Let $R$ be a commutative ring with no zero divisors, then $R$ can be embedded in an integral domain $S$.

I am facing a problem to find the monomorphism $f: R \to S$.

Will the function $f(a) ={a \over 1}$ work here?

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    $\begingroup$ In these types of questions, usually the natural maps are the right ones. $\endgroup$ – Michael Burr Mar 30 '15 at 16:14
  • $\begingroup$ what will be the map?? $\endgroup$ – user8795 Mar 30 '15 at 16:16
  • $\begingroup$ Does $R$ have identity? $\endgroup$ – Censi LI Mar 30 '15 at 16:17
  • $\begingroup$ it is not mentioned $\endgroup$ – user8795 Mar 30 '15 at 16:18
  • $\begingroup$ So your problem is how to add an identity? Otherwise I think $R$ is itself an integral domain. $\endgroup$ – Censi LI Mar 30 '15 at 16:20

How do you even know $\frac{a}{1} \in S$, since you haven't even said what "$S$" is?

Hint #1: Prove $T = R -\{0\}$ is closed under multiplication.

Hint #2: Can we choose $S = T^{-1}R$? Is this an integral domain?


I think you can simply take $S$ to be the fraction field, the construction of which is similar to the unital case: Let $T:=R\backslash\{0\}$, if $T=\emptyset$, $R$ is zero ring and cannot be embedding to integral domain; otherwise $T$ is a multiplicative set, define $S:=T^{-1}R$. Note that $S$ is a field (in particular, it has identity, namely $\frac tt$ for any $t\in T$). Furthermore, $R$ can be embedded to $S$ via $f:R\to S, r\mapsto \frac{rt}t\ (t\in T)$.


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