Let $k$ be an algebraically closed field, and let $A$ be a Noetherian $k$-algebra. If $\Omega_{A/k}=0$, then $m/m^2=0$ for every maximal ideal $m\subset A$. Is the converse true?
I would say yes, because any $k$-module of the form $m/m^2$ ($m$ maximal as above) is isomorphic, and they are also all isomorphic to $\Omega_{A/k}\otimes_Ak$; however I realized that I don't understand how is done the tensor product with $k$ over $A$ (I would understand the contrary). How is $k$ an $A$-module\algebra?
I cannot give you an actual reference because this result is on the notes of my course, however it is written that, when $A$ is not Noetherian, the proof of this fact is in Matsumura's Commutative ring theory, 25.2.