2
$\begingroup$

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?

  • 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.2012-02-14

1 Answers 1

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.

  • 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).2012-02-14