If $ab=1$ in a Ring, then $ba=1$ when $a$ or $b$ is Not a Zero Divisor

Problems and solutions of ring theory in abstract algebra

Problem 542

Let $R$ be a ring with $1\neq 0$. Let $a, b\in R$ such that $ab=1$.

(a) Prove that if $a$ is not a zero divisor, then $ba=1$.

(b) Prove that if $b$ is not a zero divisor, then $ba=1$.

 
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Definition.

An element $x\in R$ is called a zero divisor if there exists a nonzero element $y\in R$ such that $xy=0$ or $yx=0$.

So if $x$ is not a zero dividor, then $xy=0$ implies that $y=0$. Similarly, $yx=0$ implies that $y=0$.

Proof.

(a) Prove that if $a$ is not a zero divisor, then $ba=1$.

Suppose that $a$ is not a zero divisor. We compute
\begin{align*}
a(ba-1)&=aba-a && \text{by distributivity}\\
&=1\cdot a -a &&\text{by $ab=1$}\\
&=a-a=0.
\end{align*}

Since $a$ is not a zero divisor, this yields that $ba-1=0$, and hence $ba=1$.

(b) Prove that if $b$ is not a zero divisor, then $ba=1$.

Suppose that $b$ is not a zero divisor. We calculate
\begin{align*}
(ba-1)b&=bab-b && \text{by distributivity}\\
&=b\cdot 1 -b &&\text{by $ab=1$}\\
&=b-b=0.
\end{align*}

As $b$ is not a zero divisor, the equality $(ba-1)b=0$ implies that $ba-1=0$.
Hence we have $ba=1$.


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