# Properties of the finite field with $729$ elements

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.

• You should give a more illustrative title. – Morgan Rodgers Aug 24 '15 at 8:47
• You can make any polynomial monic by dividing by the leading coefficient and this won't affect whether its irreducible or not. – Matt B Aug 24 '15 at 8:58

Most of it looks good, except for the last part. You don't need $\beta$ to be a generator for $\mathbb{F}_{3}[\beta] = \mathbb{F}_{3}[\alpha]$. What you need is for the minimal polynomial of $\beta$ over $\mathbb{F}_{3}$ to have degree $6$. Equivalently, you need $\beta$ not to be in any proper subfield of $K$.
To count the number of irreducible polynomials, use the fact that the irreducible polynomials of these degrees in $\mathbb{F}_{3}[t]$ are going to factor completely in $K$. For example, an irreducible polynomial of degree $2$ is going to have two distinct conjugate roots in $\mathbb{F}_{3^{2}} \setminus \mathbb{F}_{3}$, so there should be $(9-3)/2$ of them (that are monic).
• That gives $|\mathbb F_{3^6}|-|\mathbb F_{3^3}|-|\mathbb F_{3^2}|+|\mathbb F_{3}|$ elements $\beta$ such that $\mathbb F_3[\beta] = K$? – user55268 Aug 24 '15 at 9:05
• Yes, that should give the proper answer. And these get grouped into sets of $6$ conjugate roots to give irreducible polynomials of degree $6$. – Morgan Rodgers Aug 24 '15 at 9:08