Section 48 number 26 using Example 29.19 know that there is an extension field E of ZL, containing a zero a of x- + x + 1. Byand the uniqueness of the b¡ is established.= 0, sob = b,,We give an impressive example illustrating Theorem 29.18.neThe polynomial p(x) = x² + x + L in Zzlrlis irreducible over Z2 by Theoren vesince neither element 0 nor element L of Za is a zero of p(x). By Theorem 2 By29.19 ExampleTheorem 29.18, Z2(@) has as elements 0 + Og. L+ 0a, 0 + la, and1+ 1d, ulat and1, a, and 1 +a. This gives us a new finite field of four elements! The additionmultiplication tables for this field are shown in Tables 29.20 and 29.21. For example, tocompute (1 +æ)(1+a) in Z2(@), we observe that since p(a) = a² +a +1 = 0, thena = -a – 1 = a + 1.Therefore,(1 +æ)(1+a)= L+a +a +a² = 1 +@)= 1+at!=a.Finally, we can use Theorem 29.18 to fulfill our promise of Example 29.4 andshow that R[x]/(x² + 1) is isomorphic to the field C of complex numbers. We saw inExample 29.4 that we can view R[x]/(x² + 1) as an extension field of R. Leta = x + (x² + 1).29.21 Table29.20 Table0.11+a11+a1+ a0.0.0.1111+a111+a1+a1 011+a1+ a0.1+aα11+ a1+a b. ReferTing to part (a), compare the conjugation automorpnautomorphism Va ß2 of Q(/2) with the conjugation26. Describe the value of the Frobenius automorphism ơ, on each element of the finite field of four elements givenin Example 29.19. Find the fixed field of o2.27. Describe the value of the Frobenius automorphism oz on each element of the finite field of nine elements give

To describe the value of the Frobenius automorphism σ2 on each element 0, 1, α and 1+α The field…

Answered: automorphism Ya,p. 26. Describe the…

automorphism Ya,p. 26. Describe the value of the Frobenius automorphism ơ2 on each element of the finite field of four elements give- in Example 29.19. Find the fixed field of o2. nlement

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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