Module of differentials of an extension of Dedekind domains is cyclic Let  be a Dedekind domain with fraction field , | a finite separable field extension and  the integral closure of  in . Assume that all the residue field extensions are separable.
In the proof of Serre's "Local Fields", chapter III, Proposition 14 (page 59), it is stated that in order to show that the module of differentials $Ω_{|}$ is cyclic we can assume that  is local and complete. (See  this question.)
I understand the reduction to the local case, but I don't understand the reduction to the complete case. I tried to use that the canonical map from $A_p$ to its $p$-adic completion is faithfully flat but I didn't arrive to the desired conclusion. In particular, I don't understand why, if $Ω_{'/'}$ is cyclic with with annihilator the different ideal of B' over A' [where A' and B' denote the $p$-adic completion of the DVRs A and B respectively], then $Ω_{/}$ is cyclic with annihilator the different ideal of B over A.
Can you give me some details? Thank you very much.
 A: It's easy to say "this just follows from faithful flatness", but I think the actual argument is a bit more subtle. So I'll try to be really precise.
Let $\mathfrak P\subset B$ be a prime ideal and $\mathfrak p\subset A$ its preimage. As argued in the question you linked, it suffices to show that $(\Omega_{B/A})_{\mathfrak P}$ is cyclic with annihilator the different $\mathfrak D_{B_{\mathfrak P}/A_{\mathfrak p}}$. Note that $(\Omega_{B/A})_{\mathfrak P}=\Omega_{B_{\mathfrak P}/A_{\mathfrak p}}$; see [Stacks Project, Tag 00RT]. Let $\widehat{B}_{\mathfrak P}$ and $\widehat{A}_{\mathfrak p}$ denote the completions of the DVRs $B_{\mathfrak P}$ and $A_{\mathfrak p}$. The key observation is:
Observation. $\widehat{B}_{\mathfrak P}=\widehat{A}_{\mathfrak p}[x]$ is generated by a single element. Moreover, we may assume that $x$ is already contained in $B_{\mathfrak P}$ .
Proof. The first part is well-known and appears for example in Serre's book, Chapter II Proposition 12, or in Neukirch's "Algebraic Number Theory", Chapter II Lemma (10.4). The second part isn't stated explicitly, but it becomes apparent from the proof given in Neukirch's book (and maybe also from Serre's proof, but I'm not really familiar with Serre's book). $\square$
Now we wish to show that $\Omega_{B_{\mathfrak P}/A_{\mathfrak p}}$ is generated by $\mathrm{d} x$ and that the annihilator is $\mathfrak D_{B_{\mathfrak P}/A_{\mathfrak p}}$, i.e., that the sequence
$$0\rightarrow \mathfrak D_{B_{\mathfrak P}/A_{\mathfrak p}}\rightarrow B_{\mathfrak P}\xrightarrow{\mathrm{d}x}\Omega_{B_{\mathfrak P}/A_{\mathfrak p}}\rightarrow 0 $$
is exact (where the second arrow is the inclusion of the ideal $\mathfrak D_{B_{\mathfrak P}/A_{\mathfrak p}}$ and the third arrow is the map sending $b\in B_{\mathfrak P}$ to $b\mathrm{d}x\in \Omega_{B_{\mathfrak P}/A_{\mathfrak p}}$). Since $\widehat{B}_{\mathfrak P}$ is faithfully flat over $B_{\mathfrak{P}}$, it suffices to show that $$0\rightarrow \mathfrak D_{B_{\mathfrak P}/A_{\mathfrak p}}\otimes_{B_{\mathfrak P}}\widehat{B}_{\mathfrak P}\rightarrow \widehat{B}_{\mathfrak P}\xrightarrow{\mathrm{d}x}\Omega_{B_{\mathfrak P}/A_{\mathfrak p}}\otimes_{B_{\mathfrak P}}\widehat{B}_{\mathfrak P}\rightarrow 0 $$
is exact. We already know that $\Omega_{\widehat{B}_{\mathfrak P}/\widehat{A}_{\mathfrak p}}$ is generated by $\mathrm{d}x$ with annihilator $\mathfrak D_{\widehat{B}_{\mathfrak P}/\widehat{A}_{\mathfrak p}}$. Moreover,  Chapter III Proposition 10 in Serre's book shows that $\mathfrak D_{B_{\mathfrak P}/A_{\mathfrak p}}\otimes_{B_{\mathfrak P}}\widehat{B}_{\mathfrak P}=\mathfrak D_{\widehat{B}_{\mathfrak P}/\widehat{A}_{\mathfrak p}}$. Hence we are done once we show the following lemma:
Lemma. $\Omega_{B_{\mathfrak P}/A_{\mathfrak p}}\otimes_{B_{\mathfrak P}}\widehat{B}_{\mathfrak P}=\Omega_{\widehat{B}_{\mathfrak P}/\widehat{A}_{\mathfrak p}}$.
Proof. Since $\Omega_{B_{\mathfrak P}/A_{\mathfrak p}}$ is a finitely generated $B_{\mathfrak P}$-module, we get that
$$ \Omega_{B_{\mathfrak P}/A_{\mathfrak p}}\otimes_{B_{\mathfrak P}}\widehat{B}_{\mathfrak P}=\lim_{n\geq 0}\left(\Omega_{B_{\mathfrak P}/A_{\mathfrak p}}\otimes_{B_{\mathfrak P}}B_{\mathfrak P}/\mathfrak P^n\right)$$ is its $\mathfrak P$-adic completion. Since $B_{\mathfrak P}$ and $A_{\mathfrak p}$ are DVRs, there is an integer $e\geq 1$ such that $\mathfrak P^e=\mathfrak pB_{\mathfrak P}$. Thus
$$ \lim_{n\geq 0}\left(\Omega_{B_{\mathfrak P}/A_{\mathfrak p}}\otimes_{B_{\mathfrak P}}B_{\mathfrak P}/\mathfrak P^n\right)=\lim_{m\geq 0}\left(\Omega_{B_{\mathfrak P}/A_{\mathfrak p}}\otimes_{A_{\mathfrak p}}A_{\mathfrak p}/\mathfrak p^m\right)=\lim_{n\geq 0}\Omega_{(B_{\mathfrak P}/\mathfrak p^mB_{\mathfrak P})/(A_{\mathfrak p}/\mathfrak p^m)}\,,$$
where the second equality follows from the base change property for Kähler differentials. Similarly, $\Omega_{\widehat{B}_{\mathfrak P}/\widehat{A}_{\mathfrak p}}$ is a finitely generated module over the complete DVR $\widehat{B}_{\mathfrak P}$, hence it also is $\mathfrak P$-adically complete. Thus
$$ \Omega_{\widehat{B}_{\mathfrak P}/\widehat{A}_{\mathfrak p}}\otimes_{B_{\mathfrak P}}\widehat{B}_{\mathfrak P}=\lim_{n\geq 0}\left(\Omega_{\widehat{B}_{\mathfrak P}/\widehat{A}_{\mathfrak p}}\otimes_{\widehat{B}_{\mathfrak P}}\widehat{B}_{\mathfrak P}/\mathfrak P^n\right)\,.$$
With the same trick as above, we can rewrite this as
$$ \lim_{n\geq 0}\left(\Omega_{\widehat{B}_{\mathfrak P}/\widehat{A}_{\mathfrak p}}\otimes_{\widehat{B}_{\mathfrak P}}\widehat{B}_{\mathfrak P}/\mathfrak P^n\right)=\lim_{m\geq 0}\left(\Omega_{\widehat{B}_{\mathfrak P}/\widehat{A}_{\mathfrak p}}\otimes_{\widehat{A}_{\mathfrak p}}\widehat{A}_{\mathfrak p}/\mathfrak p^m\right)=\lim_{n\geq 0}\Omega_{(\widehat{B}_{\mathfrak P}/\mathfrak p^m\widehat{B}_{\mathfrak P})/(\widehat{A}_{\mathfrak p}/\mathfrak p^m)}\,.$$
Now observe $\widehat{B}_{\mathfrak P}/\mathfrak p^m\widehat{B}_{\mathfrak P}=B_{\mathfrak P}/\mathfrak p^mB_{\mathfrak P}$ and $\widehat{A}_{\mathfrak p}/\mathfrak p^m=A_{\mathfrak p}/\mathfrak p^m$ and the lemma follows. $\square$
