Question

4. Show that the polynomial g(x) = x++x+1 is irreducible over Z2. In the quotient ring Z2[x]/(g(x)) let S = x+(g(x)), so that Z2[x]/(g(x)) = Z2(). Express 85 and (82 +1)-1 in the form a + b8 + 082 +883, where a, b, c, d e Z2.

Answer 1

(4) We have to show that gez)= 204 +2+ ] is irreducible over Zz' Z2 = {0,1}. g(x) will be irreducible over 7, if g(x) #0 for all x € 72 check het us As a E 22 or = 1 If X=0 g(0)= +0+1 If [:2=0 in 72 Thus then 1 +0 = g(0) 70 aced then (1) = 4*+1+1 = 2+1 = 18(1) #O 9000, 9(1) g(x) to for all & EZ2 992) = x4 + x + 1 is irreducible. ↑ Now consider 7,[2] (8681) het 20 + <800) = 8 as < g(x) is zero element of Z2[x] 19(x) 8 - xto 11 1 &=x Thus 12 [x] E Z2(8) because of g(t) Z [a] <9()

Now sx 85 = (a +(9(*))]5 85 [ (a+<3(x)= am +290w)) = (Q*+*+1) x +(-2+2)+<06 **+2+1 +x5 + x² + x + < 8(v)> I 25 O -22-2 (+4+1+1)* <901) 85 -22-2+ <9(0)) (0 - 1) x2 + (0-1)X +<3(2]) (2-1) *2 + (2-1) x +48 (*)) [:0=2 in 72 0.1 +1% +102 + 0.83 [:: 8=2] 85 5 IL 0.1 +18 + 1.82+0.89 Now Q-1 3 * → (8?+1)' = (a +1)' het Q = r²+1 het a + bx + C x² +0123 Qo'd = 1 La + b 1+0+2+0x3)(x+1) = 1 9x2+673 +x++d25+ a +bx+c+2 + x3 = 1 + (a+c)? + c(-x-1) + d(+ 2) + bx + a = = 1 (b+d) x3 + (Q+C+ d) +2+(-C +d+b) * +Q-C = 1 Comparing both side coefficients of 1,2,2, x3 b+ d = 0 a +c+d = 0 Solving these gives b+d - C =0 Q-C = 1 d = -1 = 2-1 = 1 (6+d)*3 a = 1 b = 1 C = 0 Thus (x2+1) '- 1 + 0 +0.7+1.43 + (8?+ 1)" = 1 + 8 +0.82 +1.83