Exercise 11.2.2

Proof.

(a)

We know (Example 11.2.3) that $N_1=p,N_2=\frac{1}{2}(p^2-p)$.

By Theorem 11.2.2,

$$\begin{array}{llll}\hfill & 3N_3+N_1=p^3,& \hfill & \\ \hfill & 4N_4+2N_2+N_1=p^4.& \hfill & \end{array}$$

Therefore

$$\begin{array}{llll}\hfill N_3& =\frac{1}{3}(p^3-p),& \hfill & \\ \hfill N_4& =\frac{1}{4}(p^4-p^2).& \hfill & \end{array}$$

(b)

For $p=2$ and $n=2,n=3$, we obtain $N_2=1,N_3=2$.

The factorisation (11.7) is here

$$x^4-x=x(x-1)\times (x^2+x+1),$$

so $x^2+x+1$ is the only irreducible polynomial over ${𝔽}_2$ of degree 2.

With the following Sage instructions

     R.<x> = GF(2)[]
     factor(x^8-x)

give

$$x\cdot (x+1)\cdot (x^3+x+1)\cdot (x^3+x^2+1).$$

So the two irreducible polynomial over ${𝔽}_2$ of degree 3 are

$$x^3+x+1,x^3+x^2+1.$$