An isomorphism between fields of fractions Let $R$ be an integral domain, let $K$ be its field of fractions, and let $S$ be a ring between $R$ and $K$.  Prove that the field of fractions of $S$ is (isomorphic to) $K$.
My attempt:
Since $S \subset K',$ define $f:K' \rightarrow S$ to be epimorphism, $g:S \rightarrow R$ to be epimorphism since $S$ between $K,R$, and $h:R \rightarrow K.$ I say if we take $t= h \circ g \circ f: K' \rightarrow K$, then this function is isomorphism? 
I am not sure if my argument true or not, so I appreciate any help.
Here is another attempt:
Let us denote the field of fraction of $R$ by $K$, and the field of fraction of $S$ by $K'$. So the functions $f_1:R \rightarrow K,\ f_2:S \rightarrow K'$ defined by $f_1(r)=\frac{r}{1},\ f_2(s)=\frac{s}{1}$ is a ring monomorphism. Next, define $f_3:R \rightarrow S$ by $f_3(r)=r$ since $R \subset S$, so it is a ring monomprphism. Now, we can define $g_1:R \rightarrow K'$ by $g_1=f_2 \circ f_3$, so this function is clearly a ring monomorphism. Hence there exists a unique ring homomorphism $\alpha: K \rightarrow K'$ such that $\alpha \circ f_1=g_1$. Now I am getting stuck here because I am not sure why this function $\alpha$ must be injective and surjective. 
Thanks for any further help.
 A: Let $Q(S)$ be the field of fractions of $S$ and let $\iota_R\colon S\to Q(S)$ be the canonical embedding. Let $\iota_R=\mathrm{id}_K\Bigm|_{R}$ be the restriction of the identity map of $K$ to $R$ (so $\iota_R$ is the embedding of $R$ into $K$) and let $j_S=\mathrm{id}_K\Bigm|_{S}$ be the restriction to $S$ (so $j_S$ is the embedding of $S$ into $K$. Finally, let $j_R\colon R\to S$ be the inclusion of $R$ into $S$.
The universal property of the field of fractions guarantees that:


*

*If $f\colon S\to F$ is any ring homomorphism with the property that $f(s)$ is invertible for every $s\neq 0$, then there exists a unique ring homomorphism $\mathfrak{f}\colon Q(S)\to F$ such that $f=\mathfrak{f}\circ \iota_S$. Moreover, if $F$ is no the trivial ring, then $f$ is one-to-one (since $0$ is not invertible in $F$, so only $0$ can map to $0$), and hence $\mathfrak{f}$, whose domain is a field, must be one-to-one.

*If $g\colon R\to F$ is any ring homomorphism with the property that $g(r)$ is invertible for every $r\neq 0$, then there exists a unique ring homomorphism $\mathfrak{g}\colon K\to F$ such that $g=\mathfrak{g}\circ \iota_R$. Moreover if $F$ is not the trivial ring, then $g$ must be one-to-one and hence $\mathfrak{g}$ must also be one-to-one.
Now, the map $\iota_S\circ j_R\colon R\to Q(S)$ induces a map $g\colon K\to Q(S)$. And the map $j_S\colon S\to K$ induces a map $f\colon Q(S)\to K$.
I claim that $f$ and $g$ are inverses of each other. 
Indeed, consider the map from $R$ to $K$ to $Q(S)$ to $K$ given by $f\circ g\circ\iota_R$. There must exist a unique morphism $h\colon K\to K$ such that $f\circ g\circ\iota_R = h\circ\iota_R$. Now, the identity map of $K$ certainly works, but so does $f\circ g$. By the uniqueness clause of the universal property, it follows that $f\circ g$ is the identity of $K$.
Symmetrically, consider the map from $S$ to $Q(S)$ to $K$ to $Q(S)$ given by $g\circ f\circ \iota_S$. There exists a unique morphism $h\colon Q(S)\to Q(S)$ such that $g\circ f\circ \iota_S = h\circ\iota_S$. Now, certainly $g\circ f$ is one such function, but so is the identity map of $Q(S)$. By the uniqueness clause, we have that $g\circ f= \mathrm{id}_{Q(S)}$.
Thus, $g\circ f = \mathrm{id}_{Q(S)}$ and $f\circ g = \mathrm{id}_{K}$, hence $f$ and $g$ are isomorphisms, so $Q(S)\cong K$, as claimed.
A: The field of fractions $K'$ of $S$ is equivalence classes $\frac{a}{b}$ for $a, b\in S$, $b\ne 0$. But $S\subset K$, so $\frac{a}{b}\in K$ since $K$ is a field. Thus $K'\subset K$. It is clear that $K\subset K'$ since $R\subset S$.
