Proposition

Let $f: A \rightarrow B$ be a homomorphism of rings and let $S$ be a multiplicatively closed subset of $A$. Let $T = f(S)$. Show that $S^{-1}B$ and $T^{-1}B$ are isomorphic as $S^{-1}A$-modules.

Solution

Let $\phi: S^{-1}B \rightarrow T^{-1}B$ be defined such that $\phi(x / s) = x / f(s)$. We claim that $\phi$ is an $S^{-1}A$-module isomorphism.

• Well-defined?
  • Let $x / s = y / t \in S^{-1}B$. Then $u(tx - sy) = 0$ for some $u \in S$. By definition, $tx = f(t)x$ and $sy = f(s)y$. Thus $u(f(t)x - f(s)y) = 0$. Moreover, $u(f(t)x - f(s)y) = f(u)(f(t)x - f(s)y)$ since $u \in S \subset A$. Then $f(u)(f(t)x - f(s)y) = 0$ where $f(u) \in T$. This implies $x / f(s) = y / f(t)$ in $T^{-1}B$.
• $S^{-1}A$-module homomorphism?
  • Let $x / s, y / t \in S^{-1}B$. $\phi(x / s + y / t) = \phi((tx + sy) / st) = (tx + sy) / f(st)$. On the other hand, $\phi(x / s) + \phi(y / t) = x / f(s) + y / f(t) = (f(t)x + f(s)y) / (f(s)f(t))$. Therefore, $\phi$ preserves the addition.
  • Let $x / s \in S^{-1}B$ and $a / t \in S^{-1}A$. Then $\phi((a / t)(x / s)) = \phi(ax / st) = ax / f(st) = ax / tf(s) = (a/t)(x/f(s)) = (a/t)\phi(x/s)$.
• Surjective?
  • Trivial.
• Injective?
  • Let $x / t \in \ker(\phi)$. Then $\phi(x / t) = 0 / 1$, so $x / f(t) = 0 / 1$ in $T^{-1}B$. This implies the existence of $f(u) \in T$ such that $f(u)(x - 0f(t)) = 0$. Thus $f(u)x = 0$. By definition, $ux = f(u)x$. Thus $u(x \cdot 1 - 0 \cdot t) = 0$, so $x / t = 0 / 1$ in $S^{-1}B$.

Hence, $\phi$ is an $S^{-1}A$-module isomorphism.