Exercise on polynomial over a finite field

Question

• Find an irreducible polynomial $p(X)$ of degree $3$ in $\mathbb{F}_3[X]$.

  $x^3 -x-1$ is irreducible over $\mathbb{F}_3$ since it has no roots in $\mathbb{F}_3$.

• Compute the multiplicative inverse of $X+1+\langle p(X) \rangle$ in $\mathbb{F}_3[X]/ \langle p(X) \rangle$.

  $x^3 -x-1=(x-1)(x^2+x)-1$

  $(x-1)(x^2)=1(\mod x^3 -x-1)$

• Suppose that $\langle r(X) \rangle \in \mathbb{F}_3[X]$ is chosen such that $r(X)+\langle p(X) \rangle$ is a primitive element of $_3[X]/\langle p(X) \rangle$. Show that:

  $(r(X)+\langle p(X) \rangle)^{-1}=(r(X)+\langle p(X) \rangle)^{25} $

Any clue for this point, please ?

Answer 1

$\mathbb{F}_3[X]/\langle p(X)\rangle$ is a field extension of degree 3 over $\mathbb{F}_3$, so it has $3^3=27$ elements. Since the multiplicative group of $\mathbb{F}_3$ is $\mathbb{F}_3 - \{0 + \langle p(X)\rangle\}$, and is cyclic (see Why is the multiplicative group of a finite field cyclic?), thus we see it's a cyclic group of order $26$, so all its elements have order 26.

That is, $\forall r(X)+\langle p(X)\rangle \in \mathbb{F}_3[X]/\langle p(X)\rangle$, if $r(X)+\langle p(X)\rangle \neq 0+\langle p(X)\rangle$, then $(r(X)+\langle p(X)\rangle)^{26} = 1+\langle p(X)\rangle$. which implies what you wanted to prove.