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....
View
Full Document
 Fall '11
 KeithCornell
 2m, 2k, Cyclic group, fQ

Click to edit the document details