I'm studying this article, and I have related to the proof of theorem 14.
Let $f$ be a linearized polynomial over $\mathbb{F}_{q^m}$, which means $f$ is of the form $$ f(x)=\sum_{i=0}^{k-1} a_0 x^{q^i}, $$ where $a_i\in \mathbb{F}_{q^m}$. In the proof I need to find two linearized polynomials $f$ and $g$ of degree at most $q^{k-1}$ such that $f$ and $g$ are equal on $k-1$ linearly independent points, i.e. $f(\alpha_i)=g(\alpha_i)$ for $i=1,...,k-1$, where the $\alpha_1,...,\alpha_{k-1}$ are linearly independent points.
My idea so far is to consider the vectorspace $V=\left<\alpha_1,...,\alpha_{k-1}\right>$ and define $f(x)=\prod_{\beta\in V}(x-\beta)$. And then define $g(x)=c\cdot f(x)$, where $c\neq1$ is some non-zero scalar in $\mathbb{F}_q$. Clearly $f(\alpha_i)=g(\alpha_i)$ for all $i$, since $\alpha_i$ are roots. This does unfortunately not work if $q=2$ because the only scalar to choose from is $1$. Does anyone have an idea how to remedy this?
Thanks a lot for your time.