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Let $F_{q^m}$ denote the finite field with $q^m$ elements. Let $\gamma$ be a primitive element of $F_{q^m}$.

What are the powers $i$ such that $F_q(\gamma^i)=F_{q^m}$?

Note that the following are equivalent, so the question could be rephrased accordingly.

  • $F_q(\gamma^i)=F_{q^m}$
  • The minimal polynomial of $\gamma^i$ over $F_q$ has degree $m$
  • There is no $s$ such that $0<s<m$ and $q^m-1 | i(q^s-1)$
geo909
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  • Note that the last bullet is equivalent to "there is no $s$ such that $0<s<m$ and $\frac{q^m-1}{q-1}|i\frac{q^s-1}{q-1}$. So it follows that when $gcd(i,\frac{q^m-1}{q-1})=1$ this is true, but I couldn't prove that this is an if and only if condition – geo909 Jul 04 '14 at 01:37
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    There is some evidence this can't be simplified: if we have an integer $n$ and an arbitrary list of divisors $d_1,\cdots,d_k$, we can view the set of divisors of $n$ which are not multiples of any $d_i$ as the downset of $n$ minus the upsets of the $d_i$ (where $\Bbb N$ is partially ordered by divisibility). If we write an isomorphism $(\Bbb N,\mid)\cong\Bbb N^{\Bbb N}$, the latter with the product order (via prime factorization), then explicitly doing this for some small examples we can see that this set is not so simple to describe in general. – anon Jul 04 '14 at 02:02
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    Have you looked at examples? Say, the fields with $8$, $16$, $32$, $64$, $9$, $27$, and $81$ elements? – Lubin Jul 04 '14 at 03:57
  • Deleted my earlier comments. A closer inspection revealed that their content/idea is covered by @blue. Why don't you consider undeleting? – Jyrki Lahtonen Jul 05 '14 at 06:47

1 Answers1

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Let $\alpha$ be in some extension of $\Bbb F_q$. If $\alpha$ has order $n\mid(q^f-1)$ for some $f$, then $\alpha\in\Bbb F_{q^f}$. We know that $\Bbb F_q(\alpha)$ is the intersection of all field extensions $\alpha$ is contained in (in some fixed algebraic closure, say), which means $\Bbb F_q(\alpha)$ is the intersection of all $\Bbb F_{q^f}$ such that $n\mid(q^f-1)$, which finally means $\Bbb F_q(\alpha)=\Bbb F_{q^f}$ where $f$ is the order of $q$ mod $n$.

We need the cyclic group generated by $\gamma^i$ to not be contained in any proper subfield of $\Bbb F_{q^m}$, as this is necessary and sufficient. Without loss of generality, $i\mid(q^m-1)$ (else $i$ is associate to a divisor's residue in $\Bbb Z/(q^m-1)\Bbb Z$). We know $\langle\gamma^i\rangle\subseteq\Bbb F_{q^d}^\times$ iff the first's order divides the second's order:

$$\frac{q^m-1}{i}\mid(q^d-1)\iff \frac{q^m-1}{q^d-1}\mid i.$$

So we want the integers $i$ that represent elements associate (mod $q^m-1$) to divisors of $q^m-1$ which are not divisible by $(q^m-1)/(q^d-1)$ for any proper divisor $d\mid m$.

There is some evidence this can't be simplified: if we have an integer $n$ and an arbitrary list of divisors $d_1,\cdots,d_k$, we can view the set of divisors of $n$ which are not multiples of any $d_i$ as the downset of $n$ minus the upsets of the $d_i$ (where $\Bbb N$ is partially ordered by divisibility). If we write an isomorphism $(\Bbb N,\mid)\cong\Bbb N^{\Bbb N}$, the latter with the product order (via prime factorization), then explicitly doing this for some small examples we can see that this set is not so simple to describe in general.

anon
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  • Thanks for your answer. Now, you say that WLOG, $i|(q^m-1)$, else $i$ is associate to a divisor's residue in $\Bbb{Z}_{q^m-1}$. Could you comment on that? I fail to understand it.. – geo909 Jul 07 '14 at 14:49
  • @geo909 For any integers $i$ and $\ell>1$, there exists a divisor $i'\mid\ell$ such that $i'$ is associate to $i$ in the ring $\Bbb Z/\ell\Bbb Z$ hence the ideals / additive subgroups they generate are the same. – anon Jul 07 '14 at 22:30