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!
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 $ba=a^n=1_R$.