Suppose F is a field s.t |F|=q. Take p to be some prime. How many monic irreducible polynomials of degree p do exist over F?

Thanks!

**Answer**

The number of such polynomials is exactly qp−qp and this is the proof:

The two main facts which we use (and which I will not prove here) are that Fqp is the splitting field of the polynomial g(x)=xqp−x,

and that every monic irreducible polynomial of degree p divides g.

Now: |Fqp:Fq|=p and therefore there could be no sub-extensions. Therefore, every irreducible polynomial that divides g must be of degree p or 1.

Since each linear polynomial over Fq divides g (since for each a∈Fq, g(a)=0), and from the fact that g has distinct roots, we have exactly q different linear polynomials that divide g.

Multiplying all the irreducible monic polynomials that divide g will give us g, and therefore summing up their degrees will give us qp.

So, if we denote the number of monic irreducible polynomials of degree p by k (which is the number we want), we get that kp+q=qp, i.e k=qp−qp.

**Attribution***Source : Link , Question Author : IBS , Answer Author : Ross Millikan*