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.

  • $\begingroup$ 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

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.

| cite | improve this answer | |
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 2
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.