# When can a ring imbed in a localization?

In Topics in Algebra, there is an exercise (3.6.5): let $R$ be a commutative, unital ring and let $S\subset R$ be non-empty and such that $s_1 s_2\in S$ if $s_1,s_2\in S$ and $0\not\in S$. Construct $R_S=R\times S/\sim$ with $(r,s)\sim(r',s')$ if there exists $s''\in S$ such that $s''(rs'-sr')=0$. I believe this construction is called a ring localization and generally denoted $S^{-1}R$, but I'll retain the notation of Herstein here.

There are six parts of the question, but the third part is the one I am hung up on (I have done all the rest). He asks "Can $R$ be imbedded in $R_S$?", an imbedding being defined as an injective ring homomorphism. I have shown that under certain circumstances (for instance, if $S$ contains no zero divisor) that an imbedding is possible. On the other hand, I have constructed a couple of explicit examples where an imbedding is not possible, one of which is: if $R=\mathbb{Z}/6\mathbb{Z}$ and $S=\{2,4\}$ then $|R|=6$ while $|R_S|=3$. I tried to come up with the general criteria for an imbedding to be possible, but I am stumped here.

My question is therefore:

is there a simple condition that I'm overlooking for $R$ to be imbeddable or not in $R_S$?

My suspicion is that $S$ having a zero divisor makes it impossible to imbed $R$ in $R_S$, but I couldn't show it. It seems very peculiar to me that the third of six parts should be the most difficult by far. Could Herstein just be looking for a simple "yes, it is possible under certain circumstances"?

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the words are equivalent and Herstein uses "imbed" –  Jonathan Jan 6 '13 at 1:00
Right, but the mathematical terminology has been changed in last 50 years since the book of Herstein was issued. –  user26857 Jan 6 '13 at 1:13

I think probably Herstein was looking for a "sometimes it works and sometimes it doesn't." For example, here is a counterexample to the claim that no zero divisors suffices...

Let $R = \mathbb{Z}[x_1^{\pm 1}, y_1, z_1, x_2^{\pm 1}, y_2, z_2, ...]/ (y_1y_2, z_1z_2, y_3y_4, z_3z_4, y_5y_6, ...)$. That is, we adjoined countably many generators and then (i) inverted countably many and (ii) made countably many zero divisors.

Now let $S = \{ y_1, y_3, y_5, ...\}$.

Then $S^{-1}R = \mathbb{Z}[ x_i^{\pm 1}, y_{2j+1}^{\pm 1}, z_k]/ (z_1 z_2, z_3z_4, ...)$ which is isomorphic to the original ring.

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Thanks and +1. I will think about your example when I have some time later today. Particularly thanks for venturing a guess as to what Herstein had in mind as an answer. –  Jonathan Jan 5 '13 at 20:46

If you are asking the following: there exists $f:R\to R_S$ an injective ring homomorphism iff $S$ contains only non-zerodivisors, then the answer is NO.

Let $R=\mathbb Z\times \mathbb Z^{\mathbb N}$ and $S=\mathbb Z\times (\mathbb Z-\{0\})^{\mathbb N}$. Then $S^{-1}R\simeq \mathbb Q^{\mathbb N}$ and we clearly have an embedding from $R$ to $S^{-1}R$. But $S$ obviously contains zerodivisors.

Remark. Maybe it's worthy to mention that I've used the following property: if $R=\prod R_i$ (an arbitrary direct product) and $S=\prod S_i$, where $S_i\subseteq R_i$ are multiplicative systems, then $S^{-1}R\simeq\prod S_i^{-1}R_i$.

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Thank you and +1, I'll study your example when I have a bit more time. It doesn't truly answer my question, but it may shed some light. –  Jonathan Jan 5 '13 at 20:44
@Jonathan Welcome! Let me just say that this example shows that there is no clear answer to your question. –  user26857 Jan 5 '13 at 20:52