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Let $R$ be any (i.e. not necessarily commutative or unital) associative ring, and let $I$ be a (two sided) ideal of $R$. Hence $I$ is a (nonunital) ring.

How can I prove: $R$ is a Noetherian/Artinian ring iff $I$ and $R/I$ are N/A rings?

We know (Atiyah & MacDonald, p.75, prp.6.3, the proof is the same for the noncommutative nonunital case) that if $0\rightarrow M'\rightarrow M \rightarrow M'' \rightarrow 0$ is an exact sequence of left $R$-modules, then $M$ is N/A iff $M'$ and $M''$ are N/A.

We have an exact sequence $0\rightarrow I\rightarrow R \rightarrow R/I \rightarrow 0$ of left $R$-modules, so $R$ is N/A iff $I$ and $R/I$ are N/A as $R$-modules. Now, since every $R$-submodule of $R/I$ is a $R/I$-submodule of $R/I$, and vice versa, we know that $R/I$ is N/A as a $R$-module iff it is such as an $R/I$-module, i.e. as a ring.

How can I prove that $I$ is N/A as a $R$-module iff it is such as an $I$-module?

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    One thing that could scare people is the fact that you are dealing with a non-unital ring. Could you please give some background on why you need $I$ to be a Noetherian/Artinian **ring** instead of just a Noetherian/Artinian **$R$-module**?2012-02-13
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    Well, I'm writing my notes, and since I've already proved that for any $R$-module $M$ with a submodule $N$, we have: $M$ is N/A iff $N$ and $M/N$ are N/A, I also wanted to prove the same for rings. I would be surprised if this did not hold. But for rings, we need an ideal to make a quotient ring $R/I$, and ideals are always nonunital rings (unless $I=R$), hence my attempt. Basically, I was reading a book on Commutative Algebra (Atiyah & MaDonald), and the asymmetry between the ring and module version of this proposition bugged me, so I wanted to fix it, by proving a more general statement.2012-02-13
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    I thought there should be some fairly easy argument, that I'm too clumsy to notice, but more experienced ring theorists would see right away. Actually, in general, the asymmetry bugs me: (1) if one wants to make a quotient group, one needs a normal subgroup, which is a subgroup; (2) if one wants to make a quotient module, one simply needs a submodule; (3) if one wants to make a quotient ring, one needs an ideal, which is **not** a subring! By modifying the definition of a ring (i.e. not requiring it to have a $1$), there is no such asymmetry anymore.2012-02-13
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    @Leon: And to make a quotient of a semigroup, you need... Actually, in all instances, what you "really" need is a congruence. It just so happens that in groups, congruences can be "coded" by subgroups.2012-02-13
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    Dear Leon: Nice question! The standard notions in the non-commutative case are (I think) those of **left** or **right** N/A rings. A ring $A$ is (say) left N/A iff its poset of left ideals is, that is if $A$ is N/A viewed as a left module over itself. Of course one can ask if the poset of two-sided ideals is N/A. This amounts to asking if $A$ is N/A as a module over $A\otimes_{\mathbb Z}A^{op}$...2012-02-14
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    Now I'll try to reply to your comment starting with "I thought there should be...". It seems to me the natural generalization of A-M's argument is the situation where $$0\to A\to B\to C\to0$$ is an exact sequence in a given **abelian category**. Symmetry is restored: the quotient by an object by a sub-object is an object.2012-02-14

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