Why is $S/Z$ a domain for the ideal $Z=\{z\in S\mid za=0,\;\forall a\in R\}$ in $S$? 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 $S$? 
 A: 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.
A: I also was stuck on this Jacobson exercise (2.17-5) and rschwieb's proof for a commutative rng helped me out.
Start by proving a statement that looks just like exercise 4 in 2.17, but it's actually reversed and I don't see why it matters whether the integer is positive.
Claim: R has no zero divisors and $a,r \in R$ both non-zero. $k$ is an integer. If $ra + kr = 0$ then $ac + kc = 0$ $\forall c \in R$.
Since R is embedded in S, we can re-write our claim in S. Our assumption is that there exists $(k,a) \in S$ and $(0,r) \in S$, such that $(0,r)(k,a) = (0k, kr + ra + 0a) = (0,0) = 0$.
Now suppose there exists some $(0,c) \in S$ such that $(k,a)(0,c) \ne 0$. By associativity, the entire product below is still zero:
$(0,r) ( (k,a) (0,c) ) = ( (0,r) (k,a) ) (0,c) = 0 (0,c) = 0$
But rewriting the left-hand side we get
$(0,r) (0, ac + kc) = (0, r(ac + kc)) = (0,0)$
Restricting to just the right part, that's a statement in R that
$r(ac+kc) = 0$. We know $r \ne 0$ and we also know that
$(k,a)(0,c) = (0, ac + kc) \ne (0,0)$
That's only possible if $ac + kc \ne 0$ in R, but in that case R must have zero-divisors. That's a contradiction, therefore we must have for all $(0,c) \in S$, 
$(k,a)(0,c) = 0 = (0, ac + kc)$
Or back in R, we have that for all $c \in R$, $ac + kc = 0$, which was our claim.
Second claim: $S/Z$ is a domain.
Start with contradiction, just like in rschwieb's answer. Suppose S/Z is not a domain. this requires that there exist $(m,a), (n,b) \in S$, neither of which is in Z, but their product is in Z. Because $(m,a)(n,b)$ is in Z, we must have for any $c \in R$ that
$(m,a)(n,b)(0,c) = 0$
This comes out to:
$(mn,ab + mb + na)(0,c) = (0, mnc + abc + mbc + nac) = 0$
or just
$mnc + abc + mbc + nac = 0$ in R.
Because $(m,a)$ is not in Z, there must exist some $r \ne 0$ in R with
$(m,a)(0,r) \ne 0$.
Multiply the equation above on the left by $r$ then factor to get,
$
\begin{align}
mnrc + rabc + mrbc + nrac
&= mr(nc) + mr(bc) + ra(bc) + ra(nc) \\
&= mr(nc + bc) + ra(bc + nc) \\
&= (mr + ra)(nc + bc)
\end{align}$
Now since $(n,b) \notin Z$ there must exist some $d \in R$ non-zero, such that,
$(n,b)(0,d) = (0,nd + bd) \ne 0$. Since $c$ above is arbitrary, choose $c$ to be that $d$. In that case, we have in $R$,
$(mr + ra)(nc + bc) = 0$
with $nc + bc \ne 0$. Since R has no zero-divisors, we must then have that
$ra + mr= 0$.
Using our claim above, this requires that
$af + mf = 0$ for all $f \in R$. In S, that means
$(m,a)(0,f) = 0$ for all $f \in R$. That's the same thing as saying that $(m,a) \in Z$. That's a contradiction, so we must have that S/Z is a domain.
