All rings are assumed to be commutative.
Let $A$ be a ring, and $B$ an $A$-algebra which is Artinian local with the maximal ideal $\mathfrak m$.
Suppose $B$ is formally smooth over $A$.
My question is whether the following is true or not:
For any $D \in Der_A(B)$, and for any $x\in \mathfrak m$, $Dx\in \mathfrak m$.
Or, saying the same thing,
$Hom_B(\Omega _{B/A},B/\mathfrak m)=0$.
I think this is true when $A$ is a field. Thanks.
Edit: Since $\Omega_{B/A}$ is a free $B$-mod (formal smoothness implies projectivity; it implies freeness), the above condition is equivalent to $\Omega_{B/A}=0$. Thus B: formally étale?
Edit: The above condition is equivalent to
$Coker(Der_A(B/\mathfrak m)\longrightarrow Der_A(B,B/\mathfrak m))=0$.
Edit: It is equivalent to saying that the canonical map $\Omega_{B/A}\otimes_B K\longrightarrow\Omega_{K/A}$, where $K=B/\mathfrak m$, is an isomorphism.
I have self-solved!
Localizing $A$ at the pullback of $\mathfrak{m}$, we may assume that $A\longrightarrow B$ is a local homomorphism ($B$ still remains formally smooth over $A$).
Let $\kappa$ denote the residue field of $A$, $\mathfrak{n}$ the maximal ideal of $A$.
Then by base-change $$B\otimes_A \kappa = B/\mathfrak{n}B$$ is formally smooth over a field $\kappa$.
Since $B$ is Artinian, a fortiori Noetherian, it follows that $B/\mathfrak{n}B$ is a regular local ring, in particular, a domain.
Therefore $\mathfrak{n}B$ is a prime ideal, and hence $\mathfrak{n}B = \mathfrak{m}$.
Thus $$D(\mathfrak{m}) = D(\mathfrak{n}B) \subset \mathfrak{n} D(B) \subset \mathfrak{n}B = \mathfrak{m}.$$