Prove that if $R$ is a non-commutative ring with $1$ and if $a,b \in R$ and if $ab =1 $ but $ba \neq 1$ then $R$ is infinite

Question

I'm not sure about this one.

I noticed that $ba$ is an idempotent, and thus $(1-ba)$ is idenmpotent... So far I have 5 elements wooh! Only need $\infty$ more! Haha but seriously i'm pretty confused, insight appreciated!

score 2 · Accepted Answer · answered Jul 23 '19 at 16:18

2

Hint: show that if $R$ is finite, then $a^n = 1$ for some natural number $n$. Now compute $b$.

answered Jul 23 '19 at 16:18

Jens Hemelaer

646

3

$a^n = 1 = ab$ so $b = a^{n-1}$ but then $ba = ab$ which is a contradiction – Jul 23 '19 at 16:22

score 1 · Answer 2 · answered Jul 23 '19 at 16:22

Consider the sequence of left ideals $R=Ra^0 \supset Ra^1 \supset Ra^2 \supset Ra^3 \supset\dots$ - we are going to show that all the inclusions are proper, meaning that there is in infinite number of ideals in $R$, thus $R$ itself cannot be finite.

Suppose that at some point the inclusion is not proper, i.e. $Ra^n = Ra^{n+1}$. Since $1 \in R$, $1 \cdot a^n \in R a^{n+1}$, meaning that there is $c \in R$ such that $a^n = ca^{n+1}$. Multiply this by $b^n$: $a^nb^n=ca^{n+1}b^n \Rightarrow 1 = ca$. So, $a$ posesses a left inverse $c$, and by assumption $b$ is the right inverse of $a$. If $a$ has both-sided inverses, then these inverses are equal $c = b$, so $ba=1$, contradicting the assumption. So, all inclusions are proper.

This actually proves a stronger result: the situation in question cannot happen in Artinian rings. This proves that, e.g., for matrices $ab=1$ implies $ba=1$.

Prove that if $R$ is a non-commutative ring with $1$ and if $a,b \in R$ and if $ab =1 $ but $ba \neq 1$ then $R$ is infinite

2 Answers2