Abstract Algebra Theory and Applications - Computer Science ...

360 CHAPTER 20 FINITE FIELDS(3) ⇒ (1). By (3), a code polynomial f(t) = a 0 + a 1 t + · · · + a n−1 t n−1 isin C exactly when f(ω i ) = 0 for i = 1, . . . , 2r. The smallest such polynomialis g(t) = lcm[m 1 (t), . . . , m 2r (t)]. Therefore, C = 〈g(t)〉.□Example 8. It is easy to verify that x 15 − 1 ∈ Z 2 [x] has a factorizationx 15 − 1 = (x + 1)(x 2 + x + 1)(x 4 + x + 1)(x 4 + x 3 + 1)(x 4 + x 3 + x 2 + x + 1),where each of the factors is an irreducible polynomial. Let ω be a root of1 + x + x 4 . The Galois field GF(2 4 ) is{a 0 + a 1 ω + a 2 ω 2 + a 3 ω 3 : a i ∈ Z 2 and 1 + ω + ω 4 = 0}.By Example 3, ω is a primitive 15th root of unity. The minimal polynomialof ω is m 1 (x) = 1 + x + x 4 . It is easy to see that ω 2 and ω 4 are also rootsof m 1 (x). The minimal polynomial of ω 3 is m 2 (x) = 1 + x + x 2 + x 3 + x 4 .Therefore,g(x) = m 1 (x)m 2 (x) = 1 + x 4 + x 6 + x 7 + x 8has roots ω, ω 2 , ω 3 , ω 4 . Since both m 1 (x) and m 2 (x) divide x 15 −1, the BCHcode is a (15, 7)-code. If x 15 − 1 = g(x)h(x), then h(x) = 1 + x 4 + x 6 + x 7 ;therefore, a parity-check matrix for this code is⎛⎜⎝Exercises0 0 0 0 0 0 0 1 1 0 1 0 0 0 10 0 0 0 0 0 1 1 0 1 0 0 0 1 00 0 0 0 0 1 1 0 1 0 0 0 1 0 00 0 0 0 1 1 0 1 0 0 0 1 0 0 00 0 0 1 1 0 1 0 0 0 1 0 0 0 00 0 1 1 0 1 0 0 0 1 0 0 0 0 00 1 1 0 1 0 0 0 1 0 0 0 0 0 01 1 0 1 0 0 0 1 0 0 0 0 0 0 01. Calculate each of the following.(a) [GF(3 6 ) : GF(3 3 )](c) [GF(625) : GF(25)]2. Calculate [GF(p m ) : GF(p n )], where n | m.3. What is the lattice of subfields for GF(p 30 )?⎞.⎟⎠(b) [GF(128) : GF(16)](d) [GF(p 12 ) : GF(p 2 )]