Characterizing the power mappings that preserve linear independence over finite fields

Question

Let $q$ be a power of a prime $p$, and $\alpha$ be a primitive element of $GF(q^n)$, for some integer $n$. For integer $l$ such that $0\leq l \leq q^n-2$ consider the following statement:

$$ S(l)= \text{ "for every } c_1, \ldots, c_t \in \left\{0, \ldots, q^n-2\right\} \text{ we have that } \alpha^{c_1}, \ldots, \alpha^{c_n} \text{ are linearly independent if and only if } \alpha^{lc_1}, \ldots, \alpha^{lc_n} \text{ are linearly independent."} $$

It is not hard to show that when $l$ is a power of the characteristic $p$, then $S(l)$ holds (see my proof below). My hunch says that the converse is also true, that is, if $S(l)$ holds then $l$ must be a power of $p$. So far I could not prove it or disprove it. I would be grateful if somebody could give me a proof or a counterexample.

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Lemma. For integer $r$, $S(p^r)$ holds.

Proof: For $y_1, \ldots, y_n \in GF(q)$ not all zero, we have that $\sum_{i=1}^n y_i\alpha^{c_i}=0$ implies $\left(\sum_{i=1}^n y_i\alpha^{c_i}\right)^{p^r}= \sum_{i=n}^{n}y_i^{p^r} \alpha^{p^rc_i}=0$. It is well known that if $y$ is a nonzero element of $GF(q)$ then so is $y^{p^r}$, hence we have proven that if $\alpha^{c_1}, \ldots, \alpha^{c_t}$ are linearly dependent, then so are $\alpha^{p^rc_1}, \ldots, \alpha^{p^rc_t}$.

For the other direction, we observe that $q=p^e$ for some $e$, and thus $q^n=p^{ne}$. Now, suppose that $\sum_{i=1}^n y_i\alpha^{p^rc_i}=0$ where $y_1, \ldots y_n \in GF(q)$ are not all zero. Then, raising to the power $p^{en-r}$ yields $\left(\sum_{i=1}^n y_i\alpha^{p^rc_i}\right)^{p^{en-r}}= \sum_{i=1}^n y_i^{p^{ne-r}}\alpha^{c_i}=0. $ Arguing as before, we have proven that if $\alpha^{p^rc_1}, \ldots, \alpha^{p^rc_t}$ are linearly dependent, then so are $\alpha^{c_1}, \ldots, \alpha^{c_t}$.

In conclusion, we showed that $\alpha^{c_1}, \ldots, \alpha^{c_t}$ are linearly dependent are linearly dependent if and only if $\alpha^{p^rc_1}, \ldots, \alpha^{p^rc_t}$ are linearly dependent, which is equivalent to $S(p^r)$.

EDIT: In the above I see $GF(q^n)$ as an $n$-dimensional vector space over $GF(q)$, and I talk about linear independence in that context.

Answer 1

Presumably you consider linear independece over the subfield $GF(q)$?

I don't know what can be said about the general case. For the time being I lead off with a few observations.

• Your argument for $S(p^r)$ seems to extend to the result: $$S(\ell)\Longleftrightarrow S(p^r\ell).$$ In other words, the truth of $S(\ell)$ only depends on the cyclotomic coset of $\ell$.
• In general the validity of $S(\ell)$ depends also on $n$. As a case in point let's consider $\ell=-1$ (or $\ell=q^n-2$ if you absolutely cannot stomach negative exponents here):
  • When $n=2$ the condition $S(-1)$ is true. This is because two non-zero vectors are linearly independent unless one is a scalar multiple of the other. But obviously any pair of non-zero elements $x,y\in GF(q^2)$ are $GF(q)$-multiples of each other if and only if the same holds for the reciprocals $1/x,1/y$.
  • But $S(-1)$ is false in general when $n=3$. Let $q=2$. The zeros of the polynomial $x^3+x^2+1$ in $GF(8)$ are linearly independent over $GF(2)$ (they form a normal basis). Their reciprocals are the zeros of the reciprocal polynomial $x^3+x+1$. But, that being a depressed cubic, the sum of those reciprocals is zero so they are linearly dependent.
• The first item in the previous bullet should generalize to the form: $S(\ell)$ is true in $GF(q^2)$ whenever $\gcd(\ell,q^2-1)=1$. This is because raising to the $\ell$th power is then a bijection of $GF(q^2)$. Furthermore, this bijection maps the subfield $GF(q)$ to itself. Consequently $x/y\in GF(q)$ if and only if $x^\ell/y^\ell\in GF(q)$.

For this to be a satisfactory answer I should be able to come up with an example where $S(\ell)$ holds, $\ell$ is not in the cyclotomic coset of $1$, and $n>2$. Unfortunately that will have to wait (and I don't know for sure whether such examples exist).