My problem is as follows:
Assume you have a vector space of dimension $(d + 1)$, with values over $GF(q)$. Every vector in this vector space can be regarded as an element of the extension field $GF(q^{d+1})$. It is well known that every element of a finite field can be expressed as a primitive element power and as a linear combination of primitive powers from $0$ up to $d$.
It turns out that the element of the field expressed as the $(i + kn)$th powers of a primitive element, taken modulo $n$ have linearly dependent vector representation (elements of vectors are the coefficients of the lin combination), where $n = (q^{d + 1} - 1)/(q - 1)$. They basically lie on the same line in the original vector space...
But i can't find a proof... I'm not really an expert of the field and i suspect the problem could be stated more formally. Any help? Thanks in advance!
Francesco
Edit: let's say $\beta = \alpha^i$, where $\beta$ is an element of the extension field and $\alpha$ is a primitive over that field. We also know that $\beta$ can be expressed as a linear combination of $1, \alpha, \alpha^2, \dots, \alpha^d$ with coefficients $(a_0, \dots, a_d)$ over $GF(q)$. Those coefficients actually determine a vector of the finite vector space of dimension $(d+1)$. Now take a second element $\beta' = \alpha^{i+kn}$, for some $k$ integer. The coefficient of the linear combination of $1, \alpha, \alpha^2, \dots, \alpha^d$ that represent $\beta'$ form a vector $(a_0', \dots, a_d')$ which is linearly dependent to $(a_0, \dots, a_d)$. So in the vector space of dimension $(d+1)$ with values over $GF(q)$, those vectors lie on the same line through the origin. I know that this is true, but I can't find a proof. So my question is actually a proof.
Hope to have make my point a bit more clear.