I'm trying to do an exercise of my homework that sais I have to prove that the iamge of $X$ in $K^{\times}=\left(\mathbb{F}_3[X]/(X^3-X^2+1)\right)^{\times}$ is a generator.
Acording to what I know, $K^{\times}$ have 26 elements. So, $\alpha^{26}$ must be 1 and $\alpha^{13}$ must be $-1$. But I've calculate this several times and I have $\alpha^{13}=X^2-1$
is the exercise wrong or it's me?
The polynomial $x^3-x^2+1$ is given to be an irreducible cubic polynomial in $\mathbb F_3[x]$, and it must be a divisor of $$x^{26}-1 = (x^{13}-1)(x^{13}+1)$$ So $x^3-x^2+1$ must be a divisor of $x^{13}-1$ or of $x^{13}+1$. If the former, then $\alpha$, a root of $x^3-x^2+1$ , has order 13 and cannot be a generator of $K^\times$. So, you must check whether $x^3-x^2+1$ divides $x^{13}-1$ or not, and you are done.