 mersenneforum.org Finite field extensions
 Register FAQ Search Today's Posts Mark Forums Read 2017-11-15, 01:22 #1 carpetpool   "Sam" Nov 2016 22×83 Posts Finite field extensions 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.   2017-11-15, 02:17   #2
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

838410 Posts Quote:
 Originally Posted by carpetpool 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.   2017-11-15, 09:42   #3
Nick

Dec 2012
The Netherlands

110111000012 Posts Quote:
 Originally Posted by carpetpool - 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.
The first is true for all non-zero $$e$$ - see Corollary 84 in our Number Theory discussion group.
For the second, see theorem 90 and proposition 88.   2017-11-15, 14:37 #4 Dr Sardonicus   Feb 2017 Nowhere 2·52·107 Posts As Nick has already pointed out, only the nonzero elements of a finite field can have a multiplicative order. Exercise: In any field, a finite multiplicative group is cyclic. Exercise: Let p be a prime number, and let K be a field of characteristic p. Show that, in K, a) (x + y)^p = x^p + y^p for every x, y in K [the "Freshman's dream"] b) If k is the field of p elements, show that x^p = x for every x in k. c) If K is a field of characteristic p, f is a positive integer, and q = p^f, then (x + y)^q = x^q + y^q Exercise: Let k be the field with p elements, f a positive integer, and K the field of q = p^f elements. Determine the number of elements in K having degree f over k.  Thread Tools Show Printable Version Email this Page Similar Threads Thread Thread Starter Forum Replies Last Post Fred PrimeNet 5 2019-04-13 16:27 paul0 Programming 6 2015-01-16 15:12 Raman Cunningham Tables 87 2012-11-14 11:24 R.D. Silverman Cunningham Tables 4 2011-04-27 09:27 fivemack Math 4 2008-03-27 17:58

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