Question

Justify all your answers. Exercise 1. In each part explain why t e Fp[a] is a unit and find its inverse. (a) t = -3+ 2a, F=Q, p= x2 – 2 (6) t = 1+a+a?, F = Z3, p= x2 +1 (c) t = 1+a+a?, F = Z2, p= x3 + x +1

Answer 1

@ since ? 2 is irreducible elementlPoly- - nomaia) o 148 ' Q ] So, FRJis a field. - Now t=-3+ax. is non-zero element so, there exist an element ť E Fp («) S. that in where 1 is the multiplicative żuntity of Fp (x). het ť= a +ba. ti t' = 1 . (-3+2x)" = a et bor. atba. arba (a +bx) (a-be) = a-be - a² (6x)2 - a-ba - a² - 6 2². Now, (a+b«) 2/-3+26) = 4 (a+bc)* = -3+28. = (-3)=-(27j2 -3-20

-3- 2 9 - 422 -3-2012 9 -4/2) - - 3-20 9-8 * = -3-22. Thus, (-3+24)") = (-3) -24 6 Again. p = x²+2. is irreducible element of 23 [x], so ppa) is again field. Now, tac? ++1 = x + 10? +1) Foto = a. Here a is not zero., so a is a unit Now, folk) = 2a Fr] as Ką2t1] = {a+be: a, be2g] het ß = arba be the insurse of a. thin. (a+686)2 = = (a +5.2) .a = 1 (x=x) 3 atbia I 8V 22 = - 8 - 2001 Thus, t'' = (*+&+1)-' =20.

b)= K2. El © P = 0x3 +06+1 is also irreducible Polynomial over Z2 Lor ] so. fpla) is a field. to = {ax?+b+&+C:4, 6,2 € 22} Now, <? ++) is not Zero, so that. 2²tati is an unit element. Now, het t = ax² tbatc. is iniurse of ++1 then K+4+)) (a<?++c) = 4. Now by Division algorithm, (&?+<+))(<+1) =(3+6+1) + [mod abiely) 71 172 tatal = 16ti),