Adv Alegbra HW Solutions 121

Adv Alegbra HW Solutions 121 - 2 mod p is solvable....

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121 3.97 Let p be a prime and let q = p n for some n 1. (i) Show that the function F : F q F q , given by F ( a ) = a p ,isan isomorphism. Solution. We have F ( 1 ) = 1 p = 1 and F ( xy ) = ( xy ) p = x p y p = F ( x ) F ( y ). Furthermore, F ( x + y ) = ( x + y ) p = x p + y p , by Proposition 1.63. Now F is an injection, for if 0 = F ( x ) = x p , then x = 0. Since F p is f nite, F is an isomorphism, by Exercise 2.13. (ii) Show that every element a F q has a p th root, i.e., there is b F q with a = b p . Solution. Each a F q has a p th root because F is surjective. (iii) Let k be a f eld of characteristic p > 0. For every positive inte- ger n , show that the ring homomorphism F n : k k , given by F n ( a ) = a p n , is injective. Solution. The proof is a straightforward induction on n 1, the base step being part (i). 3.98 Prove that every element z in a f nite f eld E is a sum of two squares. (If z = a 2 is a square, then we may write z = a 2 + 0 2 .) Solution. Absent. 3.99 If p is a prime and p 3 mod 4, prove that either a 2 2mod p is solvable or a 2 ≡−
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Unformatted text preview: 2 mod p is solvable. Solution. Since p = 3 mod 4, we have p = 4 k + 3 for some integer k , and so p 1 = ( 4 k + 3 ) 1 = 4 k + 2 = 2 ( 2 k + 1 ) ; that is, p 1 = 2 m , where m = 2 k + 1 is odd. By Theorem 2.131, F p = h 1 i H for some subgroup H of order m . Since F 2 I m = ( { 1 } H ) ( { 1 } H ) , however, either 2 H or 2 H . But Exercise 2.82 says that every element in H has a square root. If 2 has a square root, then there is an integer a with a 2 = 2 in F p . and this says that a 2 2 mod p . 3.100 (i) Prove that x 4 + 1 factors in F 2 [ x ] . Solution. x 4 + 1 factors in F 2 [ x ] because 1 is a root....
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This note was uploaded on 12/21/2011 for the course MAS 4301 taught by Professor Keithcornell during the Fall '11 term at UNF.

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