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 \(\displaystyle{\left({a},{b}\right)}\in{R}\oplus{S}\), 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