Show that $R × S$ contains a zero-divisor different from ($0_R, 0_S$)

Question

Let ($R, 0_R, 1_R, +_R, ·_R$) and ($S, 0_S, 1_S, +_S, ·_S$) be rings.

(a) Show that $R × S$ with component-wise addition and multiplication satisfies the axioms of ring.

(b) Show that $R × S$ contains a zero-divisor different from ($0_R, 0_S$).

I managed to solve part a), I proved the axioms of a ring.

Concerning part b), this is my attempt:

let ($r,s$) and ($r',s'$) be in $R×S$.

Suppose ($r,s$) is a zero-divisor thus there is ($r',s'$) $\not=$$0$ such that ($r,s$).($r',s'$) = $0$

Hence ($r.r',s.s'$) = $0$, which implies that $r.r'=0$ and $s.s'=0$. Thus $r$ is a zero-divisor of $R$ and $s$ is a zero-divisor for $S$, therefore, ($r,s$) is indeed a zero-divisor of $R×S$.

Is my attempt correcr?

Answer 1

For all elements $r\in R$ and $s\in S$, $(r,0)\cdot (0,s) = (r0,0s)= (0,0)$, since the zero element of a ring is absorbing.