Suppose that $\pi:A \to K$ is a surjective map of $K$-algebras with nilpotent kernel, for $K$ ~~a commutative unital ring~~ *field*. (Recall that this means $Ker(\pi)^n=0$ for some integer $n$.) Notice that this implies that, as a $K$-module, $A\cong K \oplus M,$ where $M=Ker(\pi).$

I would like to either prove the following, or find a counter-example:

If $a \in A$ is in every subideal $I \subseteq M,$ such that $M/I$ is a finitely generated $K$-module, then $a$ is zero.

This is trivially true when $M$ is finitely generated as a $K$-module, (since then $(0)$ satisfies this property) so one should assume that $M$ is *not*-finitely generated (as a module).

~~If necessary, I'm OK with assuming that $K$ is a field.~~

Any thoughts would be great, thanks!

EDIT: The square-zero extension of $\mathbb{Z}$ associated to the abelian group $\mathbb{Q/Z}$ provides a counter-example for commutative unital rings. I have a feeling it may be true for fields however.