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 $S$?
Completely rewritten after self-deletion
Confirm that $(m,a)\in Z$ if and only if $mr+ar=0$ for all $r\in R$.
Suppose that $(m,a)$ and $(n,b)$ are not in $Z$, and yet their product $(mn,mb+an+ab)$ is in $Z$.
Then there exists $r,s$ such that $mr+ar\neq 0$ and $ns+bs\neq 0$.
If indeed the product is in $Z$, then the equation $mnx+(mb+an+ab)x=0$ for all $x\in R$. But look: $[(m,a)(n,b)](0,rs)=mnrs+(mb+an+ab)rs=mnrs+mbrs+anrs+abrs=(ns+bs)(mr+ar)$.(Need I say more or do you see the contradiction?)Final(?) bit
Let $(m,a),(n,b)$ and $r$ and $s$ be as before.
By definition, there exists a nonzero $r\in R$ such that $mr+ar\neq 0$. Note that, for example, $mr+ar$ is a nonzero element of $R$, and since $R$ does not have zero divisors, $0\neq r(mr+ar)=rmr+rar=(rm+ra)r$ says that $rm+ra\neq 0$ as well.
However since the product annihilates $(0,s)$, we have this equation: $$ mns+mbs+ans+abs=0 $$ Multiplying on the left by $r$: $$ rmns+rmbs+rans+rabs=0 $$ and factoring gives: $$ rmns+rans+rmbs+rabs =(rm+ra)ns+(rm+ra)bs=(rm+ra)(ns+bs) =0 $$ As we have remarked, neither factor is zero, but both factors are in $R$, so this is a contradiction. Thus, $Z$ is a completely prime ideal and $S/Z$ is a domain.