代数学基础与有限域(第三讲)

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2、(Q15 (x, q = 2, 0 i 7 x0 x 2 A 1 x2 x4 x6 +x3 x2 +x2 +x3 +x4 +x6 +x4 +x5 +x5 +x7 +x7 +x7 x4 x6 x8 x10 x12 x14 1 1 1 +x 0(H 1 0 0 0 0 0 0 0 3&4 A= 1 0 1 1 1 0 1 1 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 1 1 0 1 1 1 0 1 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 7 #c i p& q A I $r 1 6, 3&

3、4 Q (x u&4(! 2 stu&w0( $ e f xy (1, 0, 0, 0, 0, 0, 0, 0 (0, 0, 1, 1, 0, 1, 0, 1 AI $ a 15 AI= 1 0 1 1 1 0 1 0 0 1 1 0 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0 1 0 0 1 0 1 1 0 1 0 1 0 1 1 0 0 0 0 0 0 0 0 0 1 0 0 0 1 1 IP&Q(RSTU&V(W ( X(Ya&b(cd 2006 c R c suvc h (x = 1 h (x Q (x- wt %( BG 1 2 15 79 h2 (x = x + x

4、2 + x5 + x6 + x7 . gcd(Q15 (x, h2 (x = x4 + x3 + 1 3&4 34 efg Q15 (x $&( ! 1 4 gcd(Q15 (x, h2 (x 1 = x4 + x + 1 f (x = (x4 + x3 + 1(x4 + x + 1 F2 Z # % 7 s E!d x4 + x + 1 x4 + x3 + 1. hijkl smn f |p do j | gt pqhBrs ij p, smn F f BR s mn f j| Z f uv |wxyz wxyz | n Hensel 3.4.6 (Hensel p s c(x, a (x,

5、 b (x, u(x, v(x %( p e e c(x ae (xbe (x mod pe , $ g h1 1, deg(u(x deg(b (x, deg(v(x deg(a (x. & &%( a (x, b (x e 1, ae (x e e e+1 e+1 u(xae (x + v (xbe (x 1 mod p. ae+1 (x = ae (x be+1 (x be (x c(x ae+1 (xbe+1 (x mod pe , mod pe , mod pe+1 , A $ l d(x = (c(x a (xb (x/p , 9 u BB | d B w d( x h j B 8

6、 & s(x, t(x Zx pe+1 B sB Y BB B e e e u(xae+1 (x + v (xbe+1 (x 1 mod p, Y BY ae+1 (x = ae (x + pe s(x be+1 (x = be (x + pe t(x (3.11 (3.12 80 IP&Q(RSTU&V(W ( ( X Ya&b(cd (3.11 2006 n 1 (3.12 c(x ae+1 (xbe+1 (x u (3.13 0 mod pe+1 2e e + 1, H (3.13 (3.13 m(nsmn (3.14 ae (xt(x + be (xs(x (c(x ae (xbe (

7、x/pe = d(x Fp mod p. (3.14 f u(xae (x + v (xbe (x = 1, op d(xu(xae (x + d(xv (xbe (x d(x (3.15 mod p (3.15 (t(x d(xu(xae (x + (s(x d(xv (xbe (x 0 mod p ae (x d(xu(x, 1 be (x h mod p i q ae (x|s(xd(xv (x, be (x|t(x t(x d(xu(x s(x d(xv (x be (x ae (x mod p hsmn g Fp ( lgn w(x, nh s(x v (xd(x + w(xae (x t(x u(xd(x w(xbe (x e+1 e+1 mod p mod p p rs s(x 1 t(x nj B mod p h q a (x 1 b (x mod p h B BB p sBmBn F f p p ( mod t mod p X e ph j e v jv h j | ( nB j 341 | s T p ?. h h j| Z f A. K. t Lenstra,