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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?

dahemar
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  • This question isn't saying that $ab = 1 \implies ba=1$ means that $R$ doesn't have zero divisors. In fact, if $a$ and $b$ are invertible (here they are inverses of each other), then neither of them are zero divisors. Otherwise, suppose $a = b^{-1}$ and $bc = 0$ for some nonzero $c \in R$, then (assume associativity), $$c = 1 \cdot c = (ab)c = a(bc) = a \cdot 0 = 0,$$ a contradiction. – oleout Jun 30 '21 at 11:37
  • @KelvinLian I think this just proves $b$ cannot be a right zero divisor if it has a left inverse. The proof linked uses the fact that b has a right inverse and is not a right zero divisor. – Astyx Jun 30 '21 at 11:42
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    @Astyx Oh yes... I'm living in a simple commutative world~ – oleout Jun 30 '21 at 11:46
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    A ring with this property, that $xy=1$ implies $yx=1$ is called Dedekind finite or directly finite. In fact, several conditions weaker than right Noetherian imply it, such as ACC on principal right ideals and being orthogonally finite. Other stronger conditions (unrelated to Noetherinity) the DCC on principal right ideals implies it, and "nilpotent elements form an ideal" implies it. – rschwieb Jun 30 '21 at 13:13

2 Answers2

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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.

GreginGre
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  • Thank you for your answer, upvoted! This looks a bit above the scope of our course (which was an introduction to ring theory) though, so I’ll wait a bit just in case there are other different answers. – dahemar Jun 30 '21 at 11:42
  • @dahemar I don't know specifically about your course but this proposition is a classical exercise on noetherian rings. – Giorgos Giapitzakis Jun 30 '21 at 11:44
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    @dahemar There is a proof of this proposition using Nakayama's lemma too if I remember correctly! You can check it out in case it comes out in future tests (: also, there should be a "dual" statement for Artinian rings. – oleout Jun 30 '21 at 11:48
  • If you don't like Noetherian modules, just apply (and prove) the Prop. in the special case needed, so left multiplication $R \to R$ with $b$. Then the kernels are ideals, namely annihilators of the powers of $b$. – Martin Brandenburg Jun 30 '21 at 11:58
  • The Nakayama proof is only available when $R$ is commutative. – Martin Brandenburg Jun 30 '21 at 11:59
  • @GiorgosGiapitzakis We haven’t seen it in class, and this kind of question in this subject’s tests should be answerable using results we have already proved (I could answer another question of the same weight in just two lines using a pair of known propositions), that’s why I was a bit surprised! But I’ll accept the answer since it is quite complete :) – dahemar Jun 30 '21 at 12:00
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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.

Martin Brandenburg
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