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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 $S$?

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Why do you think it is prime? Just a gut feeling, or is this an exercise? – Thomas Andrews Aug 1 '12 at 19:14
(Also, generally unwise to call an ideal $Z$, especially when you are explicitly using the ring $\mathbb Z$ in the problem. Obviously, not your fault if this is an exercise.) – Thomas Andrews Aug 1 '12 at 19:29
@ThomasAndrews Yes, this is an exercise for fun, it is the last exercise in 2.17 of Nathan Jacobson's Algebra, page 156. – Chelsea Dirks Aug 1 '12 at 20:35
And does the problem assert that it is prime, or just ask? (Your headline asks, but you seem stuck trying to prove.) – Thomas Andrews Aug 1 '12 at 20:40
If $R$ is not commutative, you have the wrong definition of prime, I think. – Jack Schmidt Aug 1 '12 at 20:51

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.

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Thanks, my books asserts that $S/Z$ is a domain, so I believe it is. I'll try to use your suggestions in the mean time. – Chelsea Dirks Aug 1 '12 at 20:57
Yeah, you're right, I believe you because I went and looked up the problem :) Hope you don't mind that I rewrote the deleted post into something useful! – rschwieb Aug 2 '12 at 12:19
This works when $R$ is commutative, but $(mr+ar)(ns+bs) = mnrs + mrbs + nars + arbs \neq mnrs + \underline{mbrs} + nars + \underline{abrs}$. – Jack Schmidt Aug 2 '12 at 13:31
@JackSchmidt Agh! Yeah you're right... somehow $b$ needs to be commuted out of the way. Good catch. I should become a commutative algebraist so I have an excuse... but no... I'm noncommutative :( Jacobson did not make this simple, apparently. – rschwieb Aug 2 '12 at 14:44
Thanks rschweib. The preceding exercise in Jacobson states : Let $R$ be a rng without zero divisors. Assume $R$ contains elements $a$ and $b\neq 0$ such that $ab+kb=0$ for some positive integer $k$. Then $ca+kc=0=ac+kc$ for all $c\in R$. So I think in this case if there exists some $s\neq 0$ such that $bs+ns=0$, then it follows that $b$ commutes with all of $R$ and this would work. I can't tell if such $s\neq 0$ exists or not though. – Chelsea Dirks Aug 2 '12 at 19:54

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