I asked this on SE but I did not get any answer; I got some progress but I hope here I can find some help to finish the problem out.
Let $R = F[x_1, \ldots, x_n]$ be a polynomial ring over a field $F$, and let $V$ be a $k$-dimensional vector space over $F$. Consider the $R$-module $M = V \otimes_F R$ as a graded module where $\deg(v) = 0$ for $v \in V$ and $x_i$ is a variable of degree $1$. Suppose that $(M,d)$ is a cochain complex where $d$ is a $R$-linear map of degree $+1$ and $d^2 = 0$. If $d \neq 0$ and $H^*(M) \neq 0$ then the cohomology $H^*(M)$ can't be free as a module over $R$.
This is what I got so far: Let $\{v_1, \ldots, v_k\}$ be a $F$-basis of $V$, then $d$ is determined by $d(v_i \otimes 1)$. Write $d(v_i \otimes 1) = \sum_{j=1}^k v_j \otimes P_{j,i}$ where $P_{j,i}$ is a either zero or a homogenous linear polynomial in $R$. As $d\neq 0$, $d(v_i \otimes 1) \neq 0$ for some $i$. Now if $\{m_1, \ldots, m_l\}$ is a generating set of $H^*(M)$ as $R$-module, then $m_j$ should be of the form $[w_i \otimes 1]$ and $\{w_1, \ldots, w_k\}$ is a linearly independent set in $V$. Now I am trying to use that $d(v_i \otimes 1) \neq 0$ to get a linearly dependence relation on elements from the generating set, but it might happen that $d(v_j \otimes P_{j,i}) \neq 0$ and so $v_j \notin span(w_1, \ldots, w_l\}$
