Can we see a ring $R$ as a subring of $S^{-1}R$? I know that we can consider an integral domain $D$ as a subring of its quotient field, I'm wondering why we can't consider any commutative ring with identity as a subring of $S^{-1}R$ identifying $r\in R$ as $r/1_R\in S^{-1}R$. 
Thanks in advance.
 A: This is because the natural ring homomorphism $R \to S^{-1}R$ is not always injective. It is injective when $R$ is an integral domain, which is why we can make this identification.
A: Consider the map $r\mapsto r/1$ from $R$ to $S^{-1}R$ (assuming $1\in S$). Then this is injective if and only if, from $r/1=0/1$ it follows that $r=0$.
However, saying $r/1=0/1$ means that there exists $t\in S$ such that
$$
t(1r - 0\cdot 1)=0
$$
or
$$tr=0.$$
If you can exclude this case, then the map is indeed injective: this amounts to saying that $S$ doesn't contain zero divisors. This is true, in particular, if $R$ is a domain.
If $S$ doesn't contain $1$, then you can do the same with the map $r\mapsto rs/s$, where $s$ is some fixed element of $S$ (actually the map is independent from the element you choose, as $rt/t=rs/s$ for all $r\in R$ and $s,t\in S$).
A: Here's a simple geometric example which may convince you that non-injective localizations do arise in nature and that they aren't necessarily evil.
Let $k$ be a field and $A = k[x, y]/(xy)$, which we should think of as the ring of functions on the union of the two coordinate axes in the plane. Localizing $A$ at $x$ corresponds to looking at the points where $x$ doesn't vanish, so $A_x$ should be the ring of functions on the $x$-axis less the origin, and indeed $A_x \cong k[x]_x$. The canonical map $A \to A_x$ is just restriction of functions, and the function $y$ restricts to $0$.
There should be examples more interesting than this one in Vakil's FOAG.
A: Hint: $ $ if so then $\,\Bbb Z/10\,$ would be isomorphic to a subring of $\,\Bbb Z/10[1/2] \cong \Bbb Z/5.$
Those $\,r\in R\,$ killed in $\,S^{-1}R\,$ are precisely those such that $\,rs^n = 0,$ for some $\, s\in S.$ For a proof (simpler than that in most textbooks) see this answer.
