Total quotient ring and its properties

Solution 1:

For (a), assume that $A \to S^{-1}A$ is injective. We will show that if $s \in S$, then $s$ is not a zerodivisor. In fact, if there is some $a \in A$ such that $as=0$, then $a/1=0/1$ in $S^{-1}A$, and so $a$ is in the kernel of $A \to S^{-1}A$. It follows that $a =0$ by injectivity, hence $s \in S$ is not a zerodivisor.

For (b), let $p/q \in S^{-1}A$, with $q \in S$ and $p \in A$.We have two options, depending on $p \in S$.

• If $p \notin S$, then $pr=0$ for some $r \neq 0$ (as it will be a zerodivisor), hence $r/1 \neq 0/1$ in $S^{-1}A$ (as $A \to S^{-1}A$ is injective). At last, $p/q \cdot r/1$ is zero. This shows that $p/q$ is a zerodivisor.
• If $p \in S$, then $q/p \in S^{-1}A$ as well, and clearly $p/q \cdot q/p=1/1$, i.e. $p/q$ is a unit.

At last, for (c), let $S \subset A$ be a multiplicative subset which does not contain zero (i.e. $S^{-1}A$ is not the zero ring), and consider the injection $A \to Q(A)$. Note that for every $s \in S$, the image of $s$ in $Q(A)$ is a unit, and so the universal property of localization gives us a map $S^{-1}A \to Q(A)$. The kernel of this map is given by $a/s \in S^{-1}A$ which are zero in $Q(A)$. But $a/s=0/1$ in $Q(A)$ implies that $ab=0$ for some $b \neq 0$, and so $a=0$ (as $A$ is an integral domain). Therefore, the map $S^{-1}A \to Q(A)$ is injective.