Question

Find the inverse of 1 - 2x} in Q[x]/(23 - 2) using Euclid's Algorithm.

Answer 1

Here basically we will take help of Euclid's algorithm and some basic rules of quotient ring.

Given quotient ring on]/(m32) [n]/603) Thus, qx++ba+c+ (x2) a, b, c E Q} Now we want to finid the inverse of [1-2n].

Note that 1+ (x³ -2) is the identity element of the given quotient ring.

Let I: (22) Now [.-2n] = 1-2x + I Let an²+brtc + L E E O [nb/I is the inverse of. 1-2n+J. le (Qu'+bx+c+1) (-2+3) = 141 Now (1-2) Can'thu+c) + 7 = 14) Conside (-x)(aq't bn +c) an²+bx+c - zan3 - 2bn²_2cn Now Euclide using we need to find Algorothmi p(n) and r(a) such that:

-ran3+(a-2b) 22 + (b-2c)n + = P(n) (132) +r(n) where اه degree r(n) _3. Now -2an 3 + (a-2b) n² + (6-2c)n+C -29 (m3-2) + (a.2b) n2 + (b-20) n +C-42 Now (1-2) (ana+on+c) + 1 = (2-2)*++ (6-2) + (-44 + I. and (9-2b) n° +620)*+ (-49 +1 = 1+1

comparing the coefficiente 9-26 = 0, C-4451 b-2C - 0 1 ㅋb=2C 29 4 26 1 ACEIt4. 1 ThW b=2c = 2 (i+4d 을 = 2(144) a= 4 + IGC ISG=4 41.15 크b 1 15 + 4-4) 큼 Tu 음름 n1, 15 +1= n 15 7 e inverse of [1-21].

Thus we are done!