# $\Omega_{B/A}=0$ if and only if $\Omega_{\overline{B}/k}=0$, where $\overline{B}=B\otimes_A k$.

Let $A$ be a local ring with residue field $k$. Let $B$ be an $A$-algebra such that $B$ is finitely generated as $A$-module. Let $\overline{B}:=B \otimes_A k$. Prove that $\overline{B}$ is non zero and that $\Omega_{B/A}=0$ if and only if $\Omega_{\overline{B}/k}=0$.

The fact that $\overline{B}$ is non zero holds true because it is an Artinian algebra, hence it is product of local Artin rings which are non zero. I would use the Nakayama's Lemma to prove that equivalence, but i don't know how to do it.

Thanks to everybody.

Change of basis for Kahler differentials tells you that $$\Omega_{(B\otimes_A k)/k}\cong k \otimes_A \Omega_{B/A}.$$ So you get the implication $\Omega_{B/A}=0\implies \Omega_{\overline{B}/A}=0.$
For the converse one you should use Nakayama's lemma in this spirit. Since $B$ is finitely generated $A$-module, then $\Omega_{B/A}$ is finitely generated $A$-module as well. Hence, from the above formula and Nakayam's lemma you get the converse implication.
• In fact i put in the hypothesis that B is a finitely generated A-module and so also $\Omega_{B/A}$ is finitely generated as an A-module. – Mamadness Jan 2 '18 at 13:28