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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.

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  • $\begingroup$ Note: I don't really need to say "as a module" since being finitely generated as an ideal implies being finitely generated as a module, because the ideal is assumed nilpotent. $\endgroup$ – David Carchedi Nov 7 '12 at 19:03
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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.

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    $\begingroup$ A very nice example! $\endgroup$ – Manny Reyes Nov 8 '12 at 14:42
  • $\begingroup$ Very nice indeed! It would also be interesting to see whether an example could be constructed with $K$ algebraically closed, as the above example depends on the property that a sum of squares in $\mathbb{R}$ is nonzero whenever one of the summands is nonzero. $\endgroup$ – Neil Epstein Nov 8 '12 at 15:55
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    $\begingroup$ The same example works on any field of characteristic zero, but surprisingly it's a bit harder. $\endgroup$ – Lierre Nov 8 '12 at 18:19
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    $\begingroup$ In fact it's not harder at all, whatever the characteristic. I edited the proof. In the former proof, I only used the fact that $I$ is a subring which makes the proof harder on a general field. $\endgroup$ – Lierre Nov 8 '12 at 18:45

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