# Condition on product implies that is it the trivial module?

Let $M$ be an $A$-module and let $\mathfrak{a}$ be an ideal of $A$. Suppose that $\mathfrak{m} \cdot M =0$ for every maximal ideal $\mathfrak{m}$ of $A$ such that $\mathfrak{a} \subseteq \mathfrak{m}$. Is $M$ the trivial module?

Here $\mathfrak{m} \cdot M = \{\sum_{finite} a_{i} x_{i}: a_{i} \in \mathfrak{m}, x_{i} \in M \}$

If we actually replace the condition by $M_{\mathfrak{m}}=0$ (localization of $M$ in $\mathfrak{m}$) then we can conclude that $M=\mathfrak{a}M$, this is an exercise in Atiyah's and Macdonald's book. What happens if we replace the localization condition by the condition $\mathfrak{m} \cdot M =0$ ?

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The condition $M_{\mathfrak m}$ and $\mathfrak m M = 0$ have little to nothing in common with one another.

Suppose that we take $\mathfrak a = \mathfrak m$ to be a maximal ideal of $A$, and set $M = A/\mathfrak m$. Then $\mathfrak m \cdot M = 0$ for all $\mathfrak m$ containing $\mathfrak a$ (which is just to say, for $\mathfrak m$ itself), yet $M$ is non-trivial. (And also $M \neq \mathfrak a M$, the latter being the trivial module.)

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Thank you very much. –  user6495 Apr 1 '11 at 15:37
Consider $A = K[x]$, $\mathfrak{a} = (x)$ and $M = K[x]/(x) \cong K$ ($K$ here is a field). Then since $\mathfrak{a}$ is maximal and $\mathfrak{a} M = 0$ your condition is satisfied, but $M$ is not trivial.