# Embedding a commutative ring without zero divisors (And without unity) in an integral domain.

I've got a question in this problem. I also will talk about the definitions I'm using (I'm using a Ring as an abelian aditive group with a multiplicative semigroup and distributive laws) I have a commutative ring $R$ without zero divisors. The first question. I can make a Partial Ring of Quotients of this ring. Can't I make a Field of Quotients of this ring? Why is the unity needed?

The second question is about what I was doing. I make this product: $S=R\times\mathbb{Z}$ And give it ring structure with the operations. Sum by components and product:

$(a,n)(b,m)=(ab+ma+nb,nm)$

I've proven that this is a ring, it has unity, is commutative, and may have zero divisors, even if $R$ doesn't (At least in the case $R$ is an integral domain it's very easy to find the zero divisors).

I want to prove that $I=\{s\in S \mid sa=0,\forall a\in R\}$ is a prime ideal of $S$. I've already proved it's an ideal, but I don't know how to prove it is prime. How could I do that?

I've supposed $uv\in I$ then $\forall a\in R$, $(uv)a=0$, that is $u(va)=0$ and as $S$ is commutative $v(ua)=0$...

• See the total quotient ring. a.k.a. total fraction ring. – Bill Dubuque Feb 18 '15 at 2:20
• It says that the construction "gives every non-zero-divisor of R an inverse in the larger ring". If the ring is like $2\mathbb{Z}$ that has no zero divisors, but no unity, then is the total quotient ring a field (Intuitively it would be also $\mathbb{Q}$)? If not, why would it be necessary for $R$ to have unity, to make the field of quotients? – David Molano Feb 18 '15 at 2:26

I've solved my problem, so I answer myself. We are identifying $R$ as the ring $R\times\{0\}\leq S$
If $I=\{s\in S\mid sa=0,\forall a\in R\}$
we then have it's an ideal. I had already that. Now, if $u,v\in S$ are elements such that $uv\in I$, then $(uv)a=0,\forall a\in R$. We then suppose that $v\notin I$ and mut show that $u\in I$. Now, as $v\notin I$, there exists $r\in R$ such that $(uv)r=u(vr)=0$, but $vr\neq 0$. And if $q$ is any element of $R$, we have $q(u(vr))=(qu)(vr)=0$ but as $R$ is also an ideal of $S$, we have $qu,vr\in R$ then as $R$ doesn't have zero divisors, and as $vr\neq 0$, we have $qu=uq=0$, that for all $q\in R$. Then $u\in I$ and $I$ is a prime ideal.