If $A$ is a finitely generated $k$-algebra with no nontrivial maps $m/m^2\to k$ for any $m\subset A$, is $A\cong k^n$ as modules?

Solution 1:

Yes, this is true. The condition you write is equivalent to zero-dimensional tangent spaces at every closed point of $\operatorname{Spec} A$. Since $\dim_{A/m} m/m^2$ is at least the Krull dimension of $A_m$, we have that $A_m$ is dimension zero for all maximal ideals of $A$. Since the Krull dimension of a ring is the supremum of the dimensions of it's local rings at maximal primes, we have that $A$ is of Krull dimension zero. But $A$ is also noetherian, being a finitely-generated $k$-algebra. As noetherian rings of dimension zero are artinian, and artinian $k$-algebras are finite-dimensional as $k$-vector spaces, this finishes the proof.