Prove If F is of characteristic p and p divides n, then there are fewer than n distinct nth roots of unity over F: in this case the derivative is identically 0 since n=0 in F. In fact every root of x^n-1 is multiple in this case.
Please write legibly, no blurry pictures and no cursive.
F unity.
Proof
.Let P(n )be the first primitive nth root of unity in the field; i.exp (P)= e2iπ/n.
We will prove that −P(n+1 )(2) = P(2n) By Induction
. Note that since P has Distinct Roots,D{(n + 1)/2} is an integer, where "D" is the derivative of is identically zero since (n===0) in F
therefore −P (n+1 )/2 P is in the field F
. Observe that
(P+1) 2 /P = (−1) POWER.(e2iπ P*(n+1)/ 2 =
(eiπ)*(e2iπ )N ·(N+1) 2
(e iπ)**(e iπ)(·P+1) (.P)* =(eiπ)*(eiπ)*(eiπ n )= (−1)(−1)**e2iπ
2P= D{2P}
as desired.
Prove If F is of characteristic p and p divides n, then there are fewer than n distinct nth roots...
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