Question

Let$R$and$S$ be nonzero rings(meaning that each of them contains at least one nonzero element). Show that$R\times S$ contains zero divisors.

Short answer

The zero divisors are $\left(r_1,0_s\right)$and$\left(0_R,s_1\right)$.

Step-by-step solution

Elements of Rings

If a ringhas an element$k$ , such that$k\in R$ .

Then the following equation will have a unique solutionas:

$k+x=0_R$

Proof

The given non-zero rings are $RandS$.

Let one of the elements from each ring be $r_1$and$s_1$, such that:

$r_1\ne 0_R$and$s_1\ne 0_s$

Now, using theorem, we have:

$\left(r_1,0_s\right)\left(0_R,s_1\right)=\left(r_1{0}_R,0_s{s}_1\right),\left\{\left(0_R,0_s=0_{R\times s}\right)\right\}$So, $\left(r_1,0_s\right)\in R\times S$and $\left(0_R,s_1\right)\in R\times S$

Hence, the zero-divisors are$\left(r_1,0_s\right)$and$\left(0_R,s_1\right)$ .