Let $A$ be a topological ring, we say that it is pseudo-compact if :
- there is family of ideals $\Lambda_A$ which gives a basis of neighborhoods of $0$ and such that $A/\mathfrak{a}$ is artinian for every $\mathfrak{a} \in \Lambda_A$
- $A$ is complete.
We have $A = \varprojlim\limits_{\mathfrak{a} \in \Lambda_A} A/\mathfrak{a}$
A topological module $M$ over a pseudo-compact ring $A$ is said to be profinite if :
- There is a familly of submodules $\Lambda_M$ which forms a basis of neighborhoods of $0$ and such that $M/N$ is finite over $A$ for every $N \in \Lambda_M$
- $M$ is complete
We have $M = \varprojlim\limits_{N \in \Lambda_M} M/N$
A profinite $A$ algebra is just an $A$-algebra which is profinite as an $A$ module, it is easy to see that it is in fact a pseudocompact ring.
Now let $A$ be a pseudo-compact ring and $B$ be a profinite $A$ given by a map $f : A \to B$. (We can consider this a map of formal schemes $ :Spf(B) \to Spf(A)$ if we want to be more geometric)
It is very natural to define the differentials of $B$ over $A$ to be $$\widehat{\Omega}_{B/A} := \varprojlim\limits_{\mathfrak{b},\mathfrak{a}} \Omega_{(B/\mathfrak{b})/(A/\mathfrak{a})}$$ where the limits runs over pairs of ideal over $B$ and $A$ respectively whith $f^{-1}(\mathfrak{b}) \subset \mathfrak{a}$. Indeed we are just saying that $Spf(A)$ is the glueing of real schemes $Spec(A/\mathfrak{a})$ and $Spf(B)$ is the glueing of real schemes $Spec(B/\mathfrak{b})$ over each $Spec(A/\mathfrak{a})$ and we glue our original construction of the Kähler differentials.
you can look at http://www.math.jussieu.fr/~polo/SGA3/Exp7B-23mai11.pdf for much more on the subject of pseudo-compact rings
My question : I've read that $\widehat{\Omega}_{B/A}$ represent the functor taking a profinite $B$-module to the module of continous derivations from $B$ to $M$. Although it seems intuitive and I'd be glad to accept it as a fact I haven't been able to prove it myself so I'm wondering if someone could give a proof of this (with all the gory details if possible)
To answer your question in full would take quite a bit of work (to get all details straight -- if the thing is true) which I am not willing to do. I suggest looking back at where you found this statement and look for a proof there or a reference.
But here is a baby case. Suppose we have a ring map $A \to B$ and a finitely generated ideal $I$ of $B$. Then we can endow every module with respect to the $I$-adic topology and we can consider the $I$-adic completion of a module. Now there are two things we can do to $\Omega_{B/A}$. Namely, we can take its $I$-adic completion $$ \hat \Omega_{B/A, 1} = \lim \Omega_{B/A}/I^n\Omega_{B/A} $$ and we can form $$ \hat \Omega_{B/A, 2} = \lim \Omega_{(B/I^n)/A} $$ There is a canonical map $\hat \Omega_{B/A, 1} \to \hat \Omega_{B/A, 2}$. I think your question in this case is whether this map is an isomorphism (did not carefully check this --- instead of profinite modules I am considering complete modules). Anyway, the statement is true because the kernel of the surjective map $\Omega_{B/A} \to \Omega_{(B/I^{n + 1})/A}$ is contained in $I^n\Omega_{B/A}$ by Lemma Tag 02HQ. This allows us to define the inverse map.
Hope this helps!