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Our prof gave us the following problem as an exercise:

Suppose $(R,+,\cdot)$ is a finite ring with unity (denoted $1_R$) such that for some $a,b\in R$, we have $ab=1_R$

Show that $ba=1_R$

Here's my attempt at a proof:

Proof.

Let us assume that $ba\neq 1_R$

Next, we consider $S:=\{(ba)^n\mid n\in\Bbb Z^+\}\subseteq R$

Since $R$ is a finite ring, we must have that $(ba)^m=(ba)^n$ for some distinct $m,n\in\Bbb Z^+$ (wlog say $m\gt n$). Then, we have $(ba)^{m-n}=1_R$

Note that $m\gt n$, so $m-n$ is a positive integer but $m-n\neq 1$ (hence $\gt 1$) since $ba\neq 1_R$ by assumption.

Now,

$$(ba)^{m-n}=1_R\iff b(ab)^{m-n-1}a=1_R\iff b(1_R)^{m-n-1}a=1_R\iff ba=1_R$$

which gives us a contradiction.

Is the above proof correct? I'd like the community to proofread it.

I'd also be interested in a shorter argument to prove the claim (if there is one). Thanks!

Community
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sha_256
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    How do you go from $(ba)^m=(ba)^n$ to $(ba)^{m-n}=1$ (assuming that $m>n$)? – Angina Seng Jan 05 '18 at 06:20
  • In fact $ab=1$ implies $ba=1$ in any left (or right) Noetherian ring. – Angina Seng Jan 05 '18 at 06:21
  • @LordSharktheUnknown, You're right. $R$ might not necessarily be a division ring, so my argument fails. Is there a way to fix the argument? I'd be very interested in just a hint. Thanks! – sha_256 Jan 05 '18 at 06:30
  • On the chance you want to learn more about rings with this property, google "Dedekind finite ring" and/or "directly finite ring". – rschwieb Jan 05 '18 at 14:18

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Since $a$ has a right inverse, right multiplication by $a$, $\cdot a : R\to R$ is bijective. Thus $(\cdot a)^n=\mathrm{id}_R$ for some integer $n$. Since $(\cdot a)^n = \cdot (a^n)$, this implies that $a^n = 1_R\cdot a^n = 1_R$. Thus $b=a^nb = a^{n-1}(ab)=a^{n-1}$, so $ba=a^n=1_R$.

jgon
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