Question

Let $\mathit{f}\mathbf{:}\mathit{R}\mathbf{\to }\mathit{S}$be the homomorphism of rings. If $\mathit{r}$is a zero divisor in $\mathit{R}$, is$\mathit{f}\left(r\right)$ a zero divisor in $\mathit{S}$?

Short answer

Hence proved that $f\left(r\right)$may or may not be a zero divisor in$S$ .

Step-by-step solution

Homomorphism

If any ring $R$ has elements such that,$a,b\in R$, then, the condition for homomorphism of the function of its elements is given by:

$$\begin{gathered} f\left(a+b\right)=f\left(a\right)+f\left(b\right) \\ f\left(ab\right)=f\left(a\right)f\left(b\right) \end{gathered}$$

Proof

We have both $RandS$ as non-zero rings. So, we have:

$\left(r,0_R\right)\left(s,0_S\right)=\left(r{0}_R\right),\left(s{0}_S\right)=\left(0_R,0_S\right)=0_{R\times S}$

Now, when $\left(r,0_R\right)\ne 0_{R\times S}$and$\left(s,0_S\right)\ne 0_{R\times S}$, then $\left(r,0_S\right)$ will be a zero divisor in the ring $R\times S$. But,$f\left(r,0_S\right)=r\ne 0_R$

In this case for elements $\left(1,0\right)$, we get:

$f\left(1,0\right)=1,\left(1,0\right),\left(0,1\right)=\left(0,0\right),1.0=1$

Here, we see: $\left(1,0\right)$is a zero divisor, but 1 is not.