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"?

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

  • $\begingroup$ 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. $\endgroup$ – 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 $S^{-1}R\simeq\prod S_i^{-1}R_i$.

  • $\begingroup$ 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. $\endgroup$ – Jonathan Jan 5 '13 at 20:44
  • $\begingroup$ @Jonathan Welcome! Let me just say that this example shows that there is no clear answer to your question. $\endgroup$ – user26857 Jan 5 '13 at 20:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.