I have two questions, one of those is the same here, but I'd like to use another argument and I need a check!
The text is:
i) Is it true that a primitive 3-th root of the unit over $\mathbb{F}_3$ is contained in $\mathbb{F}_9$ ?
ii) Is it true that a primitive 4-th root of the unit over $\mathbb{F}_3$ is contained in $\mathbb{F}_9$ ?
Eventually, find such roots.
My attempt:
Let $\mathbb{F}_9= \mathbb{F}_3[x] / (x^2 + 1)$ and let $\alpha = \bar{x}$. I have that $\beta = \alpha + 1$ is a primitive element of $\mathbb{F}_9$
i) Let us assume to have such an element $g$ s.t. $g^3=1$, $g \in \mathbb{F}_9$, the candidate primitive root of order $3$. Now, take $\beta \in \mathbb{F}_9$ primitive element. I know that exists $d$ s.t. $g= \beta^d$. So, I obtain $$\beta^{3d} = 1 \text{ in } \mathbb{F}_9$$
and therefore I have $3d = 8$ and hence I would find that $3\mid 8$, which is not possible. Therefore, such a root can't exist.
ii) I know that $\text{ord}_4(3) =2$. Therefore, using again $\beta$ primitive element of $\mathbb{F}_9$ I find: $$4d = 8$$
which implies $d = 2$ and hence $g=\beta^d=\beta^2 = (\alpha + 1)^2$ is such a root.