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 $ba=1$.

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 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\} […]

Torsion Submodule, Integral Domain, and Zero Divisors
Let $R$ be a ring with $1$. An element of the $R$-module $M$ is called a torsion element if $rm=0$ for some nonzero element $r\in R$.
The set of torsion elements is denoted
\[\Tor(M)=\{m \in M \mid rm=0 \text{ for some nonzero} r\in R\}.\]
(a) Prove that if $R$ is an […]

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 […]

Nilpotent Element a in a Ring and Unit Element $1-ab$
Let $R$ be a commutative ring with $1 \neq 0$.
An element $a\in R$ is called nilpotent if $a^n=0$ for some positive integer $n$.
Then prove that if $a$ is a nilpotent element of $R$, then $1-ab$ is a unit for all $b \in R$.
We give two proofs.
Proof 1.
Since $a$ […]

Rings $2\Z$ and $3\Z$ are Not Isomorphic
Prove that the rings $2\Z$ and $3\Z$ are not isomorphic.
Definition of a ring homomorphism.
Let $R$ and $S$ be rings.
A homomorphism is a map $f:R\to S$ satisfying
$f(a+b)=f(a)+f(b)$ for all $a, b \in R$, and
$f(ab)=f(a)f(b)$ for all $a, b \in R$.
A […]