# Irreducible polynomial problems

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

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## 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

ã€–(x-a)ã€—^4, for some a from F_q,

ã€–(x-a)ã€—^3(x-b), for distinct a and b from F_q ,

ã€–(x-a)ã€—^2 ã€–(x-b)ã€—^2 , for distinct a and b from F_q,

ã€–(x-a)ã€—^2(x-b)(x-c), for distinct a,b,c from F_q,

ã€–(x-a)ã€—^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,

ã€–f(x)ã€—^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 qâˆ™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^2âˆ™1/2q(q-1)=1/4 q^2 ã€–(q-1)ã€—^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-ã€–1/4 qã€—^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)

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