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Let $R$ be an Artinian Ring and suppose there exists $a,b\in R$ s.t. $I=aR=Rb$, then prove $I=bR=Ra$. (You may assume that a right Artinian Ring is Right Noetherian).

I've managed to get $Ra$, $bR$ contained in $I$ without using the fact that it's Artinian. The hint confuses me because I'm not sure how to make a useful ascending chain to use the fact that it's right Noetherian.

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Does the ring $R$ have unity? – Giovanni De Gaetano Apr 26 '12 at 10:46
yes it does have unity – Alice Apr 26 '12 at 12:22
@Arturo Maybe I´m missing something. But how can you conclude $a=bu$? It seems to me that you only have $a=ub$, and you need somehow to prove that $bR$ is a two-sided ideal to get the conclusion. (In specific the Artinianity hypothesis is missing or well hidden...) – Giovanni De Gaetano Apr 26 '12 at 15:21
@Giovanni: I got some wires crossed. Sorry. – Arturo Magidin Apr 26 '12 at 16:00
Before you go any further looking for ascending and decending chains, see if you can't make use of the maximal condition and/or minimal condition first. – rschwieb Apr 28 '12 at 2:01

By symmetry, it's enough to show that $aR = bR$.

Since $R$ is right Artinian, it is right Noetherian. So $R$ has a finite length as a right $R$-module: it has a finite composition series. By the Jordan-Holder theorem, the length $l(M)$ of any composition series of any finite length module $M$ is well-defined (does not depend on the choice of composition series), and satisfies $l(M) = l(N) + l(M/N)$ whenever $N$ is a submodule of $M$.

Now $I = Rb = aR$ so $a \in Rb$ and $b \in aR$. Thus there are $x,y \in R$ such that $a = xb$ and $b = ay$. Hence $bR = ayR \subseteq aR$. Also $bu = 0$ implies $au = xbu = 0$, so the right annihilator $rann(b)$ of $b$ is contained in $rann(a)$.

Thus $l(bR) \leq l(aR)$ and $l(rann(b)) \leq l(rann(a))$.

By the first isomorphism theorem for modules, we know that $aR \cong R / rann(a)$ and $bR \cong R / rann(b)$, so

$l(aR) + l(rann(a)) = l(R) = l(bR) + l(rann(b))$.

Hence $0 \leq l(aR) - l(bR) = l(rann(b)) - l(rann(a)) \leq 0$.

Therefore $l(aR) = l(bR)$ and $aR = bR$.

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Very nice! I thought about this some time ago and never considered annihilators. – Manny Reyes Jan 5 '14 at 12:15
Thanks. The hint given isn't very helpful! I wonder if it's true if $R$ is only assumed to be Noetherian; do you perhaps know a counterexample? – Konstantin Ardakov Jan 5 '14 at 12:18
Konstantin, I was finally able to think about your question. Much to my surprise, I believe that it may in fact be true for (right) Noetherian rings! In your notation above, notice that left multiplication by $x$ defines an endomorphism of the right module $I=aR$ because it is also a left ideal. This map is surjective since $a = xb = xay$. If $R$ is right Noetherian, then $aR$ Hopfian, whence this surjection is an automorphism. But notice that the submodule $bR \subseteq aR$ also maps onto the image $aR$ as $a = xb$. By injectivity, we must have $bR = aR$. Can it really be true? – Manny Reyes Feb 19 '14 at 19:03
Yes, I think it works. Great! – Konstantin Ardakov Feb 19 '14 at 23:53

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