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Let $R$ be a noncommutative ring and $I$ a two-sided ideal of $R$. Assume that $I$ and $R/I$ both have descending chain condition on two-sided ideals (D.C.C.), that is, if we have a chain of two-sided ideals $J_1\supset J_2\supset \cdots$, then there exists an $N\in\mathbb N$ such that $J_n = J_N$ for $n\geq N$.

Is true that this implies that $R$ also verifies D.C.C. on two-sided ideals?

And for ascending chain condition on two-sided ideals (A.C.C) we have a similar result?

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Foe the A.C.C. case, you can read my answer here: – Isaac Solomon Nov 22 '12 at 18:24
up vote 3 down vote accepted

If $R$ is a ring and $0\rightarrow M^\prime\rightarrow M\rightarrow M^{\prime\prime}\rightarrow 0$ is an exact sequence of $R$-modules, then $M$ satisfies the ACC on $R$-submodules (resp. the DCC on $R$-submodules) if and only if both $M^\prime$ and $M^{\prime\prime}$ do. So if you apply this with $M^\prime=I$, $M=R$, and $M^{\prime\prime}=R/I$, you get what you want.

Also, I wouldn't say that $I$ satisfies the DCC or ACC on ideals, because $I$ is not a ring (unless $I=R$). It is also a fact that the $R$-submodules of $R/I$ are precisely the $R/I$-submodules, which are the ideals. This is why you can apply the result I mention above.

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In my question the A.C.C. is on ideals and not in left or right ideals. So, I can not apply what you say!!!! – zacarias Nov 22 '12 at 21:02
I don't understand your comment. I just assumed the rings were commutative because I mostly deal with commutative rings, but what I say is true if you only want to consider left or right ideals, or the rings are commutative and you want to consider two-sided ideals. – Keenan Kidwell Nov 22 '12 at 21:13
The ring is not commutative and I want to consider two-side ideals. – zacarias Nov 22 '12 at 21:26
Have you tried to go through the argument in the commutative case to see if it carries over? You should edit your question to make it more clear what you're asking. – Keenan Kidwell Nov 22 '12 at 22:56

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