Equality, vanishing, and the kernel of the localisation map

The equality criterion is the definition of equality of equivalence classes. Taking $(r^{\mathrm{\prime }},s^{\mathrm{\prime }})=(0,1)$ gives $r\mathrm{/}s=0\mathrm{/}1$ exactly when $u(r\cdot 1-0\cdot s)=ur=0$ for some $u\in S$.

If $0\in S$, step 1.1 makes every fraction zero by using $u=0$. Conversely, if $S^{-1}R$ is the zero ring, then $1\mathrm{/}1=0$, so step 1.1 gives $u\in S$ with $u\cdot 1=0$, whence $u=0$ and $0\in S$.

Since ${\lambda }_S(r)=r\mathrm{/}1$, step 1.1 gives the displayed kernel. Thus ${\lambda }_S$ is injective exactly when $ur=0$ with $u\in S$ always forces $r=0$, which says precisely that every member of $S$ has trivial annihilator.

Assume $R$ is nonzero. If every member of $S$ has trivial annihilator, then $0\mathrm{\notin }S$, because $0\cdot 1=0$ and $1\mathrm{\ne }0$, and no $u\in S$ is a zero divisor by [F2]. Conversely, if $0\mathrm{\notin }S$ and $S$ contains no zero divisor, then $u\in S$ is nonzero and $ur=0$ forces $r=0$ by [F2].