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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?

MaríaCC
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1 Answers1

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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.

MaríaCC
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Dilip Sarwate
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  • Thank you or your answer. Can i make you a question? Why $x^3-x^2+1$ must be a divisor of $x^{26}-1$? – MaríaCC Mar 09 '14 at 16:17
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    @MaríaCC If $f(x) \in \mathbb F_p[x]$ is irreducible, then $\mathbb F_p[x]/f(x)$ is isomorphic to the field $\mathbb F_{p^{\deg {f}}}$ whose multiplicative group is cyclic of order $p^{\deg {f}} - 1$. – Dilip Sarwate Mar 09 '14 at 16:27