0

I have searched the site, but there is a lot of confusion regarding this topic, so I try to be as clear as possible.

Let $R$ be an integral domain and denote with $K(R)$ the field of fractions of $R$.

Now, let $S\subseteq R$, $0\notin S$, any multiplicatively closed subset of $R$.

Let $S^{-1}R$ be the localization of $R$ at $S$. Suppose that $S^{-1}R$ is a integral domain.

Is the following true or false?

$$K(R)\cong K(S^{-1}R)$$

Can this isomorphism be derived from the universal property of localization? If so, how can I proceed strictly?

user26857
  • 53,190
  • It is true, that once you've made all the nonzero elements of integral domain $R$ invertible, you've subsumed all the localizations of $R$. The key is that in a natural way $R\subset K(R)$, and localization (in the absence of zero divisors) preserves inclusion. Perhaps you could sketch the "universal property" you have in mind, to see if it suffices to adduce these facts. – hardmath Jul 13 '21 at 18:21
  • 1
    See this post or this post – Viktor Vaughn Jul 13 '21 at 18:27

0 Answers0