I am trying to solve the following problem: let $K$ be a finite field with $729$ elements.
- How many $\alpha\in K$ make $K^* = \langle \alpha\rangle$?
- How many fields $E$ are such that $K|E$ is a field extension? What number of elements have each?
- How many $\beta\in K$ satisfy $K = \mathbb F_3[\beta]$?
- How many irreducible polynomials of degree $2$, $3$ and $6$ are in $\mathbb F_3[t]$?
And I have argued as follows:
Since $K$ has $729$ elements, and $729$ is $3^6$, it follows $K\cong \mathbb F_{3^6}$, the finite field with $3^6$ elements, so every $E$ such that $K|E$ is field extension needs to satisfy $|E| = p^k$ with $k|6$, so there are, up to isomorphism, $4$ field extensions of the form $E\subseteq K$, that are $\mathbb F_3|K$, $\mathbb F_{3^2}|K$, $\mathbb F_{3^3}|K$ and $\mathbb F_{3^6}|K$.
Also, since $K^*$ is cyclic of order $p^n-1$, $K^* = \langle u\rangle$ for some $u\in K^*$, and each element $\alpha\in K^*$ can be written $\alpha = u^k$. For $\alpha\in K^*$ to satisfy $K^* = \langle \alpha \rangle$, it is needed to be $\gcd(k,p^{n}-1)=1$, since $$ \mathrm{order}(u^k) = \frac{\mathrm{order}(u)}{\gcd(k,\mathrm{order}(u))} $$ so the number of $\alpha$'s with this property is $\varphi(728) = 288$.
This should give all the elements such that $K=\mathbb F_3[\alpha]$, since every element $\beta\in K$ except $0$ satisfies $\beta = \alpha^{k}\in \mathbb F_3[\alpha]$ and $|K|=|\mathbb F_3[\alpha]|$ (is this right?)
For the last, I would only know how to calculate the number of irreducible and monic polynomials, but I don't know how to calculate the whole number of irreducible polynomials with those degrees.
I would appreciate some hints or help. Thanks in advance.