TITLE: Concerning $Frac((Frac \space D)[x])$ and $Frac(D[x])$ for an integral domain $D$
QUESTION [1 upvotes]: Is the fraction field of $\mathbb Z[x]$ a proper subfield (or isomorphic to a proper subfield) of the fraction field of $\mathbb Q [x]$ ? In general , what can we say about $Frac((Frac \space D)[x])$ and $Frac(D[x])$ for an integral domain $D$ ; is the latter a proper subfield of the former ? Please help . Thanks in advance .

REPLY [0 votes]: By here, to show that $\,R \cong {\cal F}(D[x])$ it suffices to give a ring hom  injecting $\,D[x]\,$ into $\,R\,$ such that every element of $R$ can be written as a quotient (fraction) of images of elements of $\,D[x].\,$ Here $\,R = {\cal F}(({\cal F}D)[x])\,$ so the injection arises by composing the natural injections into fraction and polynomials rings. To show quotient form: scale the top and bottom of the fraction by a  common denominator $\,d\ne 0\,$ to clear all denom's in the coeff's of the top and bottom polynomials, i.e.
$$ f,g \in ({\cal F}D)[x]\ \Rightarrow\ \exists\, d\in D\!:\ df,dg\in D[x]\ \Rightarrow\ \dfrac{f}{g} = \dfrac{df}{dg}\in {\cal F}(D[x])\qquad $$
Such a common denominator $\,d\,$ for a finite set of fractions always exists (e.g. the product of all denominators). Here, wlog, we assume the injection of $\,D[x]\,$ is  the identity map.
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