natural mapping into ring of fractions is not always injective?

Question

Let $R$ be a ring, $S$ a multiplicatively closed set in $R$, and $S^{-1}R$ the ring of fractions. Let $f:R \rightarrow S^{-1}R$ send $r \to r/1$. I know this mapping is injective for fields of fractions. But I don't think this is true for when $R$ is not an integral domain. Given $r/1=s/1$, by definition we have $t(r-s)=0$ for some $t \in S$, but no guarantee that $r-s=0$.

First, I'm looking for confirmation on whether this reasoning is correct (or not). Also, I'm having trouble thinking of a concrete counterexample, as I don't have much experience with rings of fractions. Any help would be appreciated, thanks!

EDIT: fixed my definition of equivalency in the ring of fractions

If $R$ is not an integral domain, we might have $as=0$ for some $s\in S$ and some nonzero $a\in R$, in which case $a/1=0$ in the ring of fractions. — morrowmh, Dec 29 '20 at 21:03

score 1 · Accepted Answer · answered Dec 30 '20 at 14:58

If $R$ is an integral domain, then $(r-s)t=0$ in $R$ implies $r=s$. (Since $t=0$ implies $0\in S$, so $S^{-1}R=0$).

For a counterexample, let $R=\mathbf{Z}/6\mathbf{Z}$ and $S$ be the complement of the prime ideal $2\mathbf{Z}/6\mathbf{Z}$. (i.e. $S=\{1,3,5\}$). Then $(5-3)\cdot 3=0$, so $\frac{3}{1}=\frac{5}{1}$ in $S^{-1}R$.

natural mapping into ring of fractions is not always injective?

1 Answers1