If I am looking at the following problem:
Let R be ring and $S=R\times \mathbb Z=\{(r,n)|r\in R,n\in \mathbb Z\}$. This is a ring with addition defined as $(r,n)+(s,m)=(r+s,n+m).$ and multiplication defined as :$(r,n).(s,m)=(rs+ns+mr,nm)$ .this ring has unity as $(0,1)$.
How would I show that if $(R,R_{+},R_{*})$ has zero divisors then this will imply $(S,S_{+},S_{*})$ ?
I saw similar problems but nothing showing this directly.
On
Suppse that $rs=0$, $(r,0)(s,0)=(0,0)$ so $S$ has divisors of zero if $R$ has divisors of zero.
On
The inclusion map $f:R \to S$ $r \mapsto (r,0)$ is an injective ring homomorphism (of course not unital).
Suppose $f: A \to B$ is an injective ring homomorphism between possibly non-unital rings, then if $xy=0$ in $A$ where $x$ and $y$ are non-zero, we have $f(x)f(y)=0$ in $B$ and because $f$ is injective $f(x)$ and $f(y)$ are non-zero. We see that any injective ring homomorphism preserves zero divisors.
Suppose $0_R \ne a \in R$ is a (left) zero divisor. Then there exists some $0_R \ne b \in R$ with $ab = 0_R$; now consider the elements $(a, 0_{\Bbb Z}), (b, 0_{\Bbb Z}) \in S$. Clearly $(a, 0_{\Bbb Z}) \ne (0_R, 0_{\Bbb Z}) \ne (b, 0_{\Bbb Z})$, but $(a, 0_{\Bbb Z})(b, 0_{\Bbb Z}) = (ab, 0_{\Bbb Z}) =(0_R, 0_{\Bbb Z}) = 0_S \in S$. The same argument works for right zero divisors as well; indeed one sees that $ab = 0_R \ne a, b$ shows that $b$ is a right zero divisor. So in fact, yes, the presence of zero divisors in $R$ carries over to $S$.