Given an arbitrary integral domain $R$, the fraction field $Q$ is the smallest field conaining $R$.
In the integers $Q = \mathbb{Q}$
For a polynomial ring $P = {a_nx^n+\dots+a_1}$ the fraction field $Q$ is such that:
$q \in Q \implies q=\frac{p_1}{p_2}: p_1,p_2 \in R$
Is it correct to say that to construct the fraction field from an arbitrary integral domain $R$ we do
$Q = R\times R$ with modified $+,\circ$ operations? and that $R\times R$ is homomorphic to $R$?
No. This is not even the case with $\Bbb Z$ and $\Bbb Q$.
Instead, we let $$ Q=\bigl(R\times (R\setminus\{0\})\bigr)/\sim$$ where the equivalence(!) relation $\sim$ is given by $$ (a,b)\sim (c,d)\iff ad=bc.$$ On $Q$, we define $$ [(a,b)]+[(c,d)]:=[(ad+bc,bd)]$$ and $$ [(a,b)]\cdot [(c,d)]:=[(ac,bd)]$$ (which requires us to show that these are in fact well-defined). Of course, one also needs to show that this is a field. Then one shows that $\iota\colon R\to Q$, $r\mapsto [(r,1)]$ is an injective ring homomorphism, which allows us to identify $R$ with its image under this, thereby making $R$ a subring of $Q$. If $a\in R$ and $b\in R\setminus\{0\}$, it is clear that any subfield of $Q$ that still contains $R$, must contain $\iota(r)$ and the multiplicative inverse of $\iota(b)$, hence their product $[(a,b)]$. In other words, $Q$ is indeed minimal.
