引理1:若f=gq+r,则f,g)=(r,g) poof:由(f,g)f,(f,g)g,知(f,g) 故(f,g)(,g);同理(r,g)(f,g) f=8q1+,deg≤degg deg r2 degr q3+ rs-2=rs-19s+rs degrs degrs s-1=rs95+ (f,g)=(g,1)=…=(-1,r)=C
6 , , [ ] 1 , ( , ) ( , ). f g r F X f gq r f g r g 引理 ：若 = + 则 = ( , )( , ); ( , )( , ). : ( , ) , ( , ) , ( , ) , f g r g r g f g proof f g f f g g f g r 故 同理 由 知 , , deg deg , deg deg 1 2 3 3 1 2 2 2 1 1 1 1 r r q r g r q r r r f gq r r g = + = + = + s s s s s s s s s s s s f g g r r r cr r r q r r q r r r = = = = = = + − − + − − − ( , ) ( , ) ( , ) deg deg 1 1 1 1 2 1 1
例25 fX q(X)x2+2x2-3x+x-x-2x+ 1q0 =x+1x3+x2-2xx+2x3-3x x-1 x2+2x-3 x3-x2+x+1 x2+x-2 x3-2x2+3 r2)=x-17()=x2+x-2 =(x-1)x+2) 所以(fg)=r2(x)=x-1 f=gg1 tri g=nq2+r2 8-7i92=g-(f-gq1q2 =-q+(1+q42)g=(8)7
7 g(X) f(X) x 4+ x3 - x 2 x 3+ 2x 2 -3 - 2x+ 1 q1 (X) x 4+2 x 3 - 3x - x 3 - x 2 +x + 1 - x 3 - 2x 2 + 3 r1 (x)= x2 +x -2 q2 (X) x 3 + x2 -2x x 2 +2x -3 x 2 + x -2 r2 (x)= x -1 =(x-1)(x+2) 所以 ( f, g ) = r2 (x) = x -1 , . 1 1 1 2 2 f = gq + r g = rq + r (1 ) . ( ) . 2 1 2 2 1 2 1 2 q f q q g r g r q g f gq q = − + + = − = − − = x =x -1 +1 例2.5 =( f, g )