# Generators of $\mathbb{F}_9/\mathbb{F}_3$ that do not generate $\mathbb{F}_9^{\times}$

Find a generator of the extension $\mathbb{F}_9/\mathbb{F}_3$ that does not generate the multiplicative group $\mathbb{F}_9^{\times}$. how many such elements exist? what are their minimal polynomials?

To start we should choose a description of $\mathbb{F}_9$, say $\mathbb{F}_3[x]/(x^2+1)$. (we could also choose the set of roots of $x^9-x$.)

$0,1,2$ are not generators since they are in $\mathbb{F}_3$. By direct computation, all other elements are generators.

$x$ has multiplicative orbit $x,x^2=2,2x,1$, and by adding we get all $\mathbb{F}_9$, so $x$ is a generator. the orbit doesn't contain $x+1$, so $x$ is our required element. In the same way we get $2x$.

The multiplicative group of a field is always cyclic, and it is of order $9-1=8$, so we know the requirement of not generating $\mathbb{F}_9^{\times}=C_8$ rules out $\phi(8)=4$ elements out of the $9$. So $x+1$,$x+2$,$2x+1$,$2x+2$ do generate $\mathbb{F}_9^{\times}$.

But what about the minimal polynomials of $x$,$2x$? these look like polynomials, but should be now treated as elements, and it confuses me.

To avoid conflicts we write $f=X^2+1 \in \mathbb{F}_3[X]$ which is irreducible. If we take $x$ as a root of $f$ we now can write $$\mathbb{F}_9 \cong \mathbb{F}_3[X]/\langle f\rangle \cong\mathbb{F}_3[x].$$ (Your notation $\mathbb{F}_3[x]/(x^2+1)$ is a bit confusing.)
Now you have the elemts $x,2x$ in a field extension $K$ of $\mathbb{F}_3$. As $x$ is a root of $f$ and $f$ is irreducible $f=X^2+1 \in \mathbb{F}_3[X]$ is the the minimal polynomial of $x$. It also holds that $f(2x)=0$ and so $f$ is also the minimal polynomial of $2x$.
Of course you can also name $x=\alpha$ (root of $f$) and write $$\mathbb{F}_9 \cong \mathbb{F}_3[X]/\langle f\rangle \cong\mathbb{F}_3[\alpha].$$ But the minimal polynomial still remains $f=X^2+1 \in \mathbb{F}_3[X]$ so there is no confusing.