Let $\mathbb{F}_p$ be the finite field of $p$ elements (where $p$ is a prime) and let $P(x)\in \mathbb{F}_p[x]$ be an irreducible polynomial. I have to prove that $\bar{x}$ is a primitive element of $\mathbb{F}_p[x]/(P(x))$.
I know that $\mathbb{F}_p[x]/(P(x))$ is a finite field of $p^k$ elements where $k=\deg(P(x))$, so $\bar{x}^{p^k-1}=\bar{1}$ and $\bar{x}$ is thus a root of unity. But I don't see how to prove it is primitive.
I would appreciate any help, thank you!
Best Answer
The statement you have been asked to prove is false; consider $p=3$ and $P=x^2+1$. This polynomial is irreducible over $\Bbb{F}_3$ because it has no roots in $\Bbb{F}_3$. And clearly $\bar{x}$ satisfies $\bar{x}^2=-1$ and hence $\bar{x}^4=1$, so $\bar{x}$ does not generate the multiplicative group of order $3^2-1=8$.