Abstract Algebra Theory and Applications - Computer Science ...

404 HINTS AND SOLUTIONSChapter 19. Fields1. (a) x 4 − 2 3 x2 − 629 . (c) x4 − 2x 2 + 25.2. (a) {1, √ 2, √ 3, √ 6 }. (c) {1, i, √ 2, √ 2 i}. (e) {1, 2 1/6 , 2 1/3 , 2 1/2 , 2 2/3 , 2 5/6 }.3. (a) Q( √ 3, √ 7 ).5. Use the fact that the elements of Z 2 [x]/〈x 3 + x + 1〉 are 0, 1, α, 1 + α, α 2 ,1 + α 2 , α + α 2 , 1 + α + α 2 and the fact that α 3 + α + 1 = 0.8. False.14. Suppose that E is algebraic over F and K is algebraic over E. Let α ∈ K.It suffices to show that α is algebraic over some finite extension of F . Sinceα is algebraic over E, it must be the zero of some polynomial p(x) = β 0 +β 1 x + · · · + β n x n in E[x]. Hence α is algebraic over F (β 0 , . . . , β n ).22. Q( √ 3, √ 7 ) ⊃ Q( √ 3 + √ 7 ) since {1, √ 3, √ 7, √ 21 } is a basis for Q( √ 3, √ 7 )over Q. Since [Q( √ 3, √ 7 ) : Q] = 4, [Q( √ 3 + √ 7 ) : Q] = 2 or 4. Since thedegree of the minimal polynomial of √ 3+ √ 7 is 4, Q( √ 3, √ 7 ) = Q( √ 3+ √ 7 ).27. Let β ∈ F (α) not in F . Then β = p(α)/q(α), where p and q are polynomialsin α with q(α) ≠ 0 and coefficients in F . If β is algebraic over F , thenthere exists a polynomial f(x) ∈ F [x] such that f(β) = 0. Let f(x) =a 0 + a 1 x + · · · + a n x n . Then( ) ( ) ( ) n p(α)p(α)p(α)0 = f(β) = f = a 0 + a 1 + · · · + a n .q(α)q(α)q(α)Now multiply both sides by q(α) n to show that there is a polynomial in F [x]that has α as a zero.Chapter 20. Finite Fields1. (a) 2. (c) 2.4. There are eight elements in Z 2 (α). Exhibit two more zeros of x 3 + x 2 + 1other than α in these eight elements.5. Find an irreducible polynomial p(x) in Z 3 [x] of degree 3 and show thatZ 3 [x]/〈p(x)〉 has 27 elements.7. (a) x 5 − 1 = (x + 1)(x 4 + x 3 + x 2 + x + 1).(c) x 9 − 1 = (x + 1)(x 2 + x + 1)(x 6 + x 3 + 1).8. True.11. (a) Use the fact that x 7 − 1 = (x + 1)(x 3 + x + 1)(x 3 + x 2 + 1).12. False.17. If p(x) ∈ F [x], then p(x) ∈ E[x].