Explore BrainMass

Irreducible polynomial problems

Not what you're looking for? Search our solutions OR ask your own Custom question.

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

Determine the number of monic irreducible polynomials of degree 4 in F_q without using the Moebius Inversion Formula.

https://brainmass.com/math/number-theory/irreducible-polynomial-problems-471901

SOLUTION This solution is FREE courtesy of BrainMass!

Let us first calculate the number of reducible monic polynomials of degree 4 over F_q. Each of them has one of the following forms
〖(x-a)〗^4, for some a from F_q,
〖(x-a)〗^3(x-b), for distinct a and b from F_q ,
〖(x-a)〗^2 〖(x-b)〗^2 , for distinct a and b from F_q,
〖(x-a)〗^2(x-b)(x-c), for distinct a,b,c from F_q,
〖(x-a)〗^2f(x), where f(x) is an irreducible monic polynomial over F_q of the degree 2,
(x-a)f(x), where f(x) is an irreducible monic polynomial over F_q of the degree 3,
(x-a)(x-b)f(x), where f(x) is an irreducible monic polynomial over F_q of the degree 2,
(x-a)(x-b)(x-c)(x-d), for distinct a,b,c,d from from F_q,
f(x)g(x), for distinct irreducible monic polynomials f(x) and g(x) over F_q of the degree 2,
〖f(x)〗^2, for an irreducible monic polynomial f(x) over F_q of the degree 2,
Let us calculate the number of polynomials of each form 1)-10).
It is q,
It is A_q^2=q(q-1),
It is C_q^2=1/2q(q-1),
It is (q(q-1)(q-2))/2,
As is known the number of irreducible monic polynomials of the degree 2 is 1/2q(q-1) (see the link below)
http://www.maths.susx.ac.uk/Staff/JWPH/TEACH/CODING11/soln3.pdf

So, the number of polynomials of the form 5) is q∙1/2q(q-1)= 1/2 q^2(q-1).
As is known the number of irreducible monic polynomials of the degree 3 is 1/3 (q^3-q) (see the link above). So, the number of polynomials of the form 6) is 1/3 q^2(q^2-1),
It is C_q^2∙1/2q(q-1)=1/4 q^2 〖(q-1)〗^2,
It is C_q^4=(q(q-1)(q-2)(q-3))/24,
It is C_n^2, where n=1/2q(q-1), i.e. C_n^2=1/8 q^4-〖1/4 q〗^3-1/8 q^2+1/4 q,
It is 1/2q(q-1),

totaling we obtain (3q^4+q^2)/4.

Since the number of all monic polynomials of the degree 4 is q^4, we obtain that the number of monic irreducible polynomials is
q^4- (3q^4+q^2)/4.=1/4(q^4-q^2)

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!