Let $A$ be a local ring with residue field $k$. Define $\overline{B}:=B \otimes_A k$ where B is an A-algebra which is a finitely generated $A$-module. Prove that $\Omega_{\overline{B}/k}=0$ if and only if $\overline{B}$ is the product of finitely many finite, separable field extensions of $k$.
My attempt: Start with the "only if" part. Since $\overline{B}$ is a finite dimensional $k$-vector space, then,by the structure theorem of artinian rings we have $\overline{B} \cong \prod_{i=1}^n S_i$ where each $S_i$ is a local artinian $k$-algebra. Call $\mathfrak{m}_i$ the maximal ideal of $S_i$. Use now the second exact sequence for kahler differentials and get
$\mathfrak{m}_i/\mathfrak{m}_i^2 \to \Omega_{\frac{S_i}{k}} \otimes_{S_i} S_i/\mathfrak{m_i} \to \Omega_{\frac{S_i/\mathfrak{m}_i}{k}} \to 0$
because $S_i/ \mathfrak{m_i}\cong k$, the sequence becomes
$\mathfrak{m}_i/\mathfrak{m}_i^2 \to \Omega_{\frac{S_i}{k}} \otimes_{S_i} k \to 0 \to 0$
and the first two terms of the sequence are isomorphic. But by hypothesis the second term of the sequence is zero and hence $\mathfrak{m_i}/\mathfrak{m_i}^2=0$; by Nakayama's Lemma we get $\mathfrak{m_i}=0$. It follows that $S_i$ is a field and it is a finite extension of $k$. Each of the extensions $k \subseteq S_i$ is separable because $\Omega_{\overline{B}/k} \cong \prod_{i=1}^n \Omega_{S_i/k}=0$ if and only if $\Omega_{S_i/k}=0$ if and only if $k \subseteq S_i$ is separable for each $i$.
EDIT:the isomorphism $\mathfrak{m_i}/\mathfrak{m_i}^2 \cong 0$ is justified by the fact that $(\mathfrak{m_i}/\mathfrak{m_i}^2)^* \cong \mathrm{Der}_k (S_i, k) \cong \mathrm{Hom}_{S_i}(\Omega_{S_i/k},k) \cong \mathrm{Hom}_{k}(\Omega_{S_i/k}\otimes_{S_i} k, k)$, i.e. the dual $k$-spaces of the first two terms in the sequence are isomorphic.
The "if part" it's clear. Do you think it works? Thanks to everybody!