# Detecting elements of nilpotent extensions via finitely generated ones

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.

• 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. – David Carchedi Nov 7 '12 at 19:03

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.
• 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. – Neil Epstein Nov 8 '12 at 15:55
• 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. – Lierre Nov 8 '12 at 18:45