Let $\mathbb{F}_9 = \mathbb{F}_3[\beta]$ where $\beta $ is a root of $x^2+1$ in some extension of $\mathbb{F}_3$. Then $1+\beta$ is a primitive element of $\mathbb{F}_9$. Express all elements of $\mathbb{F}_9=\mathbb{F}_3[\beta]$ as powers of $1+\beta$.
I know that $\mathbb{F}_9={\{0,1,2,\beta,1+\beta,2+\beta,1+2\beta,2+2\beta}\}$, where $\beta^2+1=0$ and the primitive elements are $1+\beta, 2+\beta,1+2\beta,2+2\beta$. But how will I able to show all elements of $\mathbb{F}_9$ as powers of $1+\beta$?
As simple as repeatedly multiplying and reducing $\beta^2=2$. Let $\theta=1+\beta$, then $$\theta^2=1+2\beta+\beta^2=2\beta$$ $$\theta^3=2\beta+2\beta^2=1+2\beta$$ $$\theta^4=1+3\beta+2\beta^2=2$$ $$\theta^5=2\theta=2+2\beta$$ $$\theta^6=2\theta^2=\beta$$ $$\theta^7=2\theta^3=2+\beta$$ $$\theta^8=2\theta^4=1$$ Note that $0$ can never be a power of $\theta$, which is expected anyway.