Atiyah-Macdonald problem 3.8

Solution 1:

Injectivity: $a/s=0/1$ in $T^{-1}A$ iff $\exists\ t\in T$ such that $ta=0$. We want to show that $a/s=0/1$ in $S^{-1}A$, that is, there is $s'\in S$ such that $s'a=0$. If not, then $\operatorname{Ann}(a)\cap S=\emptyset$ and let $P$ be a prime ideal of $A$ such that $P\supseteq\operatorname{Ann}(a)$ and $P\cap S=\emptyset$. But $t\in P\cap T$, a contradiction.

Surjectivity: for $t\in T$ there is $a\in A$ such that $at\in S$. Otherwise, $Rt\cap S=\emptyset$ and therefore there is a prime ideal $P\supseteq RT$ such that $P\cap S=\emptyset$. But then $P\cap T=\emptyset$, a contradiction.

Solution 2:

I wonder if this can be done "element-less" by universal arrows. In fact, the arrow $\phi$ is just the arrow given by universal property, since $S\subseteq T$. Then, given any prime $\mathfrak{p}$ such that $\mathfrak{p}\cap S=\varnothing$, and hence $\mathfrak{p}\cap T=\varnothing$, consider the multiplicative part $Q=A-\mathfrak{p}$, and $\bar{Q}$ its image inside $S^{-1}A$: by exercise 3.3 we know that $$Q^{-1}A=(SQ)^{-1}A\cong\bar{Q}^{-1}S^{-1}A=(S^{-1}A)_\mathfrak{p}$$ The same holds for $Q$ and $T$: $$Q^{-1}A=(TQ)^{-1}A\cong\bar{Q}^{-1}T^{-1}A=(T^{-1}A)_\mathfrak{p}$$ And the isomorphisms are the canonical arrows. Thus we get that $$(S^{-1}A)_\mathfrak{p}\cong (T^{-1}A)_\mathfrak{p}$$ Via the canonical arrows. The arrow is the one induced by $\phi$ localizing, and being locally mono/epi at every prime it must be so at the global level, thus $\phi$ is an iso.