Let $B \to A$ be a surjection where $B$ and $A$ are Artin local rings which are $k$-algebras and both having residue field $k$. Let $M$ be the kernel of the surjection and $M^2 = 0$. This induces an $A$-module structure on $M$.
Is $M$ a flat $A$-module?
I know it would be enough to prove $I\otimes M \to M$ is injective for any ideal $I \subset A$. How can I finish the argument?
No, $M$ need not be flat over $A$. Indeed, the action of $A$ on $M$ is inherited from that of $B$, so it is clear that $M$ is a finitely-generated $A$-module. As $A$ is local (and Noetherian), $M$ is then flat if and only if it is free.
Let $B=k[\![x,y]\!]/(x^2,y^2)$ and let $A=B/(xy)\cong k[\![x,y]\!]/(x^2,xy,y^2)$ with the natural surjection $B \to A$, so $M=(xy) \in B$. Note $M^2=0$; in fact, $(x,y)M=0$. As noted above, the action of $A$ on $M$ is inherited from $B$, and so in particular, the maximal ideal of $A$ annihilates $M$. This means $M \cong k$ as an $A$-module. But $k$ cannot be a free $A$-module, as this would force $A$ to be a field, which it is not.