If $R$ is a noetherian ring, then $ab=1$ implies $ba=1$ $\forall a,b\in R$ The following question was part of an exam at my university: if $R$ is a noetherian ring, then $ab=1$ implies $ba=1$ $\forall a,b\in R$.
As far as I know, the result would hold if $a$ and $b$ aren’t zero divisors. However, I don’t see why this should hold for a general noetherian ring, because $R$ being noetherian doesn’t imply that it doesn’t contain zero divisors, right?
 A: Here is the proof by GreginGre in the language of elementary ring theory.
Let $ab=1$. Let $I_n = \{x \in R : x b^n = 0\}$. This is a left ideal, and $I_n \subseteq I_{n+1}$. Since $R$ is left Noetherian, we have $I_n = I_{n+1}$ for some $n$. Thus, $x b^n b=0$ implies $x b^n = 0$. But every element $r \in R$ has the form $x b^n$, since $r = r a^n b^n$. Thus, $rb = 0$ implies $r=0$. Since $(ba-1)b=0$, we get $ba-1=0$, and we are done.
PS: I find it quite fascinating that a finiteness condition (Noetherian) implies an algebraic implication $ab=1 \implies ba=1$. For Artinian rings see here.
A: It follows from the following result :
Prop. Let $R$ be a ring, and let $M$ be a noetherian (say left) $R$-module. Then any surjective endomorphism of $M$ is bijective.
Now assume that $ab=1$ in $R$, and consider the endomorphism $x\in R\mapsto xb\in R$.
Then it is surjective, since $r=(ra)b$ for all $r\in R$. By the proposition above, it is bijective, hence injective. Now we have $(ba-1)b=bab-b=
b-b=0$ and finally  $ba-1=0$ by injectivity.
Proof of the proposition. Let $u:M\to M$ be a surjective endomorphism. The sequence of submodules $\ker(u)\subset \ker(u^2)\subset\cdots \subset \ker(u^n)\subset \cdots$ is nondecreasing, hence stationary since $M$ is noetherian. So there is some $n\geq 1$ such that $\ker(u^n)=\ker(u^{n+1})$. Let $x\in\ker(u)$. Since $u$ is surjective, so is $u^n$, and we can pick $y\in M$ such that $x=u^n(y)$. Now $u^{n+1}(y)=u(x)=0$, so $y\in\ker(u^{n+1})=\ker(u^n)$, and $x=u^n(y)=0$. Therefore, $u$ is also injective and we are done.
