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


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.


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.



Detecting elements of nilpotent extensions via finitely generated ones
What do you think of the following example :
$$ A = \frac{K[u,x\_0, x\_1, ...]}{(u^2) + (x\_i^2-u)\_{i\in\Bbb N} + (x\_i x\_j)\_{i\neq j}}$$


If $M$ is the ideal $(u, x\_0, x\_1, \dotsc)$, then $A = K \oplus M$ and $M^4 = (0)$.


Let $I$ be a proper ideal of $A$ such that $A/I$ is finite dimensional over $K$.
It is clear that there exists a linear combination of a finite number of the $x\_i$'s which is in $I$. Say $y = \sum\_{i\in S} a\_i x\_i \in I$, with $S$ a finite set and at least a $a\_i$, say $a\_0$, non-zero. Then $x\_0 y = a\_0 u$, so that $u$ is in $I$.
However $u\neq 0$.


Note : If $I$ is a subring of $A$ such that $A/I$ is finite dimensional over $K$ then $u\in I$ still holds.


