Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
1  
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 Lampret 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 Lampret Feb 13 '12 at 21:42
3  
@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
1  
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
show 1 more comment

1 Answer 1

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.

share|improve this answer
    
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 Lampret 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 Lampret Feb 13 '12 at 23:28
1  
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.