How large can the dimension of a 'Span of powers of a finite field basis' be?

Question

Let $p$ be a prime. For finite field $\mathbb{F}_{p^k}$ and $d\in\mathbb{Z}^+$, I am considering the following quantity, where we interpret the field $\mathbb{F}_{p^k}$ also as a $\mathbb{F}_p$-vector space.

$$M_{p^k}^{(d)}:=\max_{\{\beta_1,\cdots,\beta_k\}\text{ is a basis of } \mathbb{F}_{p^k}}\Big\{\dim\langle\beta_1^d, \cdots, \beta_k^d \rangle\Big\}$$

For fixed $p^k$ and $d$, is there any known results on exact value or bounds of this quantity? I am both interested in the general case and the special case of $d=2$. Any idea or comment will be very helpful. The following are some very basic facts that I have observed, but I could not go further.

• When $d$ is a power of $p$, it holds that $M_{p^k}^{(d)}=k$, regarding a normal basis and Frobenius map.
• $M_{p^k}^{(d)}\ge \lfloor \frac{k-1}{d} \rfloor + 1$, regarding first $(\lfloor \frac{k-1}{d} \rfloor + 1)$ elements of a primitive element basis of the form $\{1, \beta, \beta^2, \cdots, \beta^{k-1}\}$.
• If $\gcd (d,p^k-1)=1$, then $M_{p^k}^{(d)}=k$, regarding a primitive element basis of the form $\{1, \beta, \beta^2, \cdots, \beta^{k-1}\}$. Indeed, $\beta^d$ is a primitive element, and $\{1, \beta^d, \beta^{2d}, \cdots, \beta^{(k-1)d}\}$ is again a basis. (from the discussion with Donggeon Yhee)

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Some experimental results:

A primitive element basis of the form $\{1, \beta, \beta^2, \cdots, \beta^{k-1}\}$ usually gives the optimal basis.

$p=2$ case:

• $M_{2^k}^{(3)}=k$ for $k\le 100$, except $M_{2^2}^{(3)}=1$
• $M_{2^k}^{(5)}=k$ for $k\le 100$, except $M_{2^4}^{(5)}=2$
• $M_{2^k}^{(6)}=k$ for $k\le 100$, except $M_{2^2}^{(6)}=1$
• $M_{2^k}^{(7)}=k$ for $k\le 100$, except $M_{2^3}^{(7)}=1$
• $M_{2^k}^{(9)}=k$ for $k\le 100$, except $M_{2^2}^{(9)}=1$ and $M_{2^6}^{(9)}\ge 3$
• $M_{2^k}^{(10)}=k$ for $k\le 100$, except $M_{2^4}^{(10)}=2$
• $M_{2^k}^{(25)}=k$ for $k\le 100$, except $M_{2^4}^{(25)}=2$
• $M_{2^k}^{(27)}=k$ for $k\le 100$, except $M_{2^2}^{(27)}=1$ and $M_{2^6}^{(27)}\ge 3$

$p=3$ case:

• $M_{3^k}^{(2)}=k$ for $k\le 100$
• $M_{3^k}^{(4)}=k$ for $k\le 100$, except $M_{3^2}^{(4)} = 1$
• $M_{3^k}^{(5)}=k$ for $k\le 100$
• $M_{3^k}^{(6)}=k$ for $k\le 100$

$p=5$ case:

• $M_{5^k}^{(2)}=k$ for $k\le 100$
• $M_{5^k}^{(3)}=k$ for $k\le 100$
• $M_{5^k}^{(4)}=k$ for $k\le 100$
• $M_{5^k}^{(6)}=k$ for $k\le 100$, except $M_{5^2}^{(6)} = 1$

Answer 1

Let $\mathbb{F}$ be a finite field and $\mathbb{E}$ be an extension of a prime degree (say $q$). Let $d$ be co-prime to $|\mathbb{F}|^q-1$.

$\langle\beta_1^d,...,\beta_q^d\rangle$ is a $\mathbb{F}$-subspace of $\mathbb{E}$, whose dimension over $\mathbb{F}$ is $q$. In other words, $\{\beta_1^d,...,\beta_q^d\}$ is again a basis of $\mathbb{E}$ over $\mathbb{F}$.

Recall your question.

At first, assume $gcd(d, p^k-1)=1$. Let $k=q_1q_2...q_m$ be a prime factorization and $k_j:=q_1q_2...q_j$.

A basis of $\mathbb{F}_{p^k}$ can be given as $\mathcal{B}_m\times \mathcal{B}_{m-1}\times ...\times \mathcal{B}_1$ where $\mathcal{B}_j$ is a basis of $\mathbb{F}_{p^{k_j}}$ over $\mathbb{F}_{p^{k_{j-1}}}$. Thus, $M_{p^k}^{(d)}=k$ in the case.

Secondly, we consider $gcd(d,p^k-1)=n\neq 1$. Then $\{\beta_1^d,...,\beta_k^d\}$ lies in a proper subspace of $\mathbb{F}_{p^k}=\{0, x~|~x^{p^k-1}=1 \pmod{p}\}$. Then we reduce your question to $d'|d$ and $k'|k$ and $gcd(d',p^{k'}-1)=1$. Here $d'=\frac{d}{n}$ and $k'$ be the maximum divisor of $k$ so that $n|\frac{p^k-1}{p^{k'}-1}$. ("maximum" or "maximal" seems to be more proper than "largest"...)

typo : Sorry for the first assumption on $d$. $q-1$ is replaced by $|\mathbb{F}|^q-1$.