# about the cohomology of a tensor module.

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.

• 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. – Pedro Tamaroff Feb 2 at 11:30
• It would also be instructive for you to come up with an example where $C$ is infinite and your conclusion is false. – Pedro Tamaroff Feb 2 at 14:02
• @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. – Vitolo Feb 2 at 16:32
• If you found an answer then please do post it here! – Pedro Tamaroff Feb 2 at 17:25