# Profinite projective and free modules

I am studying cohomology of profinite groups and the following question came to my mind: suppose we have $$G$$ a pro-$$p$$ group which is Poincaré Dual of dimension $$d$$. This means that $$\mathbb{Z}_p$$ as trivial $$\mathbb{Z}_p [\![ G ]\!]$$-module admits a projective resolution of length exactly $$d$$ and degree-wise finitely generated and also that $$H^{s}(G; \mathbb{Z}_p [\![ G ]\!])=0$$ when $$s \neq d$$ and $$H^{d}(G; \mathbb{Z}_p [\![ G ]\!])=\mathbb{Z}_p$$.

Can we deduce from this that for any $$P$$ profinite projective $$\mathbb{Z}_p [\![ G ]\!]$$-module then $$H^{s}(G; \mathbb{Z}_p [\![ G ]\!])=0$$ for $$s \neq 0$$?

The idea would be that we can construct every projective module via retractions and limits of free modules. But when I tried to articulate this reasoning I realized that I had many doubts how free and projective modules are in general in this context.

If $$G$$ were discrete and we were considering the usual modules (without topology) then I know that all free modules are in the form $$\bigoplus_I \mathbb{Z}[G]$$ and that a module is projective iff it is a summand of a free module. These facts are covered by classical homological algebra.

In the case $$G$$ is pro-$$p$$ and we consider profinite modules with a continuous action then I am not sure if these facts still hold. It is easy to see that if $$I$$ is not finite then the direct sum $$\bigoplus_I \mathbb{Z}_p[\![ G ]\!]$$ has not a profinite topology (if considered as subspace of the product $$\prod_I \mathbb{Z}_p[\![ G ]\!]$$). If $$I$$ is finite then $$\bigoplus_I \mathbb{Z}_p[\![ G ]\!] \cong \prod_I \mathbb{Z}_p[\![ G ]\!]$$ and the topology is profinite, so I would think that if we limit ourselves to just finitely generated modules the story goes as in the discrete case.

But what if we want to consider also not finitely generated modules? Do we have the existence of arbitrary free and projective modules, and if yes can we provide an explicit description?

Another point which I have some perplexity: what is the correct notion of free module in this context? In the discrete case we think of free $$\mathbb{Z}[G]$$-modules as the image of the left adjoint to the forgetful functor $$\mathbb{Z}[G]-\text{Mod}\rightarrow \text{Sets}$$. If we want to adapt this to the profinite case we should be careful to choose the right category of modules: I chose the category of profinite topological modules with a continuous $$\mathbb{Z}_p[\![ G ]\!]$$-action because of convenience, but nothing stops me from taking a more suitable category.

Also I do not know if the forgetful functor should land in the category of sets or we should end in the category of topological spaces to retain some information about the topology.

• You should look at the book Profinite Groups of Ribes and Zalesskii. You will find the correct notion of free profinite modules and direct sum there for infinite generating sets. They work with generating sets converging to $0$ for free modules and similar things for coproducts (direct sums) – Benjamin Steinberg May 15 at 15:03
• I did not know of this book, I will surely look into it. Thanks for the recommendation. – N.B. May 15 at 15:13
• springer.com/gp/book/9783642016417 – Benjamin Steinberg May 15 at 15:22