nh 2, f(x)khng c nhn t bc nht. V th q(x) khng c nhn t bc nht. Suy ra

q(x) l bt kh quy trn Zp. Ly K l mt nghim ca q(x). Tngt nh chng minh Mnh 3.2.4, ta suy ra tp T = {a+ b | a, b Zp}l mt trng con ca K, trng ny c p2 phn t. V 6= 0 nn T .Theo Khng nh 1, ta c n 6= 1 vi mi n = 1, 2, 3, 4, 5, 6 v 7 = 1.V th cp ca trong nhm nhn T l 7. Theo nh l Lagrange, 7 l

c ca cp ca nhm nhn T . Ch rng T c cp p2 1. Tuy nhinp2 1 khng chia ht cho 7 v n ng d vi 1 hoc 3 theo mun 7.iu ny l v l.

Khng nh 4: f(x) khng c nhn t bc ba trong Zp[x]. Tht vy,gi s f(x) c nhn t bc ba q(x) Zp[x]. Theo cc Khng nh 2, 3,a thc q(x) bt kh quy trn Zp. Tng t nh chng minh Khng nh3, ta c th chng minh c tp

T = {a+ b + c2 | a, b, c Zp}l mt trng con ca K, trng ny c p3 phn t. V 6= 0 nn T .Theo Khng nh 1, ta c n 6= 1 vi mi n = 1, 2, 3, 4, 5, 6 v 7 = 1.V th cp ca trong nhm nhn T l 7. Theo nh l Lagrange, 7 l

c ca cp ca nhm nhn T . Nhm nhn T c cp p3 1. Ch rngp3 1 khng chia ht cho 7 v n ng d vi 5 theo mun 7, v l.Vy f(x) khng c nhn t bc mt, bc hai hay bc ba, do n bt

kh quy trn Zp.

3.2.6 Mnh . a thc f(x) = x10 + x9 + . . . + x + 1 l bt kh quy

trn trng Zp vi mi s nguyn t p sao cho p 2, 6, 7, 8 (mod 11).Chng minh. Cho p l s nguyn t sao cho p 2, 6, 7, 8 (mod 11). Theonh l Kronecker, tn ti mt trng K cha Zp v cha cc nghim ca


43

f(x). Gi K l mt nghim ca f(x). Tng t nh chng minhMnh 3.2.5, ta c th ch ra rng n 6= 1 vi mi n = 1, 2, . . . , 10 v11 = 1. R rng 0 khng l nghim ca f(x), do 6= 0.Cho d {1, 2, 3, 4, 5}. Gi s f(x) c nhn t q(x) bt kh quy bc d.Gi K l mt nghim ca q(x). t

T ={ d1

i=0

aii | ai Zp, i

}.

Do deg q(x) = q nn T l mt trng. Xt T nh mt Zpkhng gian vct. Do q(x) bt kh quy nn ta c th ch ra rng {1, , . . . , d1} l mtc s ca T . Do dimZp T = d v v th T c p

dphn t. Do 6= 0 nn

T . V n 6= 1 vi mi n = 1, 2, . . . , 10 v 11 = 1 nn c cp 11trong nhm nhn T . Theo nh l Lagrange, 11 l c ca pd 1. Theogi thit, ta c

(i) p 1 ng d vi mt trong cc s 1, 5, 6, 7 theo mun 11.(ii) p2 1 ng d vi mt trong cc s 3, 2, 4, 8 theo mun 11.(ii) p3 1 ng d vi mt trong cc s 7, 6, 1, 5 theo mun 11.(iii) p4 1 ng d vi mt trong cc s 4, 8, 2, 3 theo mun 11.(iv) p5 1 ng d vi 9 theo mun 11.Nh vy, pd 1 khng l bi ca 11 vi mi d 6 5. Suy ra f(x) khngc nhn t bt kh quy bc d vi mi d 6 5. V th f(x) bt kh quy trnZp.


44

Kt lun

Trong lun vn ny, chng ti trnh by cc ni dung sau y v a

thc bt kh quy trn mt trng:

Trnh by mt s kin thc c s v a thc bt kh quy. S dng a thc bt kh quy chng minh nh l Kronecker vs tn ti trng phn r ca mt a thc (nh l 1.3.2) v chng minh

nh l ca Galois v s tn ti mt trng hu hn (nh l 1.3.5).

a ra mt s phng php xt tnh bt kh quy ca a thc trnQ nh phng php tm nghim hu t, phng php dng B Gauss(nh l 2.2.1), phng php dng tiu chun Eisenstein (nh l 2.3.1)

v phng php rt gn theo mun mt s nguyn t (nh l 2.4.1).

S dng nh l Kronecker v trng phn r (nh l 1.3.2) v nhl Lagrange v cp ca nhm hu hn (nh l 3.1.7) xt tnh bt kh

quy ca mt vi a thc trn trng Zp vi p l s nguyn t.


Ti liu tham kho

[Bo] N. C. Bonciocat, Upper bound for the number of factors for a class

of polynomials with rational coefficients, Acta Arithmetica, (2) 113

(2004), 175-187.

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

[DO] H. L. Dorwart and O. Ore, Criteria for the irreducibility of polyno-

mials, Ann. Math, (2) 34 (1934), 81-94.

[G] P. Garrett, Abstract Algebra, Chapman - Hall/CRC, 2007.

[Rot] J. Rotman, Galois Theory, Springer (2001), Second Edition.

[Sc] A. Schinzel, Polynomials with special regards to reducibility, Cam-

bridge University Press, 2000.

[S] I. Seres, Irreducibility of polynomials, Journal of Algebra, 2 (1965),

283-286.

45


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