Here is the question I want the solution of part $(c)$ in it:
Let $R$ be an integral domain with field of fractions $K,$ and let $S \subset R \setminus \{0\}$ be a multiplicatively closed subset.
$(a)$ Show that $S^{-1}R = \{r/s \mid r \in R, s \in S\}$ is a subring of $K$ containing $R.$ $S^{-1}R$ is called the localization of $R$ at $S.$Note: If $S = \{ x^n \mid n \in \mathbb Z \}$ for prime $x \in R,$ then $S^{-1}R = R[x^{-1}].$
$(b)$ Show that if $\mathfrak{p} \subset R$ is a prime ideal, then $S = R \setminus \mathfrak{p}$ is multiplicatively closed set. In this case the ring $S^{-1}R$ is denoted by $R_{\mathfrak{p}}.$
$(c)$ Since $R$ is an integral domain, $(0)$ is a prime ideal. What is $R_{(0)}$
My question is:
I know the solutions of letters $(a)$ and $(b),$ and I am guessing that from $(b),$ I can tell that $R_{(0)} = S^{-1}R = \{r/s \mid r \in R, s \in S\} $ and it is a subring of the field of fractions $K.$ I know also from $(b)$ that $S = R\setminus (0) = R - \{0\}.$ but how can I show that $R_{(0)} = K$? could anyone show me a rigorous proof to this equality please?
Also, because $(0)$ is a prime ideal, then $R/(0)$ is an integral domain, but how can I collect all his information together to form a proof?
Is the proof is just by showing that $1 \in S$?
My definition for the field of fraction of $R$:
It is the smallest field containing $R$ as a subring.