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Is it true that if the finite field $\mathbb{F}_q$ have the following realisation:

$$\mathbb{F}_q \simeq \mathbb{F_p[\alpha]}/(f(\alpha))$$

then $\alpha$ is the generator of its multiplicative group $\mathbb{F}^{\times}_q$ ?

I tried the following. First note that $|\mathbb{F}^{\times}_q| = q - 1$, so if $$q - 1 = d_1^{e_1}\dots d_k^{e_k}$$ $\alpha$ is the generator $\iff$ $\forall i=1\dots k(\alpha^{d_i'} \neq 1 \mod f(\alpha)) $ where $d_i' = d_i^{e_i}$

But does there exist an example, where $(\alpha^{d_i'} = 1 \mod f(\alpha))$ for some $i \in \{1\dots k\}$ ?

envy grunt
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1 Answers1

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Answering the first question:

Is it true that if the finite field $\mathbb{F}_q$ have the following realisation:

$$\mathbb{F}_q \simeq \mathbb{F_p[\alpha]}/(f(\alpha))$$

then $\alpha$ is the generator of its multiplicative group $\mathbb{F}^{\times}_q$ ?

No. This is easiest to see in the particular case when $q = p^4$. In this case, each element of $\mathbb F_q \setminus \mathbb F_{p^2}$ generates the $\mathbb F_p$-algebra $\mathbb F_q$ (since otherwise, the $\mathbb F_p$-algebra it would generate would be a proper intermediate field between $\mathbb F_p$ and $\mathbb F_q$; but the only such intermediate field is $\mathbb F_{p^2}$), and thus qualifies as an $\alpha$. Hence, there are $q - p^2 = p^4 - p^2$ many possible $\alpha$s in $\mathbb F_q$. But the number of generators of $\mathbb F_q^\times$ is $\phi\left(q-1\right) = \phi\left(p^4-1\right)$, which is usually smaller than $p^4 - p^2$. (For example, if $p = 2$, then $\phi\left(p^4 - 1\right) = \phi\left(15\right) = 8$ is smaller than $p^4 - p^2 = 12$.)

I'm curious about the general case; is there an infinite family of $q$s for which every generator of the $\mathbb F_p$-algebra $\mathbb F_q$ is also a generator of the multiplicative group $\mathbb F_q^\times$, or is this merely a small-number phenomenon?

darij grinberg
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  • But $\alpha$ generates $\mathbb{F}_{2^4} \simeq \mathbb{F}_2[\alpha] / (\alpha^4 + \alpha + 1)$, we can conclude it from the simple brute force. Or I just misunderstood you? – envy grunt Nov 04 '19 at 00:27
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    @envygrunt: I'm not sure I understand you well. $\alpha$ is not fixed in my answer; only the fields are fixed. – darij grinberg Nov 04 '19 at 00:34
  • Oh, I see, my bad. I just wanted to find an explicit example for $q = p^4$. Your answer seems correct – envy grunt Nov 04 '19 at 01:06
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    What if we take $f(x) = x^4 + x^3 + x^2 + x + 1$ as irreducible polynomial over $\mathbb{F}2$ and $\alpha : f(\alpha) = 0$, so it's generates $\mathbb{F}{16}$ and $\text{ord} (\alpha) = 5$, looks like counterexample, isn't it ? – envy grunt Nov 04 '19 at 01:09
  • If $f$ is irreducible (haven't checked), then yes, it is a counterexample. – darij grinberg Nov 04 '19 at 01:15
  • thank you for response – envy grunt Nov 04 '19 at 01:37
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    It is mostly a small number phenomenon, but can happen for relatively large fields. By Zsigmondy's theorem, with a described set of exceptions, there even exists a prime $q$ such that $q\mid p^n-1$, but $q\nmid p^k-1$ for any $k<n$. Thus a root of unity of order $q$ generate the field $\Bbb{F}_{p^n}$ as an $\Bbb{F}_p$-algebra but not as a group. That is, unless $p^n-1=q$, when we are in the Mersenne prime case, and we have many counterexamples (but don't know if there will be infinitely many). – Jyrki Lahtonen Nov 04 '19 at 04:16
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    @envygrunt Correct, $f(x)=x^4+x^3+x^2+x+1$ over $\Bbb{F}_2$ is a counterexample. A "smaller" one if $f(x)=x^2+1$ over $\Bbb{F}_3$. Its zeros are obviously roots of unity of order four, and reside in the quadratic extension $\Bbb{F}_9$, but won't generate it as a multiplicative group. – Jyrki Lahtonen Nov 04 '19 at 04:19
  • Anyway, when $p>2$, an element of order $p^{n-1}+p^{n-2}+\cdots+p+1$ in $\Bbb{F}{p^n}$ will generate $\Bbb{F}{p^n}$ as an algebra over $\Bbb{F}_p$, but won't generate the multiplicative group. – Jyrki Lahtonen Nov 06 '19 at 05:39