View Single Post
Old 2017-11-15, 02:17   #2
science_man_88
 
science_man_88's Avatar
 
"Forget I exist"
Jul 2009
Dumbassville

100000110000002 Posts
Default

Quote:
Originally Posted by carpetpool View Post
When working with finite fields F(q) of order q (q is an odd prime), F(q) has exactly q elements [0, 1, 2, 3......q-3, q-2, q-1]. Addition, subtraction, multiplication, and division are performed on these elements. If e is an element in F(q), then e^(q-1) = 1 by Fermat's Little Theorem.

For the construction of F(q^2), choose a quadratic non residue r mod q. Then let s be a symbol such that s^2 = r in the same sense as i is a symbol such that i^2 = -1.

Elements are of the form e = a*s+b in F(q^2) where a and b are reduced integers mod q. Addition, subtraction, multiplication, and division are defined in F(q^2). Like in F(q), there are exactly q^2 elements in F(q^2).

- Show that for any element e in F(q^2), e^(q^2-1) = 1.
- Show that there are exactly phi(q^2-1) primitive elements e such that q^2-1 is the smallest integer m such that e^m = 1.

Can someone please provide further information on this? Thank you.
first point follows from the factorization of q^2-1 as (q-1)*(q+1) does it not ??
as to the second point I'm not sure.
science_man_88 is offline   Reply With Quote