Let R and S be commutative rings. Prove that (a, b) is a zero-divisor in R o+ S if and only if a or b is a zero-divisor or exactly one of a or b is 0.

Let (a,b) is a zero-divisor in ${\displaystyle R\oplus S}$
Since (a,b) is zero-divisor, there exists (c,d) in ${\displaystyle R\oplus S}$ such that
$(a,b)(c,d)=(0,0)$
${\displaystyle \Rightarrow }$$ac=0,bd=0$
Since $ac=0$ ${\displaystyle \Rightarrow }$ a is a zero divisor in R or $a=0$
Also, $bd=0$ ${\displaystyle \Rightarrow }$ b is zero divisor in R or $b=0$
Conversely:
Let a and b are zero divisors.
Since a is zero divisor,there exist element ${\displaystyle c\in R}$ such that ac=0
Also, b is zero divisor, there exist element ${\displaystyle d\in S}$ such that bd=0
${\displaystyle \Rightarrow }$ $ac\times bd=0$
${\displaystyle \Rightarrow }$ $(a,b)\times (c,d)=(0,0)$
Hence, for ${\displaystyle (a,b)\in R\oplus S}$, there exists (c,d) such that $(a,b)\times (c,d)=(0,0)$
By definition (a,b) is zero diisor in ${\displaystyle R\oplus S}$
Hence, proved