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Let $F$ be a field and let $C_*$ be a graded vector space over $F$ such that $C_i = 0$ for $i<0$ and $i>N$ for some integer $N$. Consider $R=F[t]$ as a graded ring (with the usual grading) and $M = C_* \otimes_F R$ be a graded $R$-module with also the usual tensor product grading (the total complex)

Suppose that $(M,d)$ is a cochain complex : $d: M_* \rightarrow M_{*+1}$ is a $R$-linear map such that $d^2 = 0$ and such that no element of the form $x \otimes 1$ belongs to $im(d)$. Then $H^*(M)$ is a free $R$-module if and only if $d = 0$.

The reverse direction is easy: if $d = 0$, then $H^*(M) = M$, as $C_*$ has a $F$-basis it induces an $R$-basis on $M$.

Now I am trying to show that if $d \neq 0$, $H^*(M)$ can't be free. As $d$ is a $R$-linear map, we may assume that $d(c \otimes 1) \neq 0$ for some $c \in C_*$. If $d(c \otimes 1) = x \otimes p$ for $x \in C_*$ and $p \in R$, then $d(x \otimes 1) = 0$ (otherwise $0 \neq p\cdot d(x \otimes 1) = d(x \otimes p) = d^2(c \otimes 1) = 0$). Therefore, the cohomology class $ [x \otimes 1] \neq 0$ is a torsion element of $H^*(M)$. For the general case we assume $d(c \otimes 1) = c_1 \otimes p_1 + x_2 \otimes p_2$. I am trying to rule a linear dependence of a chosen basis of $H^*(M)$ using that $d(c \otimes 1) \neq 0$ but I am stuck here.

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  • $\begingroup$ You are not using that $C$ is of finite length. I suggest trying to pick a largest degree in $C$ such that $d(c\otimes 1)\neq 0$, and this maybe helps into getting a contradiction. $\endgroup$ – Pedro Tamaroff Feb 2 at 11:30
  • $\begingroup$ It would also be instructive for you to come up with an example where $C$ is infinite and your conclusion is false. $\endgroup$ – Pedro Tamaroff Feb 2 at 14:02
  • $\begingroup$ @PedroTamaroff Thanks, I actually was able to prove the statement using that $C$ is finitely generated and that free $\Leftrightarrow$ torsion free as $R$ is a PID. So it was enough to construct a torsion element in $H^*(M)$. Now I am trying to generalize to modules over a polynomial ring in several variables. $\endgroup$ – Vitolo Feb 2 at 16:32
  • $\begingroup$ If you found an answer then please do post it here! $\endgroup$ – Pedro Tamaroff Feb 2 at 17:25

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