Ideal in an Artinian Ring $I=aR=Rb$, prove $I=Ra=bR$

Solution 1:

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 $aR = bR$.