Proposition

Let $A$ be a ring, let $S$ and $T$ be two multiplicatively closed subsets of $A$, and let $U$ be the image of $T$ in $S^{-1}A$. Show that the rings $(ST)^{-1}A$ and $U^{-1}(S^{-1}A)$ are isomorphic.

Solution

$ST$ and $U$ are both multiplicatively closed, so it makes sense to talk about $(ST)^{-1}A$ and $U^{-1}(S^{-1}A)$.

Let $\phi:U^{-1}(S^{-1}A) \rightarrow (ST)^{-1}A$ be defined such that $\phi((a / s)/(t / 1)) = a / st$. We claim that $\phi$ is a ring isomorphism.

• Well-defined?
  • Let $(a / s) / (t / 1) = (a’ / s’) / (t’ / 1)$. Then there exists a $u \in U$ such that $u((a / s)(t’ / 1) - (a’ / s’)(t / 1)) = 0$. By using the fact that $U$ is the image of $T$, the above expression simplifies to $uat’s’ - ua’ts = 0$ for some $u \in T$. Since $T \subset ST$, $u(at’s’ - a’ts) = 0$ implies $a / st = a’ / s’t’$ in $(ST)^{-1}A$.
• Ring homomorphism?
  • Clearly, $\phi$ maps the multiplicative identity to the multiplicative identity.
  • \[\begin{align*} \phi((a / s) / (t / 1) + (a' / s') / (t' / 1)) &= \phi(((at's' + a'ts) / ss') / (tt' / 1)) \\ &= (at's' + a'ts) / (ss'tt') \\ &= (a / st) + (a' / s't') \\ &= \phi((a / s) / (t / 1)) + \phi((a' / s') / (t' / 1)). \end{align*}\]
  • \[\begin{align*} \phi((a / s) / (t / 1) \cdot (a' / s') / (t' / 1)) &= \phi((aa' / ss') / (tt' / 1)) \\ &= aa' / ss'tt' \\ &= (a/st) \cdot (a'/s't') \\ &= \phi((a / s) / (t / 1)) \cdot \phi((a' / s') / (t' / 1)). \end{align*}\]
• Surjective?
  • Trivial
• Injective?
  • Let $(a / s) / (t / 1) \in \ker(\phi)$. Then $a / st = 0 / 1$, so there exists $s’t’ \in ST$ such that $s’t’(a - 0) = 0$. Thus $s’t’a = 0$, s $t’(as’ - 0s) = 0$. This implies $a / s = 0 / s’$, so $(a / s) / (t / 1) = 0$.

Therefore, $\phi$ is a ring isomorphism.