lý thuyết trường và galois

8/2/2019 L thuyt trng v Galois

1/108

AI HOC QUOC GIA TP HO CH MINHTRNG AI HOC KHOA HOC T NHIEN

BUI XUAN HAI

LY THUYETTRNG & GALOIS

NHA XUAT BAN AI HOC QUOC GIA TP HO CH MINH-2007-


2/108


3/108

Bang ky hieu

C - tap hp cac so phc

R - tap hp cac so thcQ - tap hp cac so hu ty

Z - tap hp cac so nguyen

N = {1, 2, 3,...} - tap hp cac so t nhienZ+ = {0, 1, 2, 3,...} - tap hp cac so nguyen khong ama, b = {a, a + 1, a + 2, . . . , b}, trong o a, b Z va a < bZn - tap hp cac so nguyen modulo n

Fq - trng Galois gom q phan t

Aut(L) - nhom cac t ang cau cua trng L

Gal(L/K) - nhom Galois cua m rong trng L/K

Sn - nhom oi xng bac n

An - nhom thay phien bac n

[L : K] - bac cua m rong

3


4/108

Li gii thieu

Bai toan tm nghiem cua nhng phng trnh ai so a c quantam t thi co ai. Trong nhng tam bang at set cua ngi Babylonco t 1600 nam trc Cong nguyen a co ghi nhng bai toan a

ve viec giai phng trnh bac hai. Cung t nhng tam bang at setnay ngi ta nhan thay rang nhng ngi Babylon a biet cach tmnghiem cua phng trnh bac hai mac du ho cha biet cach bieu dienchung bang nhng ky hieu ai so. Ngi La Ma co ai cung tng bietcach giai phng trnh bac hai bang phng phap hnh hoc, tuy hocung cha tm ra cong thc ai so e bieu dien chung. Tnh hnhcung tng t nh vay oi vi nhng phng trnh bac ba.

Phai en thi ky Phuc hng cac nha toan hoc Bologna (Y) mikham pha ra rang viec giai phng trnh bac ba cuoi cung c a

ve ba dang c ban sau ay:

X3

+ pX = q, X3

= pX+ q, va X3

+ q = pX.

S d ho phai xet ba trng hp rieng biet nay v ho khong muoncong nhan viec ton tai cac so am. Mot ngi trong so cac nha toanhoc thi o co ten la Scipio del Ferro t nhan la ong ta co li giai choca ba trng hp neu tren. Tuy nhien, nhng li giai nay c gib mat va ong ta ch day cho mot ngi hoc tro cua mnh co ten laFior biet phng phap giai mot trong ba dang phng trnh noi tren.Nhng tin tc ve viec a co ngi biet giai cac phng trnh bac ba alot ra ngoai va ieu nay kch thch s tm toi cua nhieu ngi. Cuoicung th mot ngi co ten la Niccolo Fontana (b danh Tartaglia) a

tai phat minh ra li giai vao nam 1535. Fontana a cong khai phngphap cua mnh nham canh tranh vi Fior, nhng ong t choi congbo chi tiet li giai. Nhng sau o Fontana, b mot nha vat ly hoc coten Girolamo Cardano thuyet phuc, a ong y trao li giai cho ongnay sau khi nhan c li ha la moi chuyen van se c gi b mat.The nhng, vao nam 1545 Cardano cho xuat ban cuon Ars Magna,

4


5/108

trong o ong trnh bay ay u toan bo li giai cua Fontana, kem theoli cam n tran trong gi en ngi a phat minh ra no. Tat nhienla Fontana co ly do chnh ang e gian d. The nhng, cung nhco dp may nay ma phat minh quan trong o a c ra mat quanchung.

Cung trong Ars Magna, Cardano con trnh bay mot phng phapcua Ludovico Ferrari ve viec giai phng trnh bac bon bang cach a

ve giai phng trnh bac ba.Mot tnh chat noi bat de nhan thay trong tat ca cac cong thc

c phat minh la cac nghiem eu c bieu dien bang cac bieu thccan thc, ngha la cac bieu thc ai so nhan c bang cach tac onglen cac he so cac phep toan cong, tr, nhan, chia va khai can.

Nh vay, en thi iem nay co the noi rang tat ca cac phngtrnh bac 4 eu giai c bang can thc. Mot cau hoi c at ramot cach rat t nhien la: Lieu phng trnh bac 5 co giai c bangcan thc hay khong? Cau hoi nay a thu hut s quan tam nghiencu cua rat nhieu ngi. Co the ke ra ay mot so trng hp sau:

Tschirnhaus a ra li giai nhng b Leibniz ch ra la sai lam. Eulera ra li giai sai nhng ong thi lai tm c phng phap mi egiai phng trnh bac 4. Lagrange cung nghien cu van e nay vatm ra cach thong nhat e giai quyet bai toan cho cac phng trnhbac 4. Tuy nhien ong noi rang phng phap cua ong se sai neuap dung cho phng trnh bac 5. Vay, phai chang phng trnh bac5 khong giai c bang can thc? Nam 1813, Ruffini cong bo motchng minh vi nhieu sai sot rang phng trnh bac 5 khong giaic bang can thc. Cuoi cung, vao nam 1824 Abel a chng minhmot cach thuyet phuc rang phng trnh bac 5 tong quat khong giaic bang can thc.

Van e bay gi lai la viec can giai quyet cau hoi: Lam cach naoe nhan biet mot phng trnh ai so cho trc la giai c hay

khong bang can thc? Abel ang nghien cu van e nay dang dth mat vao nam 1829. Khi ay, co mot ngi Phap con rat tre ten laEvariste Galois, sinh vien Trng Cao ang s pham (Ecole Normale

5


6/108

Superieuse) Paris, say sa nghien cu van e nay. Galois a gi tiba ban thao cho Vien Han lam khoa hoc Paris, nhng tat ca eu khongc xem xet mot cach nghiem tuc va ri vao quen lang. Nam 1832,chang thanh nien tai nang xuat chung Evariste Galois chet trong motcuoc au sung khi mi tuoi 21. May man thay, vao nam 1843,

Joseph Liouville a nhan thay gia tr cua nhng cong trnh cua Galoisva ong a viet th gi Vien Han lam khoa hoc Paris, trong o nhan

manh rang ong a tm thay trong cac bai viet cua Galois mot li giairat sau sac cho mot bai toan tuyet ep: Lam the nao e nhan bietmot phng trnh ai so la giai c hay khong bang can thc.

Ngay nay, phng phap ma Galois e xuat e tm nghiem chomot phng trnh ai so a c mang ten ong, goi la Ly thuyetGalois. Giao trnh Ly thuyet Trng & Galois nham cung cap nhngkien thc c ban ve Ly thuyet m rong trng va trnh bay nhng

y tng oc ao cua Galois e giai bai toan a noi pha tren. Tronggiao trnh cung trnh bay mot so ng dung cua Ly thuyet Trng &Galois, chang han nh bai toan dng hnh bang compa va thc kehay nh phep chng minh nh ly can ban cua ai so bang cach gan

nh ch s dung cac cong cu ai so. Cu the hn, ay co s ket hpgia viec s dung nh ly can ban cua thuyet Galois va nh ly Sylowve nhom hu han. Tuy nhien, trong chng minh chung toi van canen s tr giup mot chut cua Giai tch bang cach s dung nh ly

ve cac gia tr trung gian. ay cung la s minh chng cho ranh giirat tng oi cua cac chuyen nganh trong Toan hoc. e tien li choban oc, hai tiet au chung toi a trnh bay cac khai niem c ban

ve cac nhom hoan v va giai c. Ve mat hnh thc, viec oc cuongiao trnh nay khong oi hoi phai biet trc nhng kien thc cua aiso ai cng. Tuy nhien, e co the hieu sau van e, ngi oc cungcan phai co chut t kinh nghiem lam viec vi toan cao cap. Viec giai

bai tap la mot phan quan trong khi oc cuon giao trnh nay. Cungco mot so bai tap kho, nhng mot ban oc nghiem tuc chac chan segiai c tat ca cac bai tap nay.

Cuoi cung, nh la mot qui luat, du tac gia co cham chu en auth van co the con co nhng sai sot. V vay chung toi mong nhan

6


7/108

c cac y kien gop y, phe bnh cua ban oc e hoan thien cuon giaotrnh trong nhng lan xuat ban sau. Moi y kien xin gi ve a chhom th ien t

[email protected]

TP Ho Ch Minh 10/06/2007

PGS.TS Bui Xuan Hai

7


8/108

8


9/108

Muc luc

Bang ky hieu 3

Li gii thieu 4

1. Nhom hoan v . . . . . . . . . . . . . . . . . . . . . . . 112. Nhom giai c . . . . . . . . . . . . . . . . . . . . . . . 213. M rong hu han va m rong ai so . . . . . . . . . . . 254. Dng hnh bang compa va thc ke . . . . . . . . . . . 385. Nhom Galois va trng con co nh . . . . . . . . . . . . 466. Trng phan ra va nhng m rong chuan tac . . . . . . 527. M rong tach c . . . . . . . . . . . . . . . . . . . . 598. Bao chuan tac . . . . . . . . . . . . . . . . . . . . . . . 66

9. nh ly can ban cua thuyet Galois . . . . . . . . . . . . 7610. Cac a thc bac nho . . . . . . . . . . . . . . . . . . . 7911. a thc X4 2 . . . . . . . . . . . . . . . . . . . . . . 8212. nh ly can ban cua ai so . . . . . . . . . . . . . . . . 87

9


10/108

13. S giai c bang can thc . . . . . . . . . . . . . . . 9014. a thc bac 5 khong giai c bang can thc . . . . . 9715. Trng hu han . . . . . . . . . . . . . . . . . . . . . . 99

Tai lieu tham khao 105

10


11/108

1. Nhom hoan v

Tap hp tat ca cac song anh t tap hu han X gom n phan tvao chnh no la mot nhom vi phep nhan la phep hp noi anh xa.Ta ky hieu nhom nay la Sn va goi no la nhom oi xng bac n trentap X. Moi nhom con cua nhom Sn c goi la motnhom hoan v

bac n tren tap X. e n gian, ta co the gia thietX = {1, 2, . . . , n}.Moi phan t cua Sn c goi la mothoan v bac n. Mot hoan v bacn thng c viet di dang ma tran nh sau:

=

1 2 . . . ni1 i2 . . . in

trong o ik = (k), k 1, n.

nh ngha 1.1. Phan t

Sn c goi la mot k-chu trnh hay

mot chu trnh o dai k, neu ton tai mot tap con {i1, i2, . . . , ik} {1, 2, . . . , n} sao cho (i1) = i2, (i2) = i3, . . . , (ik1) = ik, (ik) =i1, va (j) = j, j {i1, i2, . . . , ik}.

Khi o phan t se c viet mot cach n gian la (i1i2 . . . ik).

Ta goi mot chu trnh o dai 2 la motchuyen v.

Neu Sn th ta ky hieu || la cap cua . Hien nhien neu la motk-chu trnh th || = k. Noi rieng, moi chuyen v eu la motphan t cap 2.

nh ngha 1.2. Ta noi = (i1i2 . . . ik) va = (j1j2 . . . jl) lahai chu trnh oc lap hay khong giao nhau, neu {i1, i2, . . . , ik} {j1, j2, . . . , jl} = .

Hai chu trnh oc lap se giao hoan vi nhau. T o suy ra cacchu trnh oi mot oc lap vi nhau cung giao hoan vi nhau. nh lydi ay ong mot vai tro quan trong trong ly thuyet nhom hoan v.

11


12/108

nh ly 1.3. Moi phep hoan v khac e (e la n v cua nhom Sn)eu phan tch thanh tch cac chu trnh oi mot oc lap vi nhau. S

phan tch nay la duy nhat sai khac mot th t cac chu trnh oc lap.

Chng minh. Xete = Sn va X = {1, 2, . . . , n}. Ta noi hai phant x, y X la tng ng vi nhau (ky hieu x y), neu ton taimot so nguyen m sao cho x = my. Ro rang quan he ma ta vanh ngha la mot quan he tng ng tren tap X. Ta goi moi mot

lp tng ng cua no la mot qu ao cua . Neu x X th ta kyhieu x la qu ao cua cha x. Vay

x = {mx| m Z}.

V X la tap hu han nen ton tai mot c Z+ sao cho x = cx.Goi m la so nguyen dng nho nhat thoa x = mx. Khi o

x = {x , x , . . . , m1x}.

Xet chu trnh o dai m

= (x x . . .

m1

x).

Khi o

y =

y , y xy, y x.

Gia s co tat ca k qu ao la C1, C2, . . . , C k . ng vi cac quao o ta co cac chu trnh 1, 2, . . . , k (xay dng theo cach a lam tren). Vay

iy =

y, y Ciy, y Ci.

Ro rang, neu i = j th Ci Cj = , do o i va j la cac chutrnh oc lap vi nhau. Ta se chng minh = 12 . . . k.

12


13/108

y X, at z = y . Khi o y va z cung nam trong mot quao Ci nao o cua vi iy = z. Ro rang j = i ta co jy = y

va jz = z. Do o 12 . . . k(y) = z = y. Ta a chng minh = 12 . . . k.

Bay gi, gia s = 12 . . . l la mot s phan tch khac cua thanh tch cac chu trnh oi mot oc lap vi nhau. Ro rang nhngphan t tham gia vao chu trnh j tao thanh mot qu ao Ci cua ,

do o j trung vi i (ng vi qu ao Ci). Do o hai s phan tchnh vay ch khac nhau th t cac chu trnh oi mot oc lap vinhau.

He qua 1.4. Cap cua mot hoan v bang BSCNN cua cac chieu daicua cac chu trnh trong s phan tch hoan v thanh tch cua nhngchu trnh oi mot oc lap vi nhau.

Chng minh. (Xem Bai tap 1.3).

nh ngha 1.5. Ta noi cac hoan v va co cung mot cau truc

chu trnh, neu = 1 . . . k va = 1 . . . k la nhng s phan tch va thanh tch cac chu trnh oc lap sao cho bang s anh so laith t cac chu trnh oc lap neu can, ta luon co i va i la nhngchu trnh co cung o dai (i 1, k).

Menh e 1.6. Neu Sn la mot chu trnh o dai k th Sn,

1 cung la mot chu trnh o dai k.

Chng minh. Kiem chng de dang rang, neu = (x1 x2 . . . xk) th 1 = ( x1 x2 . . . xk).

He qua 1.7. , Sn, 1

va la nhng hoan v co cung motcau truc chu trnh.

Chng minh. Phan tch thanh tch cac chu trnh oi mot oc lapvi nhau: = 12 . . . k . Khi o ta co

1 = ( 11)( 2

1) . . . ( k1).

13


14/108

Nhan xet rang i1, i 1, k la nhng chu trnh oi mot oclap vi nhau. Bay gi ap dung Menh e 1.6, suy ra pcm

Menh e 1.8. Moi hoan v khac e eu phan tch thanh tch cacchuyen v.

Chng minh. Moi chu trnh (i1 i2 . . . ik) eu co the c phan tchnh sau thanh tch cac chuyen v:

(i1 i2 . . . ik) = (i1 ik)(i1 ik1) . . . (i1 i2).

Bay gi ap dung nh ly 1.3 de dang suy ra pcm

nh ngha 1.9. Cho Sn. Ta noi cap so(i, j) la motnghch thecua , neu i < j nhng (i) > (j). Ta noi la mothoan v chan(tng ng hoan v le), neu so cac nghch the cua la mot so chan(tng ng la mot so le).

Moi chuyen v eu la mot hoan v le. That vay, gia s i < j. The

th chuyen v (i j) se co cac nghch the sau: (i, j) va tat ca cac capnghch the dang (i, k), (k, j), vi i k j.

nh ngha 1.10. Sn, at

sgn() =

1, neu la hoan v chan1, neu la hoan v le .

nh ly 1.11. Sn va mot chuyen v(ij) th vi = (ij) ta cosgn() = sgn().Chng minh. 10. Trng hp j = i + 1: Khi o ro rang ta co nhngieu sau ay:

a) Neu (i, j) la nghch the cua th (i, j) khong la nghch thecua va ngc lai.

14


15/108

b) Neu h, k = i, j th (h, k) la nghch the cua khi va ch khi(h, k) la nghch the cua .

c) Neu h < i th (h, i) la nghch the cua khi va ch khi (h, j)la nghch the cua .

d) Neu j < k th (j,k) la nghch the cua khi va ch khi (i, k)la nghch the cua .

Nhng ieu tren ay chng to so cac nghch the cua va hnkem nhau mot n v. Do o sgn() = sgn().

20. Trng hp j i = s + 1, s > 0 : Luc nay ta co the phan tchchuyen v (i, j) nh sau:

(ij) = (i + 1 i)(i + 2 i + 1) . . . (i + s i + s 1)(i + s j)(i + s 1 i + s)

(i + s 2 i + s 1) . . . (i + 1 i + 2)(i i + 1).Ngha la (ij) phan tch thanh tch cua 2s + 1 chuyen v co dang

hai phan t ke tiep nhau. Khi o, theo trng hp 10 ta thay tnh

chan, le cua se nhan c t tnh chan, le cua khi ta thay oi no2s + 1 lan. Vay sgn() = sgn().

oi vi nhng so nguyen ta cung se nh ngha mot ham dau nhsau:

nh ngha 1.12. k Z, nh ngha

sgn(k) :=

1, neu k la so chan

1, neu k la so le .

T nh ngha hien nhien ta co cong thc

sgn(k + l) = sgn(k)sgn(l).

15


16/108

nh ly 1.13. Neu Sn phan tch thanh tch k chuyen v thsgn() = sgn(k).

Chng minh. Ta se chng minh bang qui nap theo k. Khi k = 1 th la mot chuyen v, nen sgn() = 1 = sgn(k). Gia s k 1 va

= 12 . . . k+1

la s phan tch

thanh tch cuak + 1

chuyen v.

at := 12 . . . k. Theo gia thiet qui nap ta co sgn() =sgn(k). Khi o, theo nh ly 1.11 ta co

sgn() = sgn( k+1) = sgn() = sgn(k) = sgn(k + 1).

He qua 1.14. Neu la mot chu trnh o dai k th

sgn() = sgn(k 1).

He qua 1.15. , Sn,sgn() = sgn()sgn().T He qua 1.15 suy ra anh xa

sgn : Sn {1, 1} sgn()

la mot ong cau nhom. Ky hieu An la nhan cua ong cau nay, nghala

An = { Sn| sgn() = 1}.

Khi o An la nhom con chuan tac ch so2 cua Sn. Ta goi An lanhom thay phien hay nhom oi dau. Hien nhien ta co

|An| = |Sn|2

=n!

2.

16


17/108

Ta noi G la nhom n neu G = 1 va G khong co nhom conchuan tac thc s khac 1.

nh ly 1.16. n 5, An la nhom n.Chng minh. Lu y rang An c sinh ra bi cac chu trnh o dai3 (xem Bai tap 1.10). Gia s 1 = NAn. Trc het ta chng minh

neu N cha mot chu trnh o dai 3 th N cha tat ca cac chu trnho dai 3, do o N = An. Sau o ta se chng minh N nhat nh chamot chu trnh o dai 3. ieu nay tat nhien se ket thuc viec chngminh An la nhom n.

Vay, gia s N cha mot chu trnh o dai 3. Khong mat tnhtong quat, co the gia s N cha chu trnh (123).k > 3, chu trnh(23k) An, do o

(23k)1(123)(23k) = (1k2) N.

T o suy ra (12k) = (1k2)2

N. Vay (12k)

N,

k

= 1, 2.

Bay gi, bang cach tng t nh tren, t ieu kien (1k2) N suyra (1kl) N, l = 1, k. Nhng (1kl) = (kl1), nen cung bang cachnh tren suy ra (klj ) N, j = k, l. Tom lai, ta a chng minh Ncha moi chu trnh o dai 3. Vay N = An.

Ta con phai chng minh N cha t nhat mot chu trnh o dai 3.ieu nay se c chng minh qua mot so bc.

10. Gia s N cha phan t dang x = a b c . . . , trong o a ,b ,c lacac chu trnh oc lap va a = (a1 . . . am)(m 4). att = (a1a2a3).Ta co t1xt N. V t giao hoan vi b , c , . . ., nen

t1xt = (t1at)b c . . . = z N.

Ta co

zx1 = (t1at)a1 = t1(ata1) = (a1a3a2)(a2a3a4) = (a1a3a4) N.

17


18/108

Vay, trong trng hp nay N cha mot chu trnh o dai 3.

20. Gia s N cha mot phan t ma trong s phan tch thanh tchcac chu trnh oc lap co t nhat hai chu trnh o dai 3. Khong mattnh tong quat, co the gia s N cha phan t x = (123)(456)y, trongo y la mot hoan v co nh cac phan t 1, 2, 3, 4, 5, 6. att = (234),ta co (t1xt)x1 = (15243) N. Theo 10, N cha mot chu trnho dai 3.

30. Trong trng hp con lai, ta gia s N khong cha nhngphan t nh a neu trong 10 va 20. Khi o moi mot phan t cua Nse hoac la tch cua mot chu trnh o dai 3 vi cac chuyen v hoacla tch cua nhng chuyen v trong s phan tch cua no thanh tchnhng chu trnh oc lap.

Neu x = (123)p N, p2 = 1, th x2 = (132) la mot chu trnh odai 3. Vay, gia s N cha phan t x dang x = (12)(34)p, trong o pco nh 1, 2, 3, 4. atu = (145), ta co (u1xu)x1 N.

Neu x(5) = 5 th (u1xu)x1 = u1(xux1) = (154)(235) =

(15243) N,nen theo

1

0

Ncha mot chu trnh o dai

3.

Neu x(5) = k = 5 th trong N co phan t (154)(23k), nen theo20, N cha mot chu trnh o dai 3.

Vay, nh ly a c chng minh xong.

A5 chnh la nhom n khong giao hoan co cap nho nhat va Galoischnh la ngi au tien chng minh ieu nay.

Cuoi cung ta se chng minh mot ket qua lien quan en cac phant sinh cua nhom oi xng Sn.

Bo e 1.17. Nhom oi xng Sn c sinh ra bi cac chu trnh(12 . . . n) va (12).

Chng minh. at = (12 . . . n) va = (12). Goi G la nhom concua Sn sinh bi va . Khi o G cha 1 = (23), do o Gcha (23)1 = (34), . . .. Vay, mot cach tong quat, G cha tat cacac chuyen v dang (j j + 1). T o ta thay G cha cac chuyen v

18


19/108

(13) = (12)(23)(12), (14) = (13(34)(13), . . .. Tong quat, G cha tatca cac chyen v dang (1j), j 2, n. Nhng i, j = 1, ta co

(ij) = (1i)(1j)(1i),

nen G cha moi chuyen v. Vay G = Sn.

Bai tap

Bai 1.1. Cho S9 thoa ieu kien (i) = 10 i, i 1, 9. Hayphan tch thanh tch cac chu trnh oc lap.

Bai 1.2. Cho a, b la hai phan t giao hoan vi nhau trong nhom G.Gia s a co cap m, b co cap n va m, n la nhng so nguyen to cungnhau. Chng minh rang phan t ab co cap bang mn. Hay cho v duchng to rang, neu m va n khong nguyen to cung nhau th cap cuaab co the khong bang boi so chung nho nhat cua m va n.

Bai 1.3. Chng minh rang, neu = 12 . . . k la mot s phantch hoan v thanh tch cac chu trnh oc lap th cap cua bangboi so chung nho nhat cua cac cap cua cac chu trnh i, i 1, k.

Bai 1.4. Co bao nhieu phan t cap 2 trong nhom oi xng Sn?

Bai 1.5. Trong nhom oi xng S9 cho cac phan t va thoa

= 1

1 2 3 4 5 6 7 8 93 5 9 2 7 8 4 6 1

.

Neu la mot hoan v le th la hoan v chan hay le.

19


20/108

Bai 1.6. Hay phan tch

=

1 2 3 4 5 6 7 8 91 4 5 8 7 6 2 9 3

thanh tch cac chuyen v.

Bai 1.7. Cho = (134)(2357)(1846). Hay tnh cap cua 98.

Bai 1.8. Cho =

1 2 3 4 5 6 71 4 5 7 3 1 2

. Hay tnh 1.

Bai 1.9. Chng minh rang trong nhom oi xng S10 moi phan tcap 14 eu la hoan v le.

Bai 1.10. Chng minh rang, moi hoan v chan eu co the viet thanhtch cua nhng chu trnh o dai 3.

Bai 1.11. Chng minh rang An la nhom con ch so 2 duy nhattrong nhom oi xng Sn.

Bai 1.12. Hay tm tat ca cac 2-nhom con Sylow cua S4.

Bai 1.13. Hay tm tat ca cac nhom con cap 3 cua S4.

Bai 1.14. Chng minh rang trong nhom oi xng S5 khong conhom con cap 15.

20


21/108

2. Nhom giai c

nh ngha 2.1. Ta noi nhom G giai c, neu ton tai day cacnhom con

1 = G0 G1 . . . Gn = G (1)

sao cho

(i) GiGi+1, i 0, n 1;(ii) Gi+1/Gi la nhom aben, i 1, n 1.Ta noi (1) la day chuan tac cua nhom giai c G.

V du 1. 1) Moi nhom giao hoan G eu giai c, vi day chuan tac1G.

2) Nhom oi xng S3 giai c vi day chuan tac

1HS3, vi H = (123).

3) Nhom nh dien

D8 = a, b| a4 = b2 = 1, ab = b1a

giai c vi day chuan tac

1aD8.

4) Nhom oi xng S4 giai c vi day chuan tac

1V4A4S4,

trong o V4 = {1, (12)(34), (13)(24), (14)(23)} la nhom Klein.

21


22/108

nh ly 2.2. Cho nhom G, H G va NG. Khi o:(i) Neu G giai c thH giai c.

(ii) Neu G giai c thG/N giai c.

(iii) Neu N va G/N giai c thG giai c.

Chng minh. (i) Gia s

1 = G0G1 . . .Gr = G

la mot day giai c oi vi G. Vi i 1, r, atHi = H Gi. Ta sechng minh

1 = H0H1 . . .Hr = H

la mot day giai c oi vi H.

i 0, r 1, ta coHi+1/Hi = Gi+1 H/Gi H = Gi+1 H/Gi (Gi+1 H)

Gi(Gi+1

H)/Gi.

Nhng Gi(Gi+1 H)/Gi la nhom con cua nhom aben Gi+1/Gi,nen Hi+1/Hi la nhom aben. Do o H la nhom giai c.

(ii) Gia s G la nhom giai c vi day chuan tac nh trong (i).Khi o ta co day

N/N = G0N/NG1N/N . . .GrN/N = G/N.

Theo nh ly ve s ang cau, i 1, r 1 ta coGi+1N/GiN = Gi+1(GiN)/GiN Gi+1/Gi+1 (GiN)

Gi+1/Gi/(Gi+1)GiN))/Gi,ma nhom cuoi cung la nhom thng cua nhom aben Gi+1/Gi nenGi+1N/GiN la nhom aben. Vay G/N la nhom giai c.

(iii) Theo gia thiet ta co hai day chuan tac

1 = N0N1 . . .Nr

22


23/108

vaN/N = G0/NG1/N . . .Gs/N = G/N.

Khi o day

1 = N0N1 . . .Nr = N = G0G1 . . .Gs = G

la day chuan tac oi vi G. Vay G giai c.

Ta noi nhom G la m rong cua nhom A nh nhom B neu G conhom con chuan tac N sao cho N A va G/N B. T nh ly2.2, co the noi lp cac nhom giai c kn oi vi viec lay nhom con,nhom thng va m rong nhom. Lp cac nhom aben ch kn oi vi

viec lay nhom con va nhom thng nhng khong kn oi vi viec laym rong nhom. Vay, lp cac nhom giai c thc s rong hn lpcac nhom aben.

Moi nhom xyclic cap nguyen to la nhom n. Hn na, no lanhom aben nen giai c. Vay, cac nhom xyclic cap nguyen to la n

va giai c. Qua nh ly di ay ta thay ieu ngc lai cung ung.

nh ly 2.3. Nhom giai c la n khi va ch khi no la nhom giaic cap nguyen to.

Chng minh. Gia s G la nhom n giai c vi day chuan tac

1 = G0G1 . . .Gn = G.

Khong lam mat tnh tong quat co the gia thietGi+1 = Gi, i 0, m 1. Khi o Gn1 la nhom con chuan tac cua G Nhng G lanhom n va Gn1 = Gn = G, nen Gn1 = 1. Do G la nhom giaic nen G = Gn/Gn1 la nhom aben. Vay G la nhom xyclic cap

nguyen to.Di ay la mot nh ly rat quan trong ma ta se s dung ti khi

nghien cu ly thuyet Galois.

nh ly 2.4. n 5, Sn la nhom khong giai c.

23


24/108

Chng minh. Neu Sn giai c th An giai c. Vi n 5,theonh ly 1.16, An la nhom n. Nhng khi o theo nh ly 2.3, |An|la mot so nguyen to, ma vi n 5 th |An| = n!2 khong phai laso nguyen to, nen ta co mot mau thuan. Vay Sn khong giai c,n 5.

Bai tap

Bai 2.1. Chng minh rang Sn , vi 2 n 4 la nhom giai c.

Bai 2.2. Chng minh rang neu H va K la hai nhom con giai ccua nhom G va HG th HK la nhom giai c. T o suy ra, neuG la nhom hu han th trong G ton tai duy nhat mot nhom con giaic toi ai M. Hn na G/M khong cha nhom con chuan tac giaic khong tam thng.

Bai 2.3. Cho cac so nguyen to p va q. Hay chng minh cac ieusau ay:

(a) Moi nhom cap pq eu giai c;

(b) Moi nhom cap p2q eu giai c;

(c) Neu p < q th moi nhom cap pqn eu giai c.

Bai 2.4. Chng minh rang moi nhom giai c hu han khong abencap ln hn 1 eu cha mot nhom con chuan tac aben khong tam

thng.

Bai 2.5. Cho p, q va r la cac so nguyen to khac nhau. Chng minhrang moi nhom cap pqr eu giai c.

24


25/108

3. M rong hu han va m rong ai so

Neu K la trng con cua trng F th ta vietK F va goi F lam rong cua K hoac F la m rong tren K. Ta cung dung ky hieuF/K e ch F la m rong cua K va noi K la trng c s cua mrong. Neu F/K la m rong trng th co the xem F la mot khonggian vect tren K. Ky hieu [F : K] c dung e ch so chieu cuakhong gian vect nay. Ta goi [F : K] la bac cua m rong F/K. Neu[F : K] < th ta noi F la m rong hu han tren K hoac ngangon hn, ta noi F/K hu han. Trong trng hp ngc lai F cgoi la m rong vo han tren K.

nh ly 3.1. (nh ly ve bac) Cho day m rong trng

K F L.Khi o, L/K hu han neu va ch neu L/F va F/K hu han. Hn

na, neu nhng ieu noi tren xay ra th[L : K] = [L : F][F : K].

Chng minh. Goi {ai}iI la mot c s cua khong gian vectF trenK va {bj}jJ la mot c s cua khong gian vect L tren F. Ta sechng minh {aibj}(i,j)IJ la mot c s cua khong gian vect L trenK.

Vi moi x L ta cox =

jJ

jbj, j

F.

Mat khac,j =

iI

ij ai, ij K.

25


26/108

T o suy ra

x =jJ

iI

ijai

bj =

(i,j)IJ

ijaibj.

Vay tap{

aibj

}(i,j)IJ sinh ra khong gian vectL tren K.

Bay gi ta ch con can chng minh tap nay oc lap tuyen tnhtren K. That vay, gia s

(i,j)

ijaibj = 0.

Khi o jJ

iI

ijai

bj = 0.

T o suy raiI

ijai = 0, j J, do o ij = 0, (i, j) I J.

Ap dung nh ly va chng minh de dang nhan c ket qua diay:

He qua 3.2. Cho mot day m rong trng

K = F0 F1 F2 . . . Fn = F.

Khi o, F/K hu han neu va ch neu Fi+1/Fi hu han, i 0, n 1. Hn na, neu nhng ieu tren xay ra th

[F : K] = [F : Fn1][Fn1 : Fn2] . . . [F2 : F1][F1 : K].

26


27/108

nh ngha 3.3. Cho m rong trng F/K va S la mot tap conkhac cua F.

(i) Ky hieu K[S] la giao cua tat ca cac vanh con cua F cha Kva S va goi no la vanh con cua F sinh ra biK va S.

(ii) Ky hieu K(S) la giao cua tat ca cac trng con cua F chaK va S va goi no la trng con cua F sinh ra biK va S.

Neu S la tap hu han th ta noi K(S) la m rong hu han sinhtren K.

Menh e 3.4. Cho F la m rong trng cua K va a1, . . . , an F(n > 0). Khi o

(i)K[a1, . . . , an] = {f(a1, . . . , an)| f K[X1, . . . , X n]},

(ii)K(a1, . . . , an) =f(a1, . . . , an)

g(a1, . . . , an)| f, g K[X1, . . . , X n],

g(a1, . . . , an) = 0

, ngha la K(a1, . . . , an) la trng cac thng

cua mien nguyen K[a1, . . . , an].

Chng minh. Ta se ch chng minh cho trng hp n = 1, trnghp tong quat c chng minh hoan toan tng t.

(i) Gia s a F. Nhan xet rang K[a] la vanh con nho nhat chaK va a. Tap hp {f(a)| f(X) K[X]} hien nhien kn oi vi cacphep toan cong va nhan trong F, nen la mot vanh con cua F. Rorang vanh con nay cha K va a, ong thi moi vanh con cua F chaK va a eu cha tap hp noi tren. Vay

K[a] = {f(a)| f(X) K[X]}.

(ii) Nhan xet rang trng cac thng cua mien nguyen K[a] chnh

27


28/108

la tap hp

f(a)g(a)

| g(X) K[X], g(a) = 0

.

Ro rang tap nay nam trong moi trng con cua F cha K va a. Vay

K(a) =f(a)

g(a)| g(X) K[X], g(a) = 0

.

nh ngha 3.5. Xet m rong trng F/K. Phan t F cgoi la ai so tren K neu ton tai mot a thc f(X) K[X] co bac 1 nhan lam nghiem. Noi cach khac, F la ai so tren K neuton tai nhng phan t a0, a1, . . . , an

K(n > 0), khong phai tat ca

eu bang 0 sao cho

a0 + a1 + . . . + ann = 0. (1)

Gia s F la phan t ai so tren K. Xet ong cau thay the

: K[X] F,

f(X) f().

Do ai so tren K nen ton tai a thc p(X) K[X] co bac 1sao cho Ker() = p(X). Khi o K[X]/p(X) K[] la motmien nguyen, suy ra p(X) la a thc bat kha qui. Neu gia s he socao nhat cua p(X) la 1 th p(X) c xac nh duy nhat. Ta ky hieup(X) = min(K, ) va goi no la a thc toi tieu cua phan t trenK.

28


29/108

nh ly 3.6. Cho m rong trng K F va F la mot phan tai so tren K. Gia s a thc bat kha quip(X) = min(K, ) co bacla n. Khi o, ta co nhng ieu khang nh sau ay:

(i) Neu g(X) K[X] thg() = 0 khi va ch khi g(X) chia hetcho p(X) trong K[X]. Suy ra p(X) la a thc co bac nho nhat cuaK[X] nhan lam nghiem.

(ii) K() = K[]. Cac phan t 1, , . . . , n1

tao thanh mot cs cua khong gian vectK() tren K.

Chng minh. (i) Ta co g() = 0 g(X) Ker() = p(X) f(X) chia het cho p(X) trong K[X].

(ii) Gia s f() la mot phan t khac 0 bat ky cua K[]. Khi of(X) la mot a thc khac 0 cua K[X] va f(X) khong chia het cho athc bat kha qui p(X). Do o ton tai cac a thc g(X), h(X) K[X]sao cho

f(X)g(X) +p(X)h(X) = 1.

Thay vao v tr cua X, nhan c f()g() = 1 hay f() kha

nghch trong K[]. Vay K[] la mot trng, suy ra K() = K[].Do p(X) la a thc co bac nho nhat cua K[X] nhan lam nghiem

va deg(p(X)) = n nen cac phan t 1, , . . . , n1 oc lap tuyen tnhtren K. Do K() = K[], nen mot phan t bat ky cua K() codang f(), vi f(X) K[X]. Chia f(X) cho p(X) nhan c

f(X) = p(X)g(X) + r(X), vi deg(r(X)) n 1.

Suy ra f() = r(), hay cac phan t 1, , . . . , n1 sinh ra K().Vay, {1, , . . . , n1} la mot c s cua K() tren K.

V du 2. So phc i = 1 la ai so tren Q v i2 + 1 = 0. Neua = nr, vi r Q th a la ai so tren Q v a la nghiem cua a thcxn r Q[X]. Neu = e2i/n = cos2n + isin2n , th n = 1, doo ai so tren Q. Cuoi cung, lu y rang min(Q, i) = min(R, i) =X2 + 1, nhng min(C, i) = X i. Do o ta nhan thay rang a thctoi tieu cua mot phan t ai so phu thuoc vao trng c s.

29


30/108

V du 3. 1) a thc X3 2 bat kha qui tren Q theo tieu chuanEisenstein va nhan 3

2 lam nghiem, do o min(Q, 3

2) = X3 2.

Tong quat hn, neu p la so nguyen to th a thc Xn p bat kha quitren Q, do o n

p ai so tren Q va [Q( n

p) : Q] = n.

2) So phc = cos2

3+isin

2

3la nghiem cua a thc X31

Q[X]. Phan tch

X3 1 = (X 1)(X2 + X+ 1),

ta thay la nghiem cua a thc X2 + X + 1, ma a thc nay batkha qui tren Q, nen

min(Q, ) = X2 + X + 1, do o [Q() : Q] = 2.

3) Gia s p la so nguyen to va

= exp(2i/p) = cos2

p+ isin

2

p.

Khi o la nghiem cua a thc Xp 1. Ta co

Xp 1 = (X 1)(Xp1 + Xp2 + . . . + 1).

Do = 1, nen la nghiem cua a thc

fp(X) = Xp1 + Xp2 + . . . + 1.

Ta goi fp(X) la a thc chia ng tron va co the chng minh cay la mot a thc bat kha qui tren Q. Vay fp(X) = min(Q, ).

nh ngha 3.7. Ta goi m rong K F la m rong ai soneu moiphan t cua F eu ai so tren K.

30


31/108

Menh e 3.8. Moi m rong hu han eu la m rong ai so.

Chng minh. Gia s F/K la mot m rong hu han va F la motphan t bat ky cua F. Do [F : K] < nen ton tai n > 0 e cacphan t

1, , . . . , n

phu thuoc tuyen tnh tren K. Vay ai so tren K.

Lu y rang ieu ngc lai cua Menh e 3.8 la khong ung. V du,tap hp tat ca cac so phc ai so tren Q la mot m rong vo han cuaQ (xem Bai tap 3.12). Tuy nhien, chung ta se chng minh rang moim rong ai so hu han sinh tren F eu hu han tren F. Cu the taco

Menh e 3.9. Cho F = K(1, 2, . . . , n) , trong o cac phan t1, 2, . . . , n eu ai so tren K. Khi o F la m rong hu hantren K. Hn na

[K(1, 2, . . . , n) : K] n

i=1

[K(i) : K].

Chng minh. Vi moi i 1, n ta coK(1, 2, . . . , i) = K(1, 2, . . . , i1)(i).

T o, ap dung He qua 3.2 cho day m rong

K

K(1)

K(1, 2) . . .

K(1, 2, . . . , n) = K,

ta co F la m rong hu han tren K. Bat ang thc sau cung co thec chng minh bang qui nap nh sau:

Vi n = 2, ta co

[K(1, 2) : K] = [K(1)(2) : K(1)][K(1) : K].

31


32/108

Theo nh ly 3.6, min(K(1), 2) la c cua min(K, 2), do o[K(1)(2) : K(1)] [K(2) : K]. T o suy ra bat ang thcung cho trng hp n = 2. Trng hp tong quat c suy ra dedang khi ta ap dung gia thiet qui nap.

Ket qua noi tren co the m rong cho trng hp nhng m rongsinh bi mot so bat ky cac phan t.

Menh e 3.10. Cho F = K(S) , trong o S la mot tap hp bat kynhng phan t ai so tren K. Khi o K(S) la mot m rong ai sotren K. Hn na, neu S hu han th K(S) la m rong hu hantren K.

Chng minh. Xet phan t a bat ky cua K(S). Khi o a c vietdi dang

a =f(a1, a2, . . . , an)

g(a1, a2, . . . , an), vi g(a1, a2, . . . , an) = 0,

trong o a1, a2, . . . , an la nhng phan t nao o cua tap S. ieu nayco ngha la a K(a1, a2, . . . , an). Nhng khi o theo Menh e 3.9,a la phan t ai so tren K.

Bay gi ta se chng minh cac m rong ai so co tnh chat baccau. Lu y rang trong trng hp m rong hu han th tnh chat nayc suy ra ngay t nh ly 3.1.

nh ly 3.11. Cho day m rong

K F L.

Neu L/F va F/K la nhng m rong ai so thL/K cung la motm rong ai so.

Chng minh. L, do ai so tren F nen ton tai a thcf(X) = a0 + a1X + . . . + anX

n F[X] sao cho f() = 0. atF0 = F(a0, a1, . . . , an) va L0 = F0().

32


33/108

Theo Menh e 3.9, K F0 va F0 L0 la nhng m rong huhan. Khi o K L0 la m rong hu han, do o ai so tren K.

nh ngha 3.12. Cho F la mot m rong cua K. Khi o tap hp

{a F| a ai so tren K}

c goi la motbao ong ai socua K trong F.

Menh e 3.13. Cho F la mot m rong cua K va L la bao ong aiso cua K trong F. Khi o L la mot trng va L la m rong ai soln nhat cua K trong F.

Chng minh. Gia s a va b = 0 la nhng phan t ai so bat ky cuaF tren K. Khi o, theo Menh e 3.9, K(a, b) hu han tren K . Doo, theo Menh e 3.8, K(a, b) ai so tren K. T ay suy ra a b vaa/b eu la nhng phan t ai so tren K. Vay, L la trng.

nh ngha 3.14. Cho F1 va F2 la nhng m rong cua trng K.Gia s F1 va F2 eu nam trong mot trng L nao o. Khi o ta kyhieu F1F2 la trng con cua L sinh ra bi F1 F2 tren K va goi nola tch cua F1 va F2.

V du 4. Cho K = Q. Xet cac m rong cua Q trong C. Gia s = e2i/3 la mot can bac 3 cua 1 va = 1. Ta se chng to

Q(,3

2) = Q(3

2)Q(3

2).

That vay, v Q( 3

2) va Q( 3

2) eu nam trong Q(, 3

2) nen

tch cua chung cung nam trong Q(,32). Mat khac, gia s trng

con L cua C cha 3

2 va 3

2. Khi o L cha = 3

2/ 3

2. ieuo dan en L cha Q(, 3

2). Vay Q(, 3

2) la trng con nho nhat

cua C cha Q( 3

2) va Q( 3

2). Noi cach khac

Q(,3

2) = Q(3

2)Q(3

2).

33


34/108

Ngoai ra ta con co the chng minh

Q(3

2, ) = Q(3

2 + ),

ngha la Q( 3

2, ) c sinh ra bi mot phan t tren Q. That vay,ata = + 3

2, ta co

(a )3 = 2= a3 3a2 + 3a2 3 = 2= a3 3a2 + 3a2 3 = 0.Thay 2 = 1 vao ang thc tren va giai theo , nhan c

=a3 3a 3

3a2 + 3a.

Vay Q(a), suy ra 32 = a Q(a). T o ta co

Q(32, ) Q(a) = Q( 32 + ),

suy ra Q( 3

2, ) = Q( 3

2 + ).

Bai tap

Bai 3.1. Tm cac trng con cua C sinh bi cac tap hp sau tren Q:

a) {0, 1}; b) {0}; c) {0, 1, i};d)

{i,

2}; e) R; f) R

{i}.

Bai 3.2. Hay tnh :

a) [C : Q]; b) [R(

5) : R]; c) [Q( 3

2) : Q]; d) [Q(

6) : Q];

e) [Q() : Q], vi 7 = 3; f) [Q(

3,

5,

11) : Q].

34


35/108

Bai 3.3. Ta goi m rong K F la m rong n, neu F = K(),vi F. Chng minh rang neu [F : K] la mot so nguyen to thF/K la m rong n.

Bai 3.4. Chng minh rang R khong phai la m rong n tren Q.

Ch dan: Hay lan lt chng minh nhng ieu sau ay:

a) Q la tap hp em c.b) Moi m rong n tren mot trng em c la em c.

c) R khong em c.

Bai 3.5. Chng minh rang Q(

5,

7) la m rong n tren Q. Cuthe, hay chng minh

Q(

5,

7) = Q(

5 +

7).

Bai 3.6. Hay tnh a thc toi tieu :a) min(Q, i); b) min(R, i); c) min(Q,

2);

d) minQ,

5 + 1

2

; e) min

Q,

i

3 12

.

Bai 3.7. Hay xay dng m rong Q()/Q, trong o co a thc toitieu tren Q la

(a) X2

5;

(b) X4 + X3 + X2 + X+ 1;

(c) X3 + 2.

Bai 3.8. Hay xay dng mot trng co 8 phan t.

35


36/108

Bai 3.9. Chng minh rang

[Q(4

2,

3) : Q] = 8.

Bai 3.10. Cho m rong F/K va a F sao cho [K(a) : K] la motso le. Chng minh rang K(a) = K(a2). Cho v du chng to ieunay sai neu [K(a) : K] la mot so chan.

Bai 3.11. Chng minh rang neu F la m rong ai so tren K va Rla vanh con cua F cha K th R la trng.

Bai 3.12. Cho A la bao ong ai so cua Q trong C, ngha la A latap hp tat ca nhng phan t cua C ai so tren Q. Chng minh rang[A : Q] = .

Bai 3.13. Cho K la trng va f(X) K[X].

(a) a K, chng minh rang ton tai a thc g(X) K[X] saochof(X) = g(X)(X a) + f(a).

(b) Suy ra f(a) = 0 khi va ch khi (X a)|f(X) va f(X) cokhong qua deg(f) nghiem.

Bai 3.14. (a) Cho G la mot nhom aben hu han. Chng minh rang,neu vi moi c m cua |G|, G co khong qua m phan t co cap la ccua m th G la nhom xyclic.

(b) Suy ra moi nhom con hu han cua nhom nhan cua mot trngeu la nhom xyclic.

Bai 3.15. Cho la nghiem cua a thc f(X) = X3X2+X+2 Q[X] va K = Q(). Hay bieu dien cac phan t (2 + +1)(2 )

va ( 1)1 di dang a2 + b + c, vi a ,b ,c Q.

36


37/108

Bai 3.16. Chng minh rang Q(

2) va Q(

3) khong ang cau vinhau nh cac trng, nhng ang cau vi nhau nh cac khong gian

vect.

Bai 3.17. Cho L1 va L2 la nhng m rong cua trng K. Gia s L1va L2 eu nam trong mot trng F nao o. Chng minh rang L1L2la m rong hu han tren K khi va ch khi L1 va L2 eu la nhng

m rong hu han tren K.

Bai 3.18. Cho m rong hu han F/K va L1, L2 la cac trng concua F cha K.

(a) Chng minh rang [L1L2 : K] [L1 : K][L2 : K].(b) Chng minh rang dau ang thc xay ra khi [L1 : K] va

[L2 : K] la cac so nguyen to cung nhau.

(c) Cho v du chng to [L1L2 : K] < [L1 : K][L2 : K].

Bai 3.19. Cho v du mot m rong trng F/K vi [F : K] = 3nhng F = K(a), a K.

Bai 3.20. Cho m rong hu han F = K(a) tren K. Vi F,nh ngha anh xa L : F F bi L(x) = x, x F.

(a) Chng minh rang L EndK(F).(b) Chng minh rang det(XIdF L) la a thc toi tieu cua

phan t a tren K.

(c) oi vi nhng phan t nao cua F ta co det(XIdF

L) =min(K, )?

Bai 3.21. Cho m rong hu han L/K bac m. Gia s f(X) la athc bat kha qui bac n tren K va (m, n) = 1. Chng minh rangf(X) bat kha qui tren L.

37


38/108

Bai 3.22. Chng minh rang a thc f(X) = X5 9X3 + 15X+ 6bat kha qui tren Q(

2,

3).

4. Dng hnh bang compa va thc ke

Dung Ly thuyet m rong trng co the chng minh c co nhnghnh ma ta khong the nao dng c neu ch c phep dung compa

va thc ke. Trong so nhng bai toan thu v o, co the ke en viecchia ba mot goc, dng hnh hop co the tch gap oi mot hnh hopcho trc va dng mot hnh vuong co dien tch bang dien tch mothnh tron cho trc. o chnh la nhng bai toan ma ta se e cap entrong muc nay.

nh ngha 4.1. Gia s P0 va P1 la nhng iem cua mat phang Euclid

c cho bi P0 = (0, 0) va P1 = (1, 0). Ta noi mot iem P cuamat phang la dng c bang compa va thc ke t cac iem P0 vaP1 , neu P = Pn oi vi mot day cac iem P0, P1, . . . , P n cua matphang, trong o P0 = (0, 0), P1 = (1, 0) va j > 1, iem Pj la mottrong cac iem sau:

Giao iem cua hai ng thang khac nhau, ma moi ng thangay i qua t nhat hai iem cua tap hp {P0, P1, . . . , P j1}.

Giao iem cua mot ng thang noi hai iem nao o cua tap hp{P0, P1, . . . , P j1} vi mot ng tron co tam la mot iem nao o

va i qua mot iem khac cua tap iem noi tren.

Tam cua hai ng tron khac nhau, trong o moi ng tron euco tam la mot iem nao o va i qua mot iem khac cua tap iem{P0, P1, . . . , P j1}.

Ta noi mot iem la dng c, neu no co the c dng t caciem P0 va P1 bang cach ch dung compa va thc ke.

38


39/108

nh ly 4.2. Cho (x, y) la mot iem dng c cua mat phang Eu-clid. Khi o [Q(x, y) : Q] = 2r , vir la mot so nguyen dng naoo.

Chng minh. atP = (x, y) va goi P0, P1, . . . , P n = P la mot daycac iem cua mat phang thoa cac tnh chat nh trong nh ngha 4.1.Gia s Pj = (xj, yj), atK0 = K1 := Qva Kj := Kj1(xj, yj), j 2, n. Do tnh chat cua cac iem P0, P1, . . . , P n nen vi moi j, cac so

thc xj va yj la nghiem cua mot a thc bac nhat hoac bac hai vicac he so trong Kj1. Do o

[Kj1(xj) : Kj1] =

12

va

[Kj1(xj, yj) : Kj1(xj)] =

12

, j 2, n.

T o suy ra [Kn : Q] = 2s, vi s la mot so nguyen dng naoo. Nhng [Kn : Q] = [Kn : Q(x, y)][Q(x, y) : Q] nen [Q(x, y) : Q]la c cua 2s, keo theo [Q(x, y) : Q] = 2r, vi r la mot so nguyendng nao o.

S dung nh ly 4.2 co the chng to khong the dung compa vathc ke e chia ba mot goc, e nhan oi the tch mot hnh hophoac e dng mot hnh vuong co dien tch bang dien tch mot hnhtron cho trc. e chng minh nhng ieu noi tren, trc het ta canchng minh mot so bo e n gian sau.

Bo e 4.3. Neu cac iem au mut cua mot oan thang la dng cth trung iem cua oan thang ay cung dng c.

Chng minh. Gia s P va Q la hai iem dng c. Goi A va B lagiao iem cua ng tron tam P i qua Q vi ng tron tam Q i

39


40/108

qua P. Khi o giao iem I cua cac ng thang P Q va AB chnhla trung iem cua oan thang P Q.

Bo e 4.4. Neu ba nh cua mot hnh bnh hanh la dng c thnh th t cua no cung la iem dng c.

Chng minh. Gia s A ,B,C la cac nh dng c cua mot hnh

bnh hanh, trong o AB la ng cheo cua no. Dng trung iem Icua oan thang AB, sau o dng iem D la giao cua ng thangCI va ng tron tam I i qua iem C. D chnh la nh th t cuahnh bnh hanh.

Bo e 4.5. Neu A, B la hai iem dng c va AB = a th cac iem(a, 0) va (0, a) eu dng c.

Chng minh. Dng trung iem I cua oan thang P0B, sau o dngQ la giao iem cua ng thang AI vi ng tron tam I i qua A.Cuoi cung dng iem C la giao cua ng tron tam P0 i qua Q vi

truc hoanh, ta co C = (a, 0). Dng iem D la giao cua ng trontam P0 i qua C vi truc tung, ta co D = (0, a).

Menh e 4.6. Khong the dung compa va thc ke e chia ba goc

3

(radian).

Chng minh. at a = cos

9va b = sin

9. Trc het ta se chng

to rang iem

cos

3,sin

3

=1

2,3

2

dng c bang compa va

thc ke. Thc vay, iem noi tren c dng nh sau:

Ta co cac iem xuat phat la P0 = (0, 0) va P1 = (1, 0).

40


41/108

- Dng P2 va P3 la cac giao iem cua ng tron tam P0 i quaP1 va ng tron tam P1 i qua P0.

- Dng P4 la giao iem cua cac ng thang P0P1 va P2P3.

- Dng P5 la giao iem cua ng tron tam P0 i qua P4 vi truchoanh.

- Dng P6 la giao iem cua ng tron tam P4 i qua P5 vi

ng thang P2P3.

- Dng P7 la giao iem cua truc tung vi ng tron tam P0 iqua P1.

- Dng P8 la giao iem cua ng thang P6P7 vi ng trontam P7 i qua P6.

- Dng P9 la giao iem cua truc hoanh vi ng tron tam P4 iqua P8.

- Dng P10 la trung iem cua oan thang P6P9.

- Dng P11 la giao iem cua ng thang P4P6 vi ng tron

tam P4 i qua P10. Ta co P11 =1

2,

3

2

chnh la iem can dng.

Gia s co the chia ba goc

3bang compa va thc ke. Khi o,

iem (a, b) la iem dng c. oi vi mot goc bat ky ta co congthc sau:

cos3 = 4cos3 3cos.

Cho =

9, nhan c 4a3 3a = 1

2, do o 8a3 6a 1 = 0.

Vi f(X) = X3+ 3X23, ta co f(2a1) = 8a36a1 = 0. V athc f(X) bat kha qui tren Q nen [Q(a) : Q] = [Q(2a 1) : Q] = 3.

41


42/108

ieu nay mau thuan vi ket luan cua nh ly 4.2. Vay, khong the

dng c iem

cos

9, sin

9

, ngha la ta khong the chia ba goc

3

bang compa va thc ke.

Menh e 4.7. Khong the dng c bang compa va thc ke mothnh hop co the tch gap oi the tch mot hnh hop cho trc.

Chng minh. Gia s co the dng c hai iem P va Q cach nhaumot khoang bang 3

2. Khi o, theo Bo e 4.5, dng c iem

( 3

2, 0) . Nhng a thc X3 2 bat kha qui tren Q nen [Q( 32) :Q] = 3, do o ta co mot mau thuan vi nh ly 4.2. Suy luan trenchng to t mot oan thang cho trc trong mat phang ta khong thedng c ch bang compa va thc ke mot oan thang th hai saocho hnh hop co canh la oan thang th hai co the tch gap oi hnhhop co canh la oan thang th nhat.

Menh e 4.8. Khong the dng c bang compa va thc ke mothnh vuong co cung dien tch vi mot hnh tron cho trc.

Chng minh. Ta biet rang khong phai la phan t ai so tren Q, doo

cung khong phai la phan t ai so tren Q. Vay, ta khong the

dng c hai iem cach nhau mot khoang bang

. Thc vay, neudng c hai iem nh vay th se dng c iem (

, 0). Nhng

ieu nay dan en mot mau thuan vi nh ly 4.2. Vay, ta khong thedng c mot hnh vuong co cung dien tch vi mot hnh tron chotrc.

nh ly 4.9. Cho K la tap hp tat ca cac so thc x sao cho (x, 0)la iem dng c. Khi o, K la trng con cua trng cac so thcR va iem (x, y) dng c khi va ch khi x, y K. Hn na, neux K va x > 0 thx K.Chng minh. T nh ngha iem dng c suy ra 0, 1 K. Gias x, y K, ngha la (x, 0) va (y, 0) la cac iem dng c. Theo

42


43/108

Bo e 4.3, ta dng c trung iem M cua oan thang co cac iemau mut la (x, 0) va (y, 0). Khi o, giao iem cua ng tron tamM i qua P0 vi truc hoanh se la iem (x + y, 0). Vay, x + y K.Hien nhien, neu (x, 0) la iem dng c th (x, 0) cung la iemdng c. Neu x, y K th theo Bo e 4.5, cac iem (x, 0), (0, y)

va (0, 1) eu dng c. Do o, iem (x, y1) cung dng c v nochnh la iem th t cua hnh bnh hanh co ba nh la (x, 0), (0, y)

va (0, 1) (xem Bo e 4.3). Khi o, ng thang i qua cac iem (0, y)va (x, y 1) cat truc hoanh tai iem (xy, 0). Do o xy K. Gias x K\ {0}. Neu x = 1 th x1 = 1 K. Gia s x = 1. Khio, (x, 0) va (1, 0) la cac iem dng c, keo theo (1 x, 0) va(0, 1x) la cac iem dng c. ng thang i qua cac iem (0, 1)

va (0, 1 x) se cat truc hoanh tai iem 1

x, 0

, keo theo1

x K.

Vay, K la mot trng con cua R. Hn na, neu x, y K th iem(x, y) dng c do no la nh th t cua hnh ch nhat nhan caciem (0, 0), (x, 0) va (0, y) lam ba nh. Ngc lai, gia s (x, y) la

iem dng c. Khi o, iem (x, 0) dng c. That vay, neu y = 0th ieu nay la hien nhien. Neu y = 0 th (x, 0) la giao iem cuatruc hoanh vi ng thang i qua (x, y) va iem oi xng vi noqua truc hoanh. Khi o, iem (0, y) la giao iem cua truc tung vang thang i qua (x, 0) va trung iem cua oan thang noi hai iem(0, 0) va (x, y). V (0, y) dng c nen (y, 0) cung dng c. Vay,x, y K.

Cuoi cung, gia s x K va x > 0. Khi o 12

(1 x) K, keo

theo C =

0, 12

(1 x) la iem dng c. Goi (u, 0) la giao iemvi truc hoanh cua ng tron tam C i qua iem (0, 1). Ban knh

43


44/108

cua ng tron nay la1

2(1 + x), do o, theo nh ly Pythagora ta co

1

4(1 x)2 + u2 = 1

4(1 + x)2.

T o suy ra u2 = x, hay u = x. Do (u, 0) la iem dng cnen x K.nh ly noi tren co the c s dung e giai quyet bai toan khi

nao mot a dien eu n canh co the dng c ch bang compa vathc ke neu biet trc tam va mot nh cua no. Nh ta a biet,khong the chia ba goc 3 , do o cung khong the dng c a dieneu 18 canh. Bay gi, gia s ta a dng c a dien eu n canh thbang cach chia oi mot goc, ta se dng c a dien eu 2n canh.

Vay, bai toan se dan ve van e dng a dien eu vi so le cac canh.Hn the na, ngi ta co the a bai toan ve viec dng a dien eu

vi so canh la luy tha cua mot so nguyen to le.

Nam 1796, tuoi 19, Gauss a chng minh rang a dien eu17 canh co the dng c bang compa va thc ke. S dung kythuat cua Ly thuyet Galois ngi ta a chng minh c rang a dieneu n canh la dng c bang compa va thc ke khi va ch khin = 2sp1p2 . . . pr, trong o p1, p2, . . . , pr la cac so nguyen to Fermatkhac nhau. Nhac lai rang, mot so nguyen to dang 2k + 1, trong o kla mot so nguyen dng, c goi la so nguyen to Fermat.

Neu k = uv, trong o u va v la nhng so nguyen dng va v la sole th 2k+1 = wv+1 = (w+1)(wv1wv2+. . .w+1), vi w = 2u,suy ra 2k + 1 khong phai la so nguyen to. Vay, so nguyen to Fermat

phai co dang Fm = 22m

+ 1, vi m la mot so nguyen khong am naoo. Nam 1640 Fermat a nhan thay rang Fm la so nguyen to, vi moim 4. Cac so nguyen to o la F0 = 3, F1 = 5, F2 = 17, F3 = 257

va F4 = 65537. T o Fermat a ra mot gia thuyet sai lam la moiso dang Fm eu la so nguyen to. Tuy nhien, nh bay gi chung ta abiet, F5 khong phai la so nguyen to. Hn the na, ngi ta a chng

44


45/108

minh rang vi 5 m 16, Fm khong phai la so nguyen to. Ngita cung a chng minh c rang cac so nguyen to Fm, m 4 chcho chung ta tat ca 32 a dien eu dng c bang compa va thcke vi so canh le.

Cuoi cung can lu y la bai toan dng a dien eu n canh bangcompa va thc ke tng ng vi bai toan tm cong thc e bieudien can bac n cua n v trong mat phang phc bi mot bieu thc

ai so, trong o s dung cac so hu t cung vi mot so hu han cacphep toan cong, tr, nhan, chia va khai can bac hai. Vay, ay la baitoan lien quan mat thiet vi bai toan bieu dien nghiem cua mot athc cho trc bi mot bieu thc ai so, trong o s dung cac he socua a thc di tac ong cua mot so hu han cac phep toan cong,tr, nhan, chia va khai can bac p, vi p la mot so nguyen to.

Bai tap

Bai 4.1. Chng minh rang co the dng c a dien eu 7 canhbang compa, thc ke va dung cu chia ba mot goc.

Bai 4.2. Chng minh rang co the chia ba goc bang compa vathc ke khi va ch khi a thc 4X3 3X cos() bat kha qui trenQ(cos()).

45


46/108

5. Nhom Galois va trng con co nh

nh ngha 5.1. Cho m rong trng K L. Ta noi mot t angcau cua L la motK-t ang cau neu (a) = a, a K.

nh ngha 5.2. Cho m rong K L. Ta goi nhom tat ca cac K-tang cau cua L la nhom Galois cua m rong K L va ky hieu laGal(L/K).

V du 5. Xet m rong R C. at : C C, . Tachng minh Gal(C/R) = {Id,}. That vay, v C = R(i), nen moiphan t cua Gal(C/R) eu hoan toan xac nh bi tac ong cuano len phan t i. Xet Gal(C/R). V i la nghiem cua a thcX2 + 1, nen (i) cung la nghiem cua no. Vay ch co hai kha nang:

(i) = i hoac (i) = i. Do o = Id hoac = . Vay Gal(C/R)la nhom xyclic cap 2.

V du 6. Xet m rong Q Q( 32). Gal(Q( 32)/Q), ta co

((3

2))3 = ((3

2)3) = (2) = 2.

V Q( 3

2) R, nen ( 32) R, do o ( 32) = 32. Vay = Id. Do o Gal(Q( 3

2)/Q) = {Id}.

V du 7. Cho K = F2(t) la trng cac thng cua mien nguyen F2[t]va F = F2(t2). a thc f(X) = X2t2 F[X] nhan t lam nghiem(t K). Hn na, ay la nghiem duy nhat v f(X) = (X t)2.

Vay f(X) bat kha qui tren F va do o f(X) = min(F, t), suy raF(t) = K va [K : F] = 2. Neu la mot t ang cau cua K tren Fth (t) = t, nen = Id. Vay Gal(K/F) = {Id}.

46


47/108

V du 8. Xet trng phan ra K cua a thc f(X) = X2 + X+ 1 F2[X]. Neu K la mot nghiem cua f th + 1 la nghiem con laicua f. Do o, K = F2() va moi t ang cau cua K tren F2 choan toan xac nh bi tac ong cua no len . Vay, Gal(K/F)ta co

() = () = + 1.

Do o, Gal(K/F) = {Id,}, trong o la t ang cau tren F2sao cho () = + 1.

Neu K L la mot m rong trng, th moi trng M thoa K M L c goi la mot trng con trung gian. Ro rang Gal(L/M)la mot nhom con cua Gal(L/K). Neu M N th Gal(L/M) Gal(L/N). Ta se ky hieu G(M) := Gal(L/M).

Bay gi ta xetL la mot trng bat ky va S Aut(L) la mot tapcon nao o cua nhom cac t ang cau cua L. at

F(S) := {x L|(x) = x, S}.

De dang kiem chng F(S) la mot trng con cua L. Ta goiF(S) la trng con co nh cua S. Neu K la trng con cua L vaS Gal(L/K) th F(S) la trng con trung gian cua L/K.

Bo e di ay cho chung ta mot so moi lien he gia cac nhomGalois va cac trng con co nh:

Bo e 5.3. Cho L la mot trng. Khi o ta co nhng ieu sau ay:

(i) Neu L1 L2 la cac trng con cua L thG(L2) G(L1).(ii) Neu K la trng con cua L thK FG(K).(iii) Neu S1 S2 la cac tap con cua Aut(L) thF(S2) F(S1).(iv) Neu S la tap con cua Aut(L) thS GF(S).

47


48/108

(v) Neu K = F(S) oi viS Aut(L) thK = FG(K).(vi) Neu H = G(K) oi vi trng con K cua L thH = GF(H).

Chng minh. Cac khang nh (i), (ii), (iii) va (iv) eu hien nhien.

(v) a K, G(K), do (a) = a nen a FG(K). VayK FG(K).

Ngc lai, v K = F(S) nen S G(K). T o suy ra FG(K) F(S) = K. Vay K = FG(K).(vi) Do (iv) H GF(H)). Do (ii) K FG(K). Ap dung (i) suy

ra GF G(K) G(K) = H.

Bo e 5.4. (Dedekind) Cho K, L la cac trng. Khi o moi tap hpnhng n cau khac nhau : K L la oc lap tuyen tnh tren L.Chng minh. Xet cac n cau khac nhau 1, . . . , n t K vao L.Gia s, vi a1, . . . , an L,

a11(x) + . . . + ann(x) = 0, x K. (1)Neu khong phai tat ca cac ai eu bang 0 th khong mat tnh tong

quat, co the gia thiet ai = 0, i. Gia s n la so t nhien nho nhatthoa man (1). Ta se a ra mot mau thuan.

V 1 = n, nen y K, sao cho 1(y) = n(y). Do o y = 0.Hien nhien bieu thc (1) van con ung neu ta thay x bi yx. Vay

a11(yx) + . . . + ann(yx) = 0, x K.

Do o

a11(y)1(x) + . . . + ann(y)n(x) = 0, x K. (2)

T (1) va (2) suy ra

a2(1(y) 2(y))2(x) + . . . + an(1(y) n(y))n(x) = 0. (3)

48


49/108

Trong (3) he so cua n(x) la an(1(y) n(y)) = 0, do o (3) lamot bieu thc dang (1), nhng co t hn n so hang. Ta co mot mauthuan. Vay Bo e a c chng minh.

nh ly 5.5. Gia s K la trng, G la mot nhom con cua Aut(K), K0la trng con co nh cua G. Khi o, K/K0 hu han neu va ch neuG la nhom hu han. Hn na, trong trng hp nay ta co

[K : K0] = |G|.

Chng minh. 1) Gia s [K : K0] = m < |G|, va {x1, . . . , xm} lamot c s cua K tren K0. Ton tai n > m, sao cho g1 = e, g2, . . . , gnla n phan t khac nhau trong G. He phng trnh tuyen tnh thuannhat

g1(x1)y1 + . . . + gn(x1)yn = 0;...........................................

g1(xm)y1 + . . . + gn(xm)yn = 0.

(4)

co so an nhieu hn so phng trnh nen co nghiem khong tam thng.Goi (1, . . . , n) Kn la mot nghiem khong tam thng cua (4).a K, a c viet mot cach duy nhat di dang

a = 1x1 + . . . + mxm, vi 1, . . . , m K0.

Do o

g1(a)1+. . .+gn(a)n = g1(mk=1 kxk)1+. . . gn(

mk=1 kxk)n

=

k(g1(xk)1 + . . . + gn(xk)n) = 0.

Vay, cac n cau g1, . . . , gn phu thuoc tuyen tnh tren K, mauthuan vi Bo e Dedekind. Do o m khong the nho hn |G|, keotheo G la mot nhom hu han. Vay, ta co the gia thiet |G| = n vaG = {g1 = e, g2, . . . , gn}.

49


50/108

2) Gia s [K : K0] > n. Khi o, ton tai mot tap hp gomn + 1 phan t oc lap tuyen tnh tren K0 cua K. Gia s tap o la{x1, . . . , xn+1}. Ton tai 0 = (1, . . . , n+1) Kn+1 sao cho

gj(x1)1 + . . . + gj(xn+1)n+1 = 0, j 1, n. (5)

Trong tap hp tat ca nhng bo (1, . . . , n+1) nh vay, chon mot

bo sao cho no co so cac phan t khac 0 t nhat. Khong lam mat tnhtong quat, co the gia thiet

1 = 0, . . . , r = 0, r+1 = . . . = n+1 = 0.

Khi o (5) tr thanh

gj(x1)1 + . . . + gj(xr)r = 0, j 1, n. (6)

Lay g G va tac ong g len hai ve cua (6), nhan c

ggj(x1)g(1) + . . . + ggj(xr)g(r) = 0, j 1, n. (7)V G = gG, nen (7) tr thanh

gj(x1)g(1) + . . . + gj(xr)g(r) = 0, j 1, n. (8)

Nhan (5) vi g(1), (8) vi 1 roi tr cho nhau, nhan c

gj(x2)(2g(1) g(2)1) + . . . + gj(xr)(rg(1) g(r)1) = 0.

ay la mot ho nhng bieu thc giong nh (5), nhng co t so cac

so hang hn, do o ta phai coig(1) 1g(i) = 0, i 2, r.

T o suy ra ig(1) = 1g(i), hay 11 i = g(i)g1(1) =

g(i11 ), g G. Vay i11 K0. at zi := i11 K0, ta co

50


51/108

i = 1zi, i 2, r. Neu atk := 1 K th i = kzi, i 2, r.Trong (6), neu cho j = 1, ta nhan c

x1k + x2kz2 + . . . + xrkzr = 0.

Do k = 1 = 0, nen t o suy rax1 + z2x2 + . . . + zrxr = 0,

ngha la cac phan t x1, x2, . . . , xr phu thuoc tuyen tnh tren K0,keo theo mot mau thuan. Vay [K : K0] = n = |G|.

Lu y rang, neu G la nhom hu han th nhng ieu chng minhtren van con hoan toan thch hp, ngha la khi o ta se suy ra [K : K0]hu han va [K : K0] = |G|.

He qua 5.6. Nhom Galois cua mot m rong hu han la hu han.Hn na, cap cua nhom Galois khong vt qua bac cua m rong.

Chng minh. Gia s L/K la m rong hu han va G = Gal(L/K).Goi K0 la trng con co nh cua G. Ro rang K

K0. Ap dung

nh ly 5.5, ta co|G| = [L : K0] [L : K].

Bai tapBai 5.1. Tnh cac nhom Galois sau ay:

(a) Gal(Q(

2)/Q).

(b) Gal(Q( 5

7)/Q).

(c) Gal(Q(2, 3)/Q).

Bai 5.2. Tnh nhom Galois Gal(F/Q), trong o F la trng phanra cua a thc

X4 2X3 8X 3.

51


52/108

Bai 5.3. Tnh nhom Galois cua nhng m rong sau:

(a) Q()/Q, vi = e2i/3.

(b) K/Q, trong o K la trng phan ra tren Q cua a thcX4 3X2 + 4.

6. Trng phan ra va nhng m rongchuan tac

nh ngha 6.1. Cho K la trng va f(X) K[X]. Ta noi a thcf phan ra tren K neu no phan tch thanh tch nhng nhan t tuyentnh

f(X) = a(X 1) . . . (X n),vi a, 1, . . . , n K.

Ro rang trong trng hp nay 1, . . . , n la tat ca cac nghiemcua a thc f(X).

Neu f(X) la a thc tren K va L la m rong cua K th f(X)cung la mot a thc tren L. Do o ta cung co the noi f(X) phan ratren L.

nh ngha 6.2. (1) Ta noi trng F la mottrng phan ra cua athc f(X) K[X], neu F la m rong cua K thoa cac ieu kien:

(1.1) f phan ra tren F;

(1.2) neu K F

F va f phan ra tren F

th F

= F.(2) Cho S la tap hp cac a thc bac 1 tren K. Ta noi trng

F la mot trng phan ra cua S tren K, neu F la m rong cua Kthoa cac ieu kien:

(2.1) f phan ra tren F, f S;

52


53/108

(2.2) neu K F F va f phan ra tren F, f S th F = F.Ro rang ieu kien (1.2) tng ng vi ieu kien:

(1.2) F = K(1, . . . , n), trong o 1, . . . , n la tat ca cacnghiem cua f.

ieu kien (2.2) tng ng vi ieu kien

(2.2) F = K(), trong o la tap hp tat ca cac nghiem cuatat ca cac a thc f S.Neu S{f1, . . . , f n} th trng phan ra cua S tren K chnh la

trng phan ra cua mot a thc f = f1 . . . f n tren K. Neu S la mottap bat ky (khong nhat thiet hu han) th luon luon ton tai trngphan ra tren K cua S. Hn na, trng phan ra nay duy nhat saikhac mot ang cau. Chung ta se chng minh ieu nay oi vi mot athc f(X) K[X]. Trng hp tong quat cung co the chng minhc nhng phc tap hn. Va lai, nhng ng dung sap ti trong giaotrnh nay khong can en trng hp tong quat nh vay. Do o chungta se han che viec chng minh oi vi mot a thc ma thoi. Nhng

trc khi lam ieu nay ta can en bo e sau:

Bo e 6.3. Cho p(X) la mot a thc bat kha qui tren trng K. Khio ton tai mot m rong F cua K trong o p(X) co nghiem.

Chng minh. Do p(X) bat kha qui tren K nen F = K[X]/p(X)la mot trng. Anh xa

: K K[X]a a

la mot n cau. Anh xa t nhien

: K[X] K[X]/p(X)la mot ong cau. Vay j = la mot n cau t K vao F, nen F lamot m rong cua K. at

= X+ p(X) F.

53


54/108

Ta co p() = p(X) + p(X) = 0 trong F. Vay la mot nghiem cuap(X) trong F.

nh ly 6.4. (Ve s ton tai cua trng phan ra) Cho K la mottrng va f(X) K[X]. Khi o ton tai trng phan ra cua f trenK.

Chng minh. Ta se chng minh bang qui nap theo bac n cua athc f.

Neu n = 1 th f phan ra tren K, do o K chnh la trng phan racua f tren K. Gia s f khong phan ra tren K. Khi o f co mot nhant bat kha qui f1 bac > 1. Goi 1 la mot nghiem cua f1 trong motm rong nao o cua K (ton tai do Bo e 6.3). Khi o f = (X1)g,

vi g K(1)[X] va degg = degf 1. Theo gia thiet qui nap tontai trng phan ra F cua g tren K(1). Ro rang F la trng phan racua f tren K.

Gia s : K L la mot n cau trng. Vi a K oikhi ta se dung ky hieu a e ch anh cua a qua n cau . Neuf(X) = a0 + a1X + . . . + anXn K[X] th ta ky hieu

f (X) = a0 + a1X+ . . . + a

nX

n.

Bo e 6.5. Cho ang cau trng : K L va cac m rongn K(), L(), trong o va la cac phan t ai so tng ngtren K va L. Gia s p(X) = min(K, ), q(X) = min(L, ) vap(X) = q(X). Khi o ton tai mot ang cau j : K() L() saocho j|K = va j() = .Chng minh. Moi phan t cua K() eu co dang f() vi f(X) K[X]. atj(f()) = f(), ta chng minh j chnh la ang cau can

xay dng. That vay, neu j(f()) = 0 th f() = 0, keo theo fchia het cho q(X) = p(X) trong L[X]. T o suy ra f(X) chia hetcho p(X) trong K[X], keo theo f() = 0. Vay, j la n cau. Bay gixet mot a thc bat ky h(X) L[X]. Khi o, vi f(X) = h1(X)ta co j(f()) = h(), ngha la j con la toan cau. Ta a chng minhj la ang cau.

54


55/108

Bo e 6.6. Cho ang cau trng : K K. Gia s f(X) K[X], F la trng phan ra cua f tren K va L la m rong cua K

sao cho a thc f phan ra tren L. Khi o, ton tai mot n cauj : F L sao cho j|K = .Chng minh. Ta se chng minh s ton tai cua j bang qui nap theobac cua a thc f. Theo gia thiet, a thc f phan ra tren F:

f(X) = a(X 1) . . . (X n).at p(X) = min(K, 1). Khi o p|f, t o suy ra p|f, ma fphan ra tren L, nen p cung phan ra tren L:

p = (X 1) . . . (X r), 1, . . . , r L.V p bat kha qui tren K nen p = min(K, 1). Theo Bo e 6.5,ton tai ang cau

j1 : K(1) K(1)sao cho

j1|K = va j1(1) = 1.

Ro rang F la trng phan ra cua a thc g = f /(X1) tren trngK(1). Theo gia thiet qui nap, ton tai mot n cau j : F L saocho j|K(1) = j1. Nhng khi o j|K = va ta co pcm.

nh ly 6.7. (Ve s duy nhat cua trng phan ra). Cho : K K la mot ang cau trng. Gia s F la trng phan ra cua f trenK va F la trng phan ra cua f tren K. Khi o ton tai mot angcau j : F F sao cho j|K = .Chng minh. Theo Bo e 6.6, ton tai n cau j : F

Fsao cho

j|K = . Nhng j(F) la trng phan ra cua f tren Kva j(F) Fnen j(F) = F. Vay j la ang cau.

Neu trong nh ly tren lay K = K va = IdK la phep ongnhat tren K th ta se nhan c s duy nhat cua trng phan ra cuamot a thc (sai khac mot s ang cau).

55


56/108

nh ngha 6.8. Ta goi m rong K L la m rong chuan tac, neumoi a thc f(X) K[X] bat kha qui tren K va co nghiem trongL eu phan ra tren L.

V du 9. 1) R C la m rong chuan tac.2) Q Q( 32) khong phai m rong chuan tac, v a thc x3 2

bat kha qui tren Q, co nghiem

3

2 trong Q(3

2), nhng hai nghiemcon lai la nhng nghiem phc, nen khong nam trong Q( 32).

nh ly 6.9. M rong K L la chuan tac va hu han khi va ch khiL la trng phan ra cua mot a thc nao o tren K.

Chng minh. Gia s K L la chuan tac va hu han. Khi o co thevietL di dang

L = K(1, . . . , n),

trong o i la nhng phan t ai so tren K. at

f = p1 . . . pn,

trong o pi = min(K, i).

V L chuan tac tren K nen moi a thc pi eu phan ra tren L,do o f cung phan ra tren L. V L c sinh ra bi K va mot songhiem cua f, ong thi L cha tat ca cac nghiem cua f, nen L latrng phan ra cua f tren K.

Ngc lai, gia s L la trng phan ra cua a thc g tren K. Khio, hien nhien L/K la m rong hu han. Ta ch con can chngminh L/K la m rong chuan tac. Vay, gia s a thc bat kha quif

K[X] co nghiem trong L. Ta can phai chng minh f phan ra

tren L. Goi M L la trng phan ra cua a thc f g tren L. Gia s1, 2 la hai nghiem cua f trong M. Ta se chng to

[L(1) : L] = [L(2) : L].

Xet bieu o di ay:

56


57/108

dd

K

K(1) K(2)

dd

L

L(1) L(2)

ddd

M

Hnh 2

trong o cac day cung ch quan he bao ham. Vi j = 1, 2 ta co

[L(j) : L][L : K] = [L(j) : K] = [L(j) : K(j)][K(j) : K].(1)

V 1, 2 eu la nhng nghiem cua a thc bat kha qui f tren Knen

[K(1) : K] = degf = [K(2) : K]. (2)

Do L la trng phan ra cua g tren K nen L(j) la trng phanra cua g tren K(j). Ngoai ra ta co

K(1) K(2)1 2

la mot ang cau tren K (ngha la gi nguyen cac phan t cua K),nen no thac trien thanh ang cau gia cac trng phan ra

L(1) L(2).

Do o[L(1) : K(1)] = [L(2) : K(2)]. (3)

Ket hp (1), (2) va (3) suy ra

[L(1) : L] = [L(2) : L].

57


58/108

T ang thc cuoi cung suy ra, neu 1 L th 2 L, do oL/K la m rong chuan tac.

Bai tap

Bai 6.1. Xay dng trong C cac trng phan ra tren Q cua cac athc

X3 1, X4 + 5X2 + 6 va X6 8.

Bai 6.2. Tm bac tren Q cua cac trng noi trong Bai tap 6.1.

Bai 6.3. Xay dng trng phan ra cua a thc X3 + 2X + 1 trenZ3.

Bai 6.4. Xay dng trng phan ra cua a thc

X3 + X2 + X + 2

tren trng Z3.

Bai 6.5. Co bao nhieu a thc n khi bac 2 tren trng Zp?

Bai 6.6. Liet ke tat ca cac a thc n khi bac 2 tren Z5. Tmtrong so o nhng a thc bat kha qui tren Z5. Xay dng cac trngphan ra cho cac a thc bat kha qui ay.

Bai 6.7. Cho f(X) la a thc bac n tren trng K va L la trng

phan ra cua f tren K. Chng minh rang [L : K] chia hetn!.

Bai 6.8. Cac m rong nao di ay la chuan tac:

a) Q(5)/Q; b) Q( 75)/Q.

c) Q(

5, 7

5)/Q( 7

5); d) R(7)/R.

58


59/108


60/108

Tat ca cac nghiem nay eu khac nhau, do o f5(X) la a thctach c tren Q.

Tong quat hn, vi moi so nguyen top, a thc chia ng tronfp(X) tach c tren Q.

2) Xet trng Zp, p la so nguyen to va K = Zp(u), trong o ula mot phan t sieu viet tren Zp (mot phan t c goi la phan t

sieu viettren mot trng, neu no khong phai la phan t ai so trentrng o). Xet a thc

f(X) = Xp u K[X].

Goi L la trng phan ra cua f(X) tren K, va la mot nghiemcua f trong L. Ta co

f(X) = Xp u = Xp p = (X )p.

Vay f(X) ch co mot nghiem boi duy nhat bac p la . e chngminh f(X) khong tach c tren K, ta ch can chng minh f(X)bat kha qui tren K. Vay, gia s f = g h, g ,h K[X] va bac cuag, h thc s nho hn bac cua f. Khi o

g(X) = (X )s, 0 < s < p.

T o suy ra s K. V (s, p) = 1 nen ton tai a, b Z sao chosa + bp = 1. T o suy ra

= sa+bp = (s)a(p)b = (s)a.ub K.

Do o viet c di dang

=v(u)

w(u), vi v(u), w(u) Zp[u].

60


61/108


62/108

Do o

Df = 2(X )g(X) + (X )2Dg = (X )[2g + (X )Dg].

Vay la nghiem chung cua f va Df. Do o f va Df eu chiahet cho min(K, ).

Ngc lai, gia s f khong co nghiem boi trong trng phan ra cua

no. atn = degf. Neu n = 1 th Df K, do o f va Df khongthe co c chung bac 1. Vay, gia s n > 1 va h(X) K[X] lac chung bac 1 cua f va Df. Gia s la mot nghiem cua h(X)trong trng phan ra L cua no. Khi o, la nghiem chung cua f vaDf. Do o

f(X) = (X )g(X), trong o g(X) L[X].T o

Df = (X )Dg + g,keo theo la nghiem cua g. Vay, la nghiem boi bac 2 cua f vata co mot mau thuan. ieu nay chng to f va Df khong the co cchung bac 1 trong K[X].

Bay gi ta co the a ra mot ieu kien can va u e mot a thcbat kha qui la tach c.

Menh e 7.4. (i) Neu K la trng ac trng 0 th moi a thc batkha qui tren K eu tach c tren K;

(ii) neu K la trng ac trng p > 0 th a thc bat kha quif(X) K[X] khong tach c tren K khi va ch khif(X) K[Xp],

ngha la khi va ch khi

f(X) = b0 + b1Xp + b2(X

p)2 + . . . + br(Xp)r;

(iii) neu charK = p > 0 va f(X) la mot a thc bat kha quitren K th ton tai mot so nguyen m 0 va mot a thc g K[X],bat kha qui, tach c tren K sao cho f(X) = g(Xp

m

).

62


63/108

Chng minh. Theo Bo e 7.3, a thc bat kha qui f(X) K[X]khong tach c tren K khi va ch khi f va Df co c chung bac 1. V f bat kha qui va Df co bac nho hn bac cua f nen khi ota phai co Df = 0. at

f(X) = a0 + a1X+ . . . + anXn.

Ta coDf = a1 + 2a2X+ . . . + nanX

n1 = 0 neu va ch neu kak = 0,

k 1, n.(i) Trng hp char(K) = 0: T ang thc kak = 0 suy ra

ak = 0(k 1, n). Vay, neu f(X) la a thc bat kha qui bac 1 thf tach c tren K.

(ii) Trng hp char(K) = p > 0: T ang thc kak = 0 suy raak = 0 neu p |k. atbi = aip, suy ra

f(X) = b0 + b1Xp + b2(X

p)2 + . . . + br(Xp)r.

(iii) Goi m la so nguyen khong am ln nhat sao cho f(X) K[Xp

m

]. Lu y rang som nh vay luon ton tai v f K[Xp0] va chton tai mot tap hu han nhng so r 0 e f K[Xpr], do moi athc trong K[Xp

r

] co bac khong be thua pr. Vay f(X) = g(Xpm

).Do tnh toi ai cua m, g(X) K[Xp]. Hn na g(X) bat kha qui trenK. That vay, gia s g(X) = h(X)k(X). Khi o f(X) = g(Xp

m

) =h(Xp

m

)k(Xpm

), keo theo mot mau thuan vi tnh bat kha qui cuaf(X) tren K. Theo (ii), g(X) tach c tren K.

T chng minh Menh e 7.4 (ii), ta thay trong trng hp

char(K) = p > 0 th a thc bat kha qui f khong tach c tren Kkhi va ch khi f(X) = g(Xp), vi mot a thc nao o g(X) K[X].

nh ngha 7.5. (i) Mot a thc bat ky f K[X] c goi la tachc tren K, neu moi nhan t bat kha qui cua no eu tach c trenK;

63


64/108

(ii) cho L/K la m rong trng va L la mot phan t ai sotren K. Khi o ta noi tach c tren K, neu min(K, ) tach ctren K. Ta noi m rong L/K tach c neu moi phan t L eutach c tren K.

V du 11. Cho K la trng ac trng 0. Khi o moi a thc trongK[X] eu tach c tren K. T o suy ra moi m rong ai so cua

K eu tach c tren K.Neu K la trng ac trng p > 0 th K(X) khong phai la m

rong tach c tren K(Xp), v a thc min(K(Xp), X) = tp Xpch co mot nghiem duy nhat la t = X.

Menh e 7.6. Neu L/K la m rong tach c va K M L thM/K va L/M la nhng m rong tach c.

Chng minh. Hien nhien M/K tach c. K, xetpK(X) :=min(K, ) va pM := min(M, ). Ta co pM|pK trong M[X]. V tach c tren K nen pK tach c tren K, do o pMtach c trenM. Vay L/M la m rong tach c.

Bai tap

Bai 7.1. a thc nao trong cac a thc di ay

X3+1, X2+2X1, X6+X5+X4+X3+X2+X+1, 7X5+X1

tach c tren

a) Q; b) C; c) Z2; d) Z3; e) Z5;f) Z7; g) Z19.

Bai 7.2. Cho m rong ai so L/K. Ta noi phan t a L \ K lathuan tuy khong tach c tren K neu a thc toi tieu min(K, a)

64


65/108

ch co mot nghiem. Ta noi m rong L/K la thuan tuy khong tachc neu moi phan t cua L la thuan tuy khong tach c tren K.Chng minh rang a thuan tuy khong tach c tren K khi va ch khiton tai so t nhien r 1 sao cho apr K, trong o p = charK.

Bai 7.3. Cho m rong L/K va L la phan t khong tach ctren K. Chng minh rang ton tai so nguyen m

0 sao cho p

m la

phan t tach c tren K, trong o p = charK.

Bai 7.4. Cho K la trng ac trng p > 0 va a K Kp. Chngminh rang Xp a la a thc bat kha qui tren K.

Bai 7.5. Cho K la trng ac trng p > 0 va L la m rong thuan tuykhong tach c tren K bac pn. Chng minh rang ap

n K, a L.

Bai 7.6. Cho day m rong trng K L F sao cho F/L chuantac va L/K la m rong thuan tuy khong tach c. Chng minh

rang F/K la m rong chuan tac.

Bai 7.7. Cho L, F la cac m rong cua trng K va cung nam trongmot trng ln M nao o. Chng minh rang, neu L va F cung tachc tren K th LF cung tach c tren K. ieu ngc lai co ungkhong?

Bai 7.8. Cho L, F la cac m rong cua trng K va cung nam trongmot trng ln M nao o. Chng minh rang, neu L va F cung thuantuy khong tach c tren K th LF cung thuan tuy khong tach c

tren K. ieu ngc lai co ung khong?

65


66/108

8. Bao chuan tac

nh ngha 8.1. Gia s K la trng con cua cac trng M va L.Neu

: M L

la mot n cau trng sao cho (a) = a, a K th ta noi la motK-n cau t M vao L. Ta con noi la mot n cau tren K

nh ly 8.2. Gia s L la mot m rong chuan tac hu han trenK, K M L va

: M Lla motK n cau. Khi o ton tai motK-t ang cau cua L saocho |M = .Chng minh. V L/K la chuan tac va hu han, nen theo nh ly 6.9,L la trng phan ra tren K cua mot a thc f K[X] nao o. Do la K-n cau nen f = f va K

(M). Vay, L con ong thi la

trng phan ra tren M va tren (M) cua f. Do o theo nh ly 6.7,ton tai mot ang cau : L L sao cho |M = . V la K-ncau, nen la K-t ang cau.

S dung nh ly va chng minh ta co the xay dng cac K-tang cau cua mot m rong chuan tac hu han L tren K nh sau:

Menh e 8.3. Gia s L/K la mot m rong chuan tac hu han va, la cac nghiem nam trong L cua mot a thc p(X) bat kha quitren K. Khi o ton tai motK-t ang cau cua L sao cho () = .

Chng minh. Theo Bo e 6.5, anh xa : K() K()

vi () = va gi nguyen cac phan t cua trng K la mot angcau tren K va do o la mot K-n cau t K() vao L. Theo nhly 8.2, m rong thanh motK-t ang cau cua L.

66


67/108

Neu mot m rong khong phai la chuan tac th ta se co gang mrong no thanh mot m rong chuan tac. Mot co gang nh vay a tati mot nh ngha ve bao chuan tac.

nh ngha 8.4. Cho L la mot m rong ai so tren K. Ta goi motbao chuan tac cua m rong L/K la mot m rong N cua L sao cho:

(1) N/K chuan tac.

(2) Neu L M N va M/K chuan tac th M = N.Vay N la m rong nho nhat cua L chuan tac tren K.

nh ly di ay cho ta mot ieu kien u e ton tai bao chuantac.

nh ly 8.5. Neu L/K la m rong hu han th ton tai bao chuantac N cua L/K. Hn na, N hu han tren K. Neu M la mot baochuan tac khac cua L/K th ton tai mot ang cau : M N trenK.

Chng minh. Gia s x1, . . . , xr la c s cua L tren K va mi =min(K, xi), i 1, r. Goi N la trng phan ra cua a thc f =m1 . . . mr tren L. Khi o ro rang N cung la trng phan ra cua ftren K (do L = K(x1, . . . , xr)). Do o theo nh ly 6.9, N/K la mrong chuan tac hu han. Gia s L P N va P chuan tac trenK. Moi a thc bat kha qui mi eu co mot nghiem xi L P, nendo tnh chuan tac cua P tren K, mi phan ra tren P. Suy ra f phanra tren P. V N la trng phan ra cua f nen P = N. Vay N la baochuan tac cua L/K.

Bay gi, gia s M la bao chuan tac khac cua L/K. a thc f

nh a neu tren se phan ra ca tren M va N. Vay, M va N chacac trng phan ra cua f tren K. Cac trng phan ra nay eu chaL va chuan tac tren K, do o chung tng ng bang M va N. To suy ra M N.

V du 12. Xet m rong trng Q Q( 32). a thc bat kha qui

67


68/108

X3 2 tren Q ch co mot nghiem nam trong Q( 32), do o m rongnay khong chuan tac. Goi K C la trng phan ra cua a thcX3 2 tren Q. Khi o K = Q( 32, w), vi w = e 2i3 . Vay K la baochuan tac cua m rong Q( 3

2)/Q.

Bo e 8.6. Cho mot day m rong trng

K L N M,trong o L hu han tren K va N la bao chuan tac cua L/K. Khio, neu : L M la motK-n cau th(L) N.Chng minh. Xet phan t bat ky L. V L la m rong hu hantren K nen la phan t ai so tren K. Goi p(X) = min(K, ). Khio () cung la mot nghiem cua p(X), nen () N.

Ket qua noi tren cho phep chung ta han che viec khao sat len baochuan tac cua mot m rong hu han cho trc, neu nh viec khaosat ch lien quan en nhng n cau.

nh ly di ay cho ta mot so ieu kien tng ng cua mrong chuan tac.

nh ly 8.7. oi vi m rong hu han L/K, nhng ieu kien diay tng ng:

(i) L/K chuan tac.

(ii) Ton tai m rong chuan tac N hu han tren K cha L saocho moiK-n cau : L N la motK-t ang cau cua L.

(iii) oi vi moi m rong chuan tac hu han M/K cha L, moi

K-n cau : L M la motK-t ang cau cua L.(iv) oi vi moi m rong chuan tac M/K cha L , moiK-n

cau : L M la motK-t ang cau cua L.Chng minh. (i)= (iv). Neu L chuan tac tren K th L la bao chuantac cua L/K. Gia s M/K chuan tac, L M va : L M la

68


69/108

motK n cau. Khi o, theo Bo e 8.6, (L) L. Nhng K-ncau co the xem nh mot n cau t khong gian vect hu hanchieu L tren K vao M. Do o dimK((L)) = dimK(L). T o suyra (L) = L va do o la K-t ang cau cua L.

(iv)= (iii). Hien nhien.(iii)= (ii). Theo nh ly 8.5, ton tai bao chuan tac N cua L/K,

N hu han tren K. Theo (iii), N/K chnh la m rong thoa nhng oihoi can thiet.

(ii)= (i). Gia s p(X) la mot a thc bat kha qui tren K va L la mot nghiem cua p(X). Khi o, do N chuan tac tren K nenp(x) phan ra tren N. Neu la mot nghiem khac cua p(X) th theoMenh e 8.3, ton tai motK-t ang cau cua N sao cho () = .Khi o |L : L N la motK-n cau, nen theo (ii), |L la motK-t ang cau cua L, do o = () L. Vay L/K la m rongchuan tac.

nh ly 8.8. Cho L/K la m rong hu han tach c bac n. Khio, ton tai ung n K-n cau khac nhau t L vao bao chuan tac Ncua m rong L/K.

Chng minh. Ta se chng minh bang qui nap theo bac n = [L : K].Neu n = 1 th L = K va ieu khang nh la hien nhien.

Gia s n > 1, va ieu khang nh la ung oi vi moi m ronghu han, tach c bac be thua n. xet phan t L\K, va p(X) =min(K, ). Khi o, deg(p) = r = [K() : K] > 1. V a thc batkha qui p co mot nghiem nam trong m rong chuan tac N nen pphan ra tren N. Do tach c tren K nen cac nghiem cua p eu langhiem n. Gia s cac nghiem o la 1, . . . , r. ats = nr . Theo

gia thiet qui nap, co ung s K()-n cau 1, . . . , s : L N.Theo Menh e 7.3, ton tai r K-t ang cau 1, . . . , r cua N sao choi() = i.

Cac anh xa

ij = ij, 1 i r, 1 j s

69


70/108

cho ta n = rs K-n cau L N khac nhau. That vay, gia s(i, j) = (k, l). Neu i = k hoac j = l th hien nhien ij = kl. Neui = k va j = l th ij() = i() = k() = kl(). T o suyra ij = kl. Bay gi ta se chng minh rang ngoai ra khong con coK-n cau nao khac t L N.

Vay, gia s : L N la motK-n cau bat ky. Khi o ()la mot nghiem cua p(X) trong N, nen () = i vi moti nao o.

Anh xa = 1i la motK()-n cau L N, nen theo gia thietqui nap = j vi motj nao o. Do o = ij = ij .

Nh co cac nh ly va chng minh ta co the tnh c cap cuanhom Galois cua mot m rong hu han, chuan tac va tach c. That

vay, ap dung cac nh ly 8.7 va 8.8, ta nhan c ngay ket qua diay:

He qua 8.9. Neu K L la m rong hu han, chuan tac, tach cbac n th co ung n K-t ang cau khac nhau cua L, do o

|Gal(L/K)

|= n.

Bay gi ta de dang suy ra ket qua quan trong sau ay:

nh ly 8.10. Cho K L la mot m rong hu han bac n vi nhomGalois G. Neu L/K la chuan tac va tach c thK la trng conco nh cua G.

Chng minh. Goi K0 la trng con co nh cua G va |G| = n. Theonh ly 5.5, [L : K0] = |G|. Theo He qua 8.9, n = [L : K] = |G|.Nhng do K K0 nen t o suy ra K = K0.

ieu ngc lai cua nh ly 8.10 cung ung. Tuy nhien, trc khichng minh ieu ay ta can en mot ket qua di ay:

nh ly 8.11. Cho day m rong trng

K L M

70


71/108

sao cho [M : K] hu han, [L : K] = n va L tach c tren K. Khio, ton tai nhieu nhat la n K-n cau L M.Chng minh. Goi F la bao chuan tac cua M/K. Theo nh ly 8.5,F la m rong hu han tren K. Moi K-n cau L M co thexem nh motK-n cau L F. Vay so cac K-n cau L Mkhong vt qua so cac K-n cau t L vao bao chuan tac F cua M.Bay gi ta se chng minh co ung n K- n cau L

F. That vay,

ta thay F la m rong chuan tac va hu han tren K, nen ro rang Fcha mot bao chuan tac N cua L/K:

K L N F.

T Bo e 8.6 suy ra moi motK-n cau L F eu co the xemnh motK-n cau L N. Ma theo nh ly 8.8 ch co ung nK-n cau L N, nen suy ra ch co ung n K-n cau L F.

nh ly 8.12. Cho m rong hu han L/K vi nhom Galois G =

Gal(L/K). Neu K la trng con co nh cua G th L/K la chuantac va tach c.

Chng minh. Theo nh ly 5.5, [L : K] = |G| = n. Vay co ung nK-t ang cau cua L. Tng t nh trong chng minh nh ly 8.8,ta co the chng minh c rang neu L khong tach c th co t hnn K-n cau L N, vi N la bao chuan tac cua L/K. Do o Lphai tach c tren K. Ta ch con can chng minh L la m rongchuan tac tren K. Vay, gia s M la m rong chuan tac hu han trenK cha L va : L M la mot K-n cau. Theo nh ly 8.11,co khong qua n K-n cau L M. Nhng moi mot phan t cuanhom Gal(L/K) eu co the xem nh motK-n cau L M, maco ung n phan t nh vay, nen phai la mot trong nhng phan tcua nhom Gal(L/K). Ap dung nh ly 8.7, suy ra L/K la m rongchuan tac.

nh ngha 8.13. Cho m rong hu han L/K. Ta noi L la m rongGalois tren K neu K = F(Gal(L/K)).

71


72/108

Bo e 8.14. M rong hu han L/K la m rong Galois khi va ch khi[L : K] = |Gal(L/K)|.Chng minh. Neu L/K la m rong Galois th theo nh ngha K latrng co nh cua G := Gal(L/K). Khi o, theo nh ly 5.5, ta co

[L : K] = |G|.

Ngc lai, gia s [L : K] = |G|. Goi K0 la trng con co nhcua G. Khi o, theo nh ly 5.5

[L : K0] = |G|.

V K K0, nen K = K0, keo theo L la m rong Galois tren K.

nh ly 8.15. Cho m rong hu han L/K. Khi o cac ieu kiendi ay tng ng:

(i) L/K la m rong Galois.(ii) L/K la m rong chuan tac va tach c.

(iii) L la trng phan ra cua mot a thc tach c tren K.

Chng minh. (i) = (ii). Ap dung nh ly 8.12.(ii) = (iii). Do L/K la m rong chuan tac va hu han nen theo

nh ly 5.9, L la trng phan ra tren K cua mot a thc f K[X].Gia s p la mot nhan t bat kha qui bat ky cua f. V moi nghiem cuap cung la nghiem cua f va theo gia thiet cac nghiem nay eu tachc tren K nen p tach c tren K, keo theo f tach c tren K.

(iii) = (i). Ta se chng minh bang qui nap theo bac n = [L : K].Neu n = 1 th L = K, do o ieu khang nh la hien nhien. Vay,

gia s n > 1 va ieu khang nh la ung oi vi moi m rong bacnho hn n. Gia s L la trng phan ra cua a thc f tach c trenK. V [L : K] = n > 1 nen ton tai phan t L \ K sao cho

72


73/108

la nghiem cua f. atF = K(). Khi o [F : K] > 1, keo theo[L : F] < n. Hien nhien a thc f cung tach c tren F va L latrng phan ra cua f tren F. Do o, ap dung gia thiet qui nap suyra L la m rong Galois tren F. atH := Gal(L/F) Gal(L/K).a thc bat kha qui p(X) = min(K, ) la c cua a thc f tachc tren K nen cung tach c tren K. Goi 1, . . . , r la tat cacac nghiem khac nhau cua p(X). Khi o [F : K] = r. i 1, r, anh

xa i : K() K(i), vi i va gi nguyen cac phan t cuaK la motK-ang cau. Xem L ong thi nh cac trng phan ra cuaF = K() va K(i) roi ap dung Bo e 5.7, ta tm c mot phan ti Gal(L/K) sao cho i() = i. Ta se chng minh cac lp ke traiiH oi mot khac nhau. That vay, gia s i = j va iH = jH. Khi o1i j H = Gal(L/F), keo theo 1i j() = hay i() = j(),ngha la i = j . Nhng ay la ieu mau thuan vi cach at cacphan t i. atG := Gal(L/K) va ap dung ieu mi chng minhta co

|G| = [G : H]|H| r|H| = [F : K]|H| = [F : K][L : F] = [L : K].

Mat khac, theo He qua 5.6, ta co |G| [L : K]. Vay, [L : K] =|G|. Theo Bo e 8.14, L/K la m rong Galois.

He qua 8.16. Cho L/K la mot m rong hu han. Khi o:

(i) Neu L = K(1, . . . , n) , vi cac phan t i eu tach ctren K th L tach c tren K; (ii) L tach c tren K neu va ch

neu L nam trong mot m rong Galois nao o cua K.

Chng minh. (i) Gia s L = K(1, . . . , n) va moi i eu tach ctren K. Goi F la trng phan ra tren K cua a thc

f(x) :=n

i=1

min(K, i).

Ro rang f tach c tren K , do o theo nh ly 8.15, F tachc tren K , keo theo L tach c tren K.

73


74/108

(ii) Neu L F va F/K la m rong Galois th theo nh ly 8.15,F tach c tren K. T o suy ra L tach c tren K. Ngc lai,gia s L tach c tren K. Do L/K hu han nen ta co the vietL = K(1, . . . , n), trong o moi i eu tach c tren K. atFla trng phan ra tren K cua a thc

f(x) :=

ni=1

min(K, i).

Theo nh ly 8.15, F la m rong Galois tren K.

Bo e 8.14 cho chung ta mot tieu chuan e chng minh khi naomot m rong hu han la m rong Galois. Tuy nhien, e s dungno ta can phai tnh c nhom Galois cua m rong. Ma ieu nay laikhong phai luc nao cung de dang. Di ay, ta xet mot trng hprieng khi L la m rong n tren K.

Menh e 8.17. Cho m rong L/K va a L la mot phan t ai sotren K. Khi o nhom Galois Gal(K(a)/K) la hu han va co capbang so cac nghiem khac nhau trong K(a) cua a thc min(K, a).Suy ra K(a) la m rong Galois tren K khi va ch khi a thcmin(K, a) co ung n nghiem khac nhau trong K(a) , vin la baccua min(K, a).

Chng minh. Gia s G := Gal(K(a)/K). Khi o (a) cung langhiem cua min(K, a). Hn na, neu va la hai phan t khacnhau cua G th (a) = (a). T o suy ra tng ng (a) la motn anh t G vao tap hp tat ca cac nghiem cua min(K, a) trongK(a).

Bay gi, gia s la mot nghiem bat ky cua min(K, a) trong K(a).Do K() K(a) va [K() : K] = [K(a) : K] nen K() = K(a).

Anh xa : K(a) K()

74


75/108

gi nguyen cac phan t cua K va (a) = xac nh motK-ang caugia K(a) va K(). Nhng K(a) = K() nen Gal(K(a)/K).

Vay, tng ng (a) xac nh mot song anh gia Gal(K(a)/K)va tap hp tat ca cac nghiem nam trong K(a) cua a thc min(K, a).ieu khang nh sau cung la hien nhien.

V du 13.Q( 3

2)/Q khong phai la m rong Galois. That vay, a thc

x3 2 = min(Q, 32) co ba nghiem khac nhau, nhng ch co motnghiem nam trong Q( 32), nen |Gal(Q( 32)/Q)| = 1, trong khi oth [Q( 3

2) : Q] = 3.

V du 14. Cho k la trng ac trng p > 0 va k(t) la trng cac phanthc hu t bien t tren k. Xet m rong k(t)/k(tp). Ta thay t la nghiemcua a thc Xp tp, bat kha qui tren k(tp). Mat khac, trong k(t) athc nay ch co duy nhat mot nghiem X = t v Xp tp = (X t)p.

Vay|Gal(k(t)/k(tp))| = 1 = p = deg(min(k(tp), t)).

Do o k(t)/k(tp) khong phai la m rong Galois.

V du 15. Cho K la mot trng. Gia s a K la mot phan t saocho a thc p(X) = X2 a khong co nghiem trong K. Goi E lamot m rong cua K cha nghiem cua p(X) va E la mot nghiemnh vay. Khi o, K() cha ca hai nghiem cua p(X) nen theo Menhe 8.17, |Gal(K()/K)| = 2 = [K() : K]. Do o K()/K la mrong Galois.

Bai tap

Bai 8.1. Xay dng bao chuan tac N cua nhng m rong di ay:

75


76/108

(a) Q( 5

3)/Q.

(b) Q( 7

2)/Q.

(c) Q(

2,

3)/Q.

(d) Q( 3

2,

2)/Q.

(e) Q()/Q, vi la nghiem cua a thc X3 3X2 + 3.

Bai 8.2. Tnh nhom Galois cua cac m rong trong (a)-(d) cua Bai8.1.

Bai 8.3. Tnh cac nhom Galois cua N/Q, trong o N la bao chuantac cua cac m rong trong (a)-(d) cua Bai 8.1

9. nh ly can ban cua thuyet Galois

Xet m rong trng L K vi nhom Galois G. Trong 5 ta anh ngha hai anh xa H F(H) va M G(M) t tap hp tat cacac nhom con cua nhom Galois G = Gal(L/K) vao tap hp tat cacac trng con trung gian cua m rong L/K va ngc lai. Chung tacung a nhan xet rang cac anh xa noi tren ao ngc quan he baoham va M F(G(M)), H G(F(H)). Di ay ta se chng minhmot nh ly rat quan trong, c goi la nh ly can ban cua thuyetGalois. Nhng trc het ta can en mot bo e sau ay:

Bo e 9.1. Cho M la trng con trung gian cua m rong L/K va la K-t ang cau cua L. Khi o

G((M)) = G(M)1.Chng minh. at M = (M) va lay G(M), x1 M. Khi o

x1 = (x), vi x M va ta co 1(x1) = (x) = (x) = x1.

76


77/108

Vay G(M)1 G((M)).Tng t, chng minh c 1G(M) G(1(M)). T o

suy ra 1G((M)) G(M). Vay G((M)) = G(M)1.

nh ly 9.2. (nh ly can ban cua thuyet Galois) Cho L/K la mrong hu han, chuan tac va tach c bac n vi nhom Galois G. Khio ta co nhng ieu sau ay:

(1) |G| = n.(2) Cac anh xa G va F la cac song anh ngc nhau ao ngc

quan he bao ham gia tap hp tat ca cac trng con trung gian cuam rong L/K va tap hp tat ca cac nhom con cua nhom GaloisG = Gal(L/K).

(3) Neu M la trng con trung gian cua L/K th [L : M] =|G(M)| va [M : K] = |G|/|G(M)|.

(4) Trng con trung gian M la m rong chuan tac cua K khiva ch khi

G(M)G.

(5) Neu trng con trung gian M la m rong chuan tac cua KthGal(M/K) G/G(M).Chng minh. (1) Xem He qua 8.9.

(2) Xet trng con trung gian M bat ky cua m rong L/K. VL/K tach c nen L/M cung tach c. Do L/K chuan tac nentheo nh ly 5.9, L/M cung chuan tac. Vay, theo nh ly 8.15, L/Mla m rong Galois. Do o M la trng con co nh cua G(M). Vay

F(G(M)) = M. ()

Ngc lai, gia s H la mot nhom con cua nhom Galois G. Theonh ly 7.5, |H| = [L : F(H)]. Theo chng minh pha tren, L/F(H)la m rong Galois nen [L : F(H)] = |G(F(H))|. Hn na, nh ta abiet trong 6, H G(F(H)), nen t o suy ra H = G(F(H)).

Ta a chng minh (2)

77


78/108

(3) Nh a noi trong chng minh (2), oi vi moi trn

lý thuyết trường và galois

Documents