Irreducibility of cyclotomic polynomials of prime order

Question

I am stuck with an exercise, where I have to prove the $\Phi_5[X] \in \mathbb{F}_2(x)$ is irreducible.

I know that

• $X^5-1=\Phi_5(X)(X-1)$ (shown in previous part of the exercise)
• $X^2+X+1$ is the only irreducible element in $\mathbb{F}_2(X)$ of order two

I was trying to prove that $\Phi_5[X]$ is a prime element, hence irreducible, but with no luck...

A hint would be appriciated

Answer 1

Suppose $\Phi_5(X)$ is reducible. Then its decomposition should be :

$$\Phi_5(X)=PQ $$

With $deg(P)=1,2,3$ and $deg(Q)=3,2,1$. So that if $\Phi_5(X)$ is reducible, either it is divisible by a polynomial of degree $1$ (and so it has a root) either it is divisible by two irreducible polynomials of degree $2$.

From your definition of $\Phi_5$ it should be clear that it has no root in $\mathbb{F}_2$ (I leave the calculation to you).

So that we are forced to have $\Phi_5=PQ$ where $P$ and $Q$ are irreducible polynomials of degree 2, but you just showed that $P=X^2+X+1=Q$ then it follows that $\Phi_5=(X^2+X+1)^2$, why this cannot happen ?