# Let $M$ be a noetherian $R$-module and $I=\mathrm{Ann}_R(M)$. Then $R/I$ is a noetherian ring for all ring $R$?

Matsumura $$p22$$,thm$$3.5$$ says

Let $$M$$ be a noetherian $$R$$-module and $$I=\mathrm{Ann}_R(M)$$. Then $$R/I$$ is a noetherian ring.

My question is,; is this true for arbitrary ring $$R$$,or,is this true for non-commutative rings?

I would be appreciated if you could give me some examples. Thank you for your kind help.

• The answer is “no” and a user has already pointed out that any non-right-Noetherian right primitive ring is a counterexample. Hopefully they find the time to post it soon. Dec 21 '20 at 0:54

It is not true that $$R/\text{ann}_R(M)$$ must be right Noetherian if $$M$$ is a Noetherian right $$R$$-module. Any right primitive ring that is not right Noetherian is a counterexample: take $$M$$ to be a faithful simple module (which is certainly Noetherian). Examples of such rings include endomorphism rings of infinite dimensional vector spaces, free algebras on more than one generator, and group algebras of free groups on more than one generator. (See, e.g., The Algebraic Stucture of Group Rings'' by Passman, Corollary 9.2.11 for the group algebra case.)
If you want a noncommutative result, you have to assume $$M$$ is a bimodule. For example, if $$M$$ is an $$R$$-$$R$$-bimodule that is Noetherian as both a left and a right $$R$$-module, then $$R/\text{ann}_R(M)$$ is right (or left) Noetherian, where the annihilator is the right (or left) annihilator.
To see this, suppose $$m_1,\dots,m_n$$ generate $$M$$ as a left $$R$$-module and define a right $$R$$-module map $$\phi:R\to M^n$$ by $$\phi(r)=(m_1r,\dots,m_nr)$$. Because $$m_1,\dots,m_n$$ generate $$M$$ as a left $$R$$-module, the kernel of $$\phi$$ is $$\text{ann}_R(M)$$, where the annihilator is the right annihilator of $$M$$. Thus $$R/\text{ann}_R(M)$$ emebeds in $$M^n$$ as a right module and so is right Noetherian.
We can weaken the hypotheses. If we assume $$M$$ is an $$S$$-$$R$$-bimodule for some ring $$S$$ and assume that $${}_SM$$ is finitely generated and $$M_R$$ is Noetherian, then the same proof shows $$R/\text{ann}_R(M)$$ is right Noetherian.