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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? – M Turgeon Feb 13 '12 at 21:29
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. – Leon Feb 13 '12 at 21:40
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. – Leon Feb 13 '12 at 21:42
@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. – Arturo Magidin Feb 13 '12 at 21:51
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. – Pierre-Yves Gaillard Feb 14 '12 at 16:05

What you claim isn't true. Consider the ring $R=\mathbb{R}^{\mathbb{N}}$ of sequences of real numbers, with component-wise operations. This ring isn't Artinian, since the ideals $$I_n=\{(a_i)_i\in R;\forall i\leq n\colon a_i=0\}$$ form a descending chain which doesn't stabilize. It also isn't Noetherian, since the ascending chain of ideals $$J_n=\{(a_i)_i)\in R;\forall i\geq n\colon a_i=0\}$$ also doesn't stabilize.

Now consider the ideal $J_1$. It is a field, since it is isomorphic as a ring to the reals, and is as such both Noetherian and Artinian.

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Umm, how is this a counterexample? We have: $R$ isn't N/A, $I:=J_2$ is N/A, $R/I$ isn't N/A. – Leon Feb 13 '12 at 22:29
I meant it more as an observation that ideals can have different "ideal structures" when considered either as ideals in a larger ring or as rings in their own right. This has implications for the last question in your post, but I see I was very much hasty when writing it up. Should've been a comment really. – Miha Habič Feb 13 '12 at 23:00
Oh, ok. +1 for the idea. It certainly is interesting that a nonunital ring $\mathbb{Q}\times2\mathbb{Z}$ can have a unital subring (a field) which is an ideal: $\mathbb{Q}\times0$. – Leon Feb 13 '12 at 23:28
If $R$ is any ring, with or without unity, then constructing the Dorroh extension of $R$ (underlying set $R\times\mathbb{Z}$, coordinate-wise addition, multiplication $(r,n)(s,m) = (rs + ns+mr, nm)$ gives you a ring that has $R\times\{0\}$ as an ideal; and the set of elements $(r,n)$ with $r\in R$ and $n\in k\mathbb{Z}$ (for any $k$) is a subring (in fact, an ideal). – Arturo Magidin Feb 14 '12 at 6:12

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