X goila idean nguyen t6 ngu• va chi lieu veil u, u E tich
Ea h Eb it/ ) =G ah(bfj )=
h+k=m +j=kh+i+j=m
E (alibi) = E ahbi )
h+i+j=m j+1=m ‘1.1-4-i=1
tit dd to al Agri nhan trong P la kat hop. Day (1, 0,..., 0,...)
la !Than tis don vi cim P. Vay P la 1111# vi nhdm nhan giao hoan t Cu6i cling lug phan phdi trong A cho phen ta vigt
E cti(bi + = E alb,/ + E aPj i+j=k
vol moi k = 0, 1, 2, ...,
trong - P.
Bay gib ta hay Kat day
x =
at co then quy tac nhan x2 = k3 = • • = i+j=k i+j=k ta suy ra (0, 1, 0 (2) (0, 0, 1, Q... (0, 0, 0, 1,.. (0, 0,..., 0, 1,
tit do lust phan phdi
0,
n
lh quy valc vi&t
x ° = (1, 0, Mat khac ta xet anh xa
A v P
a (a, 0„.., 0, ..)
Xnh xa nay hign nhien la mat don eau (vanh) Do do tit gib
ta clang nhat plign to a E A veil day (a, 0, ..., 0, ...) E P, tit vi vay A la mat vanh con coo. vanh P Vi mai phan to cim P la mat day
(%,ate, ...)
trong do the ai bang 0 tat ca trit mat so` hitu hen, cho nen mai
phalli to caa P co clang
(cto, a lp ..., ac, 0,...) trong cid a
0 ,
E A kh6ng nhat thigt khac 0. \riga (long nhat a vOi (a, 0, 0, ...) va viac dua vao day x cho phop ta vigt
(a., %, 0, ) (a., 0, ) + (0, a1, 0, ..) +
+ (0, a., 0, ...) = (a., 0, ) + (a i , 0, ) (0, 1, 0, ...) + + + (an, 0, ) (0, . 0, 1, 0, ) = a. + aix + + ax" =
= anx ° + st lx + + an x".
Nob( to thubng ki hieu cac phan td elm P vier dual clang + a ix + + an?
bang f(x),
Dinh nghia 1. Vinh P goi la heath da thhc ctia tin x idly
he hi trong A, hay van tat vInh da thac cua an x tren A, va hiOu la A /X/ Cite Orin tit cua vanh dd goi la da thee oda
an x lay he trong A. Trong mat da thdc
f(x) = ox + a ix + + an?.
cac i = 0, 1, n gal la cac he hi caa da thdc. Cac as goi
la cac hang hi mist da thitc, dim biOt ace = o goi la hang hi hi do.
2. Bac ctla m01 da thac
Xet mat day (an, a. . , an, )
thuoc vanh P Vi cac ar bang 0 tat ca trb mot s6 hitu hart nen
nau
thi bao gib cUng cd mat chi s6 n sao cho a.x 0 ye a, = 0.
I > n. Theo nhu tren, to vita
(an , ..., an , 0, .) = + a ix + an?
Dinh nghin 2. Elbe caa da thud khac 0
f(x) = aox° + +an _ +
NMI vhy ta chi dinh nghia bac coa mat da flit khac 0. D61 vii da thdc 0 ta bao no khOng c6 bhc.
Dinh li 1. GU/ sit f(x) ua g(x) Ia hai da Mac kheic 0. (i) Neu bac f(x) khac bac g(x), thi ta co
f(x) + g(x) m 0 ua bac (f(x) + g(x)) = max (bac f(x), bac g(x)).
Neu be f(x) = bac g(x), on neu them nha f(x) + g(x) # 0, thi ta co bac (f(x) +g(x)) s max (bat f(x), t4c g(x)).
(ii) Neu f(x) g(x) x 0, thi ta co
bac (f(x) g(x)) Lc. bac f(x) + bac g(x).
Viac chung minh khOng c6 gi kh6 khan, xin nhubng cho ban doe
Dinh li 2. Neu A la mot mien nguyen f(x) tia g(x) la hai da Mac kluic 0 cart vanh A[x], thi f(x) g(x) # 0 Mr bac (f(x) g(x)) = bac f(x) + bac g(x).
Chang mink. Gia sd f(x), g(x) E A[x] la hai da thdc khac 0
f(x) = ac + + amain (am # 0)
g(x) = bo + + (bn # 0) Theo quy tac nhan da thdc ta
f(x)g(x)= aobo + ...+(a.bk + +akb o)xk + + amboxn+m .
am va b. Winn 0, nen amb # 0 (A khOng Tide cda kliOng), do
d6 f(x) g(x) # 0 va bac (f(x) g(x)) = m + n = bac f(x) + bac g(x)1 HO qua. Neu A la mien nguyen, thi A [x] cling In mien nguyen.
3. Phep chia vii du
Trong muc 2 to da thdy ngu A la mat mien nguyen thi A[x]
clang la mat mien nguyen. Ta td dat eau hei : ngu A la mat
tutting thi A [x] co phai la mat tuning khOng ? Cau hOi duqc
tit lbi ngay hie khac, A[x] khong phai la met truing vi da three_ x chang han khong co nghich &to. Thy fly trong truing
hop nay A[x] la met mien nguyen dew Met, no la met vanh oclit
(eh V, §2) nghla la met vanh trong di( cc( phep chic voi du.
Dinh 11 3. Gid sit A la mot trilling, f(x) va g(x) x 0 la hai da three ezia vanh A[x] ; thd thz bao gib wing co hai da three day nhat q(x) va r(x) thitec A silo cho
f(x) = g(x) q(x) + r(x), noi bdc r(x) < bdc g(x) nen r(x) # O.
Cluing minh. TrUtIc hdt ta hay chling minh tinh duy nha fix) = g(x) q'(x) + bac r'(x) < be g(x)
neu r'(x) # 0 Ta suy ra
0 = g(x) (q(x) - ex)) + r(x) - r'(x).
IsIdu r(x) = r'(x), ta co g(x) (q(x) - q'(x)) = 0, vi g(x) # 0 va A[x]
la met mien nguyen, nen, suy m q(x) - q'(x) = 0 We la q(x) = q'(x). Gia sii r(x) # r'(x), vay
bac (r(x) - r'(x)) = her (g(x)(q(x) - qrx)) = bac g(x) + bac (q(x) - &fix))
(dinh If 2).
Mat khac theo gia thiet va dinh II 1
bac (r(x) - r'(x)) S max (bac r(x), bac r'(x)) < bac g(x) bac g(x) + bac (q(x) - qlx)),
&du nay mau thuan voi clang thile tren. Chu y : nau met trong
hai da thdc r(x) va r'(x) bang 0 thi ta khong thd nit deli bac
nhung then do khong anh hureng Uri viec chdng minh,
vi 10c do bac (r(x) - r'(x)), bang W.c r(x) nen r'(x) = 0 va bang
bac r'(x) ndu r(x) = 0.
Con sit ton tai cua q(x) va r(x) thi suy ra tii thuat Wan dud! day. Tim q(x) va r(x) goi la three hien phep chia f(x) cho g(x). Da Utak q(x) goi la thuong, da thttc r(x) la du cua f(x) cho g(x). Viec tim thuong va du la tic Mac ndu b4c f(x) < bac g(x). Ta
chi can dat q(x) = 0, r(x) = f(x). Trong twang hap trai lai ta dung nhan xet eau day :
Neu ta biet mot da attic h(x) sao cho fi (x) = f(x) - g(x) h(x)
cc) bac tittle stir be hen bac cua f(x) thi bai Man tra thanh don gian tam : tim thadng va du cha fi (x) cho g(x). That vay, nee
fi (x) = g(x) q t (x) + r i(x), ta guy ra
f(x) = g(x) (h(x) + q i(x)) + r i (x)
tit do
q(x) = h(x) + q i(x), r(x) = r i (x)
Trong tittle tihn, vat
f(x) = + + + ao
g(x)=bne +bn_ixn 1 + +be , br, m 0 va n 5 m to nhan xet am
rang, My h(x) = thi da thtic fi (x) =
ce bac tittle sit be him bac cfm f(x), hoac fi (x) tniang hop f (x) = 0, dv r(x) = 0 va thitang q
fi (x) x 0 ta tip tut vai fi (x), ta dude f2(x)... Day da tilde fi(x), f2 (x)... ce bac giam dam Khi ta di den mot da tittle ce bac flute sit be ban bac mitt g(x) thi da thIc d6 chinh la du r(x.). Ngu
mot da thtic cfm day bang 0 thi dv r(x) = 0. De nhin thay 1.6 hon ta hay vast ra the bade ma ta da thuc hien d6 duqc day fi (x),
fi (x) = f(x) - g(x) h(x) f2 (x) = ft (x) - g(x) h i (x)
fk (x) = fk_ 1 (x) - g(x) hk_ k (x)
vai fk(x) = 0 hoac bac fk(X) < be g(x). Ding va yea ve cite clang thtic de 1ai, ta duqc
f(x) - g(x) h(x)
bang 0. Trong
(x) = h(x). Ngu
f(x) = g(x)(h(x) + h i (x) + + hk _ 1 (x)) + fk(x), tit do
q(x) = h(x) + h i(x) + + hk _ 1 (x), r(x) = fk(x). n