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Given finite field $GF(2^b)$ with elements $v_1,\ldots,v_{2^b}$, and integer $a$ such that $2^{b} \geq 2(a+1)!$, I am trying to locate a proof that the set of column vectors $e_i = (1,v_i,v_i^3,\ldots,v_i^{2a+1})^{\operatorname{T}}$, $1 \leq i \leq 2^b$, form a spanning set for additive group $\mathbb{Z}_2^{1+b(a+1)}$ (where each $v_i$ is itself a column vector, so $e_i$ is a vector of length $1+b(a+1)$).

Firstly, is this statement actually true? I'm aware a series of vectors don't really 'span' a group in the sense of a basis, but I mean that each element of $\mathbb{Z}_2^{1+b(a+1)}$ is a linear combination of the $e_i$.

Secondly, if indeed it is true (I believe/hope it is!), is there a particularly nice and elementary or concise way to prove it? Thanks for your help!

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    Doesn't look like that is enough. The counterexample in my answer has $a=2$, $b=4$.2012-03-07

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I'm afraid the answer to you question is negative, unless you place some restrictions on the parameters.

One problem (there may be others) is related to the norm maps. As the smallest example consider the case $b=4$. The (relative) norm map $N(x)=x^5$ from $GF(16)$ to itself only takes values in the subfield $GF(4)$. Therefore the component $v_i^5$ will have only four possible values, and these four values form a proper additive subgroup. Thus those components will never span all of $GF(16)$. The problem can be described by stating that all the vectors $(1,x,x^3,x^5)$ with $x$ ranging over $GF(16)$ belong to the subgroup $GF(2)\oplus GF(16)\oplus GF(16)\oplus GF(4)$. Therefore they cannot span anything bigger.

Similar cases occur for suitable values of $a$ whenever the field $GF(2^b)$ has proper subfields, i.e. unless $b$ is a prime. I need to sleep on it to determine, whether the extra condition $2^b\ge 2(a+1)!$ rules out all but finitely many of these norm map problems.

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    Thanks for your help so far! The suggested method did indeed go on to say that the linear independence 'followed from the theory of BCH codes', but since I didn't know anything about them I was simply hoping there might be a method which avoided that road which I would find more natural. $A$nyway, it's an interesting unique case if you are correct - much of the maths I am studying is 'for (*parameter*) large enough', so it may be that this single failing is just ignored here.2012-03-07