Prove that the homomorphism $\phi:R\to S^{-1}R$ is injective if and only if $S$ contains no zero-divisors.

Question

Suppose that $S\subset R$ is a multiplicative set in $R$, where $R$ is a commutative ring with identity $1\neq 0$. Prove that the homomorphism $\phi:R\to S^{-1}R$ is injective if and only if $S$ contains no zero-divisors.

I have an idea about how to approach this question, but I am not sure exactly how to structure it. For one direction, where we assume $\phi$ is injective, this means that $\ker(\phi)=\{0\}$, which I think means that there are no zero divisors in $R$, so there are no zero divisors in $S$ because it is a subset of $R$? Am I headed in the right direction?

Edit: I believe this question is different than the one it is marked as duplicate to, because that question uses the concept of monomorphisms which are not mentioned here.

Perhaps you could enquire what in general the kernel of $\phi$ is? — Angina Seng, Jan 27 '18 at 20:31
In general, $\ker(\phi)={r\in R|\phi(r)=0}={r\in R|rs^{-1}=0\text{ for some }s\in S}$, so if this is equal to zero, then we are saying that for each $r\in R$, there exists no $s\in S$ such that $rs^{-1}=0$. — MathStudent1324, Jan 27 '18 at 20:40
Do you really mean $s^{-1}$ here? In general a multiplicatively closed set will not consist entirely of units. — Angina Seng, Jan 27 '18 at 20:43
I think I am assuming that $S$ consists only of invertible elements. — MathStudent1324, Jan 27 '18 at 20:51
If $S$ consists only of invertible elements in $R$ then $\phi$ is an isomorphism, and there is really no point in considering $S^{-1}R$. — Angina Seng, Jan 27 '18 at 20:52
My mistake, you are correct. In this case, I'm not really sure how to picture $\phi$. — MathStudent1324, Jan 27 '18 at 20:56

Prove that the homomorphism $\phi:R\to S^{-1}R$ is injective if and only if $S$ contains no zero-divisors.

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