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Let $p$ be a prime, and consider the finite field $\mathbb{F}_p$. Fix any $n\ge1$, and consider the field extension $\mathbb{F}_{p^n}/\mathbb{F}_p$. If $\alpha\in\mathbb{F}_{p^n}$ is a multiplicative generator, then $\alpha$ is also a primitive element, that is, $\mathbb{F}_{p^n}=\mathbb{F}_p(\alpha)$, and further, the set $\{1,\alpha,\ldots,\alpha^{n-1}\}$ is an $\mathbb{F}_p$-linear basis of $\mathbb{F}_{p^n}$. Therefore, we can conclude that the elements $\alpha,\ldots,\alpha^{n-1}\not\in\mathbb{F}_p$.

In the above statement, I could conclude 'non-membership' in $\mathbb{F}_p$ by using linear independence over $\mathbb{F}_p$. Is a converse of this true, that is, must 'non-membership' imply linear independence over $\mathbb{F}_p$? Specifically, for $n,m\ge1$, if $\alpha\in\mathbb{F}_{p^n}$ is a multiplicative generator such that the elements $\alpha,\ldots,\alpha^{m-1}\not\in\mathbb{F}_p$, then must $m\le n$?

The Discrete Guy
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  • If $\alpha$ is primitive then its lowest power in the prime field is that with exponent $(p^n-1)/(p-1)$. BTW, finite field people reserve the term primitive element to mean a generator of the multiplicative group. This differs from what's common elsewhere in field theory. My elaboration. – Jyrki Lahtonen Mar 29 '24 at 20:00
  • That is useful to know. In the above question, I really want the element $\alpha$ to be a multiplicative generator and satisfy the primitive element theorem. So as per your elaboration, I simply need to consider a multiplicative generator. – The Discrete Guy Apr 07 '24 at 02:23
  • Correct. But non-membership of low powers is a rather low bar for an element of a bigger field to clear. – Jyrki Lahtonen Apr 07 '24 at 05:11
  • True, I understand. However, it becomes interesting in an asymptotic setting. Assume $p\to\infty$ over primes, and another parameter $m\to\infty$. By the 'non-membership' condition that I could get from @ancientmathematician 's answer (see the comments to his answer), one only needs to take $\mathbb{F}_{p^2}$ in order to get an element $\alpha$ so that $\alpha,\ldots,\alpha^{m-1}\not\in\mathbb{F}_p$, as long as $m=o(p)$ (that is $m/p\to\infty$). In particular, one can get as many as $O(p/\log p)$ powers being not in $\mathbb{F}_p$. I don't see an easier argument for something like this. – The Discrete Guy Apr 08 '24 at 00:09
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    You can actually do a bit better than that. If $\alpha$ is a primitive element of $\Bbb{F}_{p^2}$, then none of $\alpha,\alpha^2,\ldots,\alpha^p$ belong to the prime field (but $\alpha^{p+1}$ is always in $\Bbb{F}_p$). For example, we can think of $\Bbb{F}_9$ as $\Bbb{F}_3(i), i^2=-1$. Then we can choose $\alpha=1+i$, and the fourth power $\alpha^4$ is the lowest in the prime field ($\alpha^4=2=-1$). – Jyrki Lahtonen Apr 08 '24 at 06:20
  • Oh right, yes. I missed the point that $\alpha^{p+1}$ must always be in $\mathbb{F}_p$. – The Discrete Guy Apr 08 '24 at 18:49

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Let $p=2$ and $n=3$, and let $\alpha$ be a generator of the multiplicative group, so that $\alpha$ has order $7$.

Then none of $\alpha,\alpha^2,\alpha^3,\alpha^4,\alpha^5,\alpha^6$ lie in $\mathbb{F}_2$. Yet $7>3$.

ancient mathematician
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  • Just to make sure I understand: $\alpha$ is the root of an irreducible cubic polynomial over $\mathbb F_2$, and ${1,\alpha,\alpha^2}$ is a basis for $\mathbb F_8$ over $\mathbb F_2$, correct? – J. W. Tanner Mar 28 '24 at 01:04
  • How do we know that none of $\alpha,\alpha^2,\alpha^3,\alpha^4,\alpha^5,\alpha^6$ lie in $\mathbb{F}_2$? Because if one of them were $1$ then the order wouldn’t be $7$, and if one of them were $0$ then we would have $\alpha^7=0$, not $1$ – J. W. Tanner Mar 28 '24 at 01:58
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    @J.W.Tanner (i) Yes, we can assume $\alpha, \alpha^2,\alpha^4$ are the roots of the irreducible $X^3+X+1$, and $1,\alpha,\alpha^2$ is an $\mathbb{F}_2$ basis of $\mathbb{F}_8$. (ii) The multiplicative group is cyclic of order $7$, $\alpha$ is a generator, so these are six distinct invertible elements of order $7$, so do not belong to $\mathbb{F}_2$. – ancient mathematician Mar 28 '24 at 07:36
  • Yes, I think we can abstract out the example of @ancientmathematician to conclude that, following the notations in the question, `non-membership' is characterized by the condition $(m-1)(p-1) < p^n - 1$. – The Discrete Guy Mar 29 '24 at 11:29