Inspired by a previous question (Dimension of a kernel of a linear map) and some of the answers I was given I thought wheter I can generalize the question to the following:
Compute the kernel (or at least the dimension of the kernel) of the map $f_{k,n}:\mathbb{F}_{n}[x_1,\dots,x_k]\to\mathbb{F}_{n}[x_1,\dots,x_{k+1}]$ where $\mathbb{F}$ is a field of dimension $2$, $\mathbb{F}_{n}[x_1,\dots,x_k]$ denotes the set of all the polynomials whose monomials are of the form $x_1^{a_1}\dots x_k^{a_k}$ with $a_1+\dots+a_k=n,a_i\geq 1$ and: $$f_{k,n}(p(x_1,\dots,x_k))=p(x_2,\dots,x_{k+1})+\sum\limits_{i=1}^k p(x_1,\dots,x_{i-1},x_i+x_{i+1},x_{i+2},\dots,x_{k+1})+p(x_1,\dots,x_k)$$ Hence, the elements of $\ker(f_{k,n})$ are $k$-cocycles. Now, let: $$p(x_1,\dots,x_n)=\sum_{(a_1,\dots,a_k)\in I'}x_1^{a_1}\dots x_k^{a_k}\in\ker(f_{k,n})$$ where $I'\subset I=\lbrace(a_1,\dots,a_k)\mid a_1+\dots+a_k=n,a_i\geq1\rbrace$. Now, observe that the coefficient of $x_1^{b_1}\dots x_{k+1}^{b_{k+1}}$ in $f_{k,n}(p(x_1,\dots,x_k))$ is: $$c_1\binom{b_1+b_2}{b_1}+c_2\binom{b_2+b_3}{b_2}+\dots+c_k\binom{b_k+b_{k+1}}{b_k}$$ where $c_k=1$ iff $(b_1+b_2,b_3,\dots,b_{k+1})\in I'$. In the case $k=2$ the coefficients are easy to control, since either both binomial coefficients are $0$ or both are $1$, from which we can prove that every $(a,b)\in I'$ decompose the binary decomposition of $n$ into two disjoint subsets. However, in the case $k>2$ this is not as easy, but I was expecting to have a binary decomposition of $n$ into $k$ disjoint subsets. Maybe it is difficult to solve it in general, but maybe some small cases such that $k=3$ or $k=4$ can be solved.
Thanks for your help.
