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

## 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 $ba=1$.

## 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 $ba=1$.

### More from my site

• No Nonzero Zero Divisor in a Field / Direct Product of Rings is Not a Field (a) Let $F$ be a field. Show that $F$ does not have a nonzero zero divisor. (b) Let $R$ and $S$ be nonzero rings with identities. Prove that the direct product $R\times S$ cannot be a field.   Proof. (a) Show that $F$ does not have a nonzero zero divisor. […]
• If Every Proper Ideal of a Commutative Ring is a Prime Ideal, then It is a Field. Let $R$ be a commutative ring with $1$. Prove that if every proper ideal of $R$ is a prime ideal, then $R$ is a field.   Proof. As the zero ideal $(0)$ of $R$ is a proper ideal, it is a prime ideal by assumption. Hence $R=R/\{0\}$ is an integral […]
• Is the Quotient Ring of an Integral Domain still an Integral Domain? Let $R$ be an integral domain and let $I$ be an ideal of $R$. Is the quotient ring $R/I$ an integral domain?   Definition (Integral Domain). Let $R$ be a commutative ring. An element $a$ in $R$ is called a zero divisor if there exists $b\neq 0$ in $R$ such that […]
• If a Prime Ideal Contains No Nonzero Zero Divisors, then the Ring is an Integral Domain Let $R$ be a commutative ring. Suppose that $P$ is a prime ideal of $R$ containing no nonzero zero divisor. Then show that the ring $R$ is an integral domain.   Definitions: zero divisor, integral domain An element $a$ of a commutative ring $R$ is called a zero divisor […]
• Finite Integral Domain is a Field Show that any finite integral domain $R$ is a field.   Definition. A commutative ring $R$ with $1\neq 0$ is called an integral domain if it has no zero divisors. That is, if $ab=0$ for $a, b \in R$, then either $a=0$ or $b=0$. Proof. We give two proofs. Proof […]
• Ring is a Filed if and only if the Zero Ideal is a Maximal Ideal Let $R$ be a commutative ring. Then prove that $R$ is a field if and only if $\{0\}$ is a maximal ideal of $R$.   Proof. $(\implies)$: If $R$ is a field, then $\{0\}$ is a maximal ideal Suppose that $R$ is a field and let $I$ be a non zero ideal: \[ \{0\} […]
• A Ring is Commutative if Whenever $ab=ca$, then $b=c$ Let $R$ be a ring and assume that whenever $ab=ca$ for some elements $a, b, c\in R$, we have $b=c$. Then prove that $R$ is a commutative ring.   Proof. Let $x, y$ be arbitrary elements in $R$. We want to show that $xy=yx$. Consider the […]
• The Preimage of Prime ideals are Prime Ideals Let $f: R\to R'$ be a ring homomorphism. Let $P$ be a prime ideal of the ring $R'$. Prove that the preimage $f^{-1}(P)$ is a prime ideal of $R$.   Proof. The preimage of an ideal by a ring homomorphism is an ideal. (See the post "The inverse image of an ideal by […]

#### You may also like...

This site uses Akismet to reduce spam. Learn how your comment data is processed.

##### Every Ideal of the Direct Product of Rings is the Direct Product of Ideals

Let $R$ and $S$ be rings with $1\neq 0$. Prove that every ideal of the direct product $R\times S$ is...

Close