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Let $R$ be a ring which is not Noetherian. Let $M$ be an $R$-module whose annihilator is trivial. Is it possible for $M$ to be Noetherian?

Intuitively, it seems like the answer should be no, as we should be able to somehow "embed" $R$ in $M$ via its action on $M$. However, I am not seeing a straightforward proof.

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up vote 6 down vote accepted

In the commutative case at least, if $M$ is faithful (trivial annihilator) and Noetherian, then $R$ is actually a Noetherian ring. So if $R$ is non-Noetherian, it cannot have a faithful Noetherian module.

The proof is basically as you suggest. Let $m_1,\ldots,m_n$ generate $M$ and define $j:R\rightarrow M^n$ by $j(r)=(rm_1,\ldots,rm_n)$. This is injective because the annihilator of $M$ in $R$ is zero. So $R$ is a submodule of the Noetherian module $M^n$, and thus is itself Noetherian as an $R$-module, i.e., $R$ is a Noetherian ring.

More generally, the same argument proves Theorem 3.5 of Matsumura's Commutative Ring Theory: if $M$ is Noetherian, then $R/\mathrm{ann}(M)$ is a Noetherian ring.

Though I'm rather incompetent with non-commutative rings, it seems to me that this argument works for non-commutative (unital) rings as well.

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