Suppose $R$ is a rng with no zero divisors, not necessarily commutative. I know $R$ can be embedded into a ring $S:=\mathbb{Z}\times R$ by identifying $r\in R$ with $(0,r)\in S$. The operations on $S$ are defined as $(m,a)+(n,b)=(m+n,a+b),\qquad (m,a)(n,b)=(mn,mb+na+ab)$ with $1=(1,0)$ and $0=(0,0)$ of course.
The question I'm working on is as follows (Jacobson Algebra I, 2.17.5):
Let $Z=\{z\in S\mid za=0\text{ for all } a\in R\}$. Show that $Z$ is an ideal in $S$ and $S/Z$ is a domain. Show that $a\mapsto a+Z$ is a monomorphism of $R$ into $S/Z$.
I see that the set $Z=\{z\in S\mid za=0,\;\forall a\in R\}$ is an ideal of $S$. It is clearly a left ideal. The fact that it is a right ideal follows from the fact that $R$ is an ideal in $S$ by the definition of multiplication in $S$. That is, if $z\in Z$, and $s\in S$, then for any $a$, $(zs)a=z(sa)=0$ since $sa\in R$. Since $R$ has no zero divisors, I understand why $a\mapsto a+Z$ is a monomorphism.
But why is $S/Z$ a domain? I tried to prove it by showing $Z$ is prime in $S$ by taking $s,t\in S$, with $st\in Z$. If $t\in Z$, we're done. Otherwise, there exists $a\in R$ such that $ta\neq 0$. Then $(st)a=s(ta)=0$, and since $ta\neq 0$, I think the fact that $R$ has no zero divisors would imply that $s=0$, so $s\in Z$. What makes me uneasy is that even though $s(ta)\in R$, it need not be the case that $s\in R$, so maybe this doesn't apply. What is the correct way to show $Z$ is prime in $S$?