Quotient field of the intermediate integral domain

Question

Let $R\subset T \subset F_R$, where $R,T$ are two integral domains and $F_R$ is the quotient field of $R$. I need to show that $F_T\cong F_R$.

My effort: Since $T$ embeds in a field $F_R$, it must contain an isomorphic copy of its quotient field, i.e $T\subset F_T\subset F_R$. Since $R$ is contained in the field $F_T$, we must conclude that an isomorphic copy of $F_R$ is contained in $F_T$.

Therefore we conclude that $R\subset T\subset F_R'\subset F_T\subset F_R$, where $F_R'$ is an isomorphic copy of $F_R$ inside itself. I feel I am very close to proving $F_T\cong F_R$, but lack the closing argument. Any help is appreciated.

Answer 1

Using the hint given by @lhf and my original effort, here is the sketch of a proof.

We know that $R\hookrightarrow T\hookrightarrow F_R'\hookrightarrow F_T\hookrightarrow F_R$. Show that $F_T$ is a field of quotient of $R$ by showing the universal property.

Existence: For any $R\to S$, we can extend the map to $F_R\to S$ because $F_R$ is a field of quotient of $R$. We restrict this map down to $F_T\to S$.

Uniqueness: If $f,g:F_T\to S$ are two maps such that their restriction to $R$ coincides, then the restriction of $f,g: F_R'\to S$ must coincide because $F_R'$ is a quotient field of $R$. Then the restriction of $f,g$ to $T$ must coincide. It follows that $f,g$ must coincide because $F_T$ is the quotient field of $T$.

Since both $F_T$ and $F_R$ are quotient fields of $R$ they are isomorphic.

Answer 2

Perhaps it is clear if you use the universal property of the field of fractions of a domain:

Given $D$ a domain, a field of fractions $Q$ of $D$ is a field such there is an injection $i: D \to Q$ and for every injection $j: D \to K$ of $D$ into a field $K$, there is a unique injection $q: Q \to K$ such that $j=q\circ i$.

Here all maps are ring homomorphisms.

A corollary of this universal property is that all field of fractions of a given domain are isomorphic.