# Concerning $Frac((Frac \space D)[x])$ and $Frac(D[x])$ for an integral domain $D$

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 .

## 3 Answers

Let $R$ be a domain, with $Q$ as field of fractions. It is clear that $\operatorname{Frac}(R[x])$ is a subfield of $\operatorname{Frac}(Q[x])$, because $R[x]$ is a subring of $Q[x]$.

On the other hand, if $f,g\in Q[x]$, with $f,g\ne0$, then there are $r,s\in R \setminus \{0\}$ such that $rf,sg\in R[x]$ (just use common denominators of the coefficients in $f$ and $g$).

Thus $$\frac{f}{g}=\frac{s}{r}\frac{rf}{sg}\in\operatorname{Frac}(R[x])$$

• @MartinBrandenburg Thanks, the $\ne0$ condition got lost during edits. ;-) – egreg Apr 6 '15 at 17:16

If you are willing to accept that $D\subseteq Frac(D)$ (via identification), then you have

$$Frac(D)\subseteq Frac(Frac(D))$$

(If $R\subseteq S$ then $Frac(R)\subseteq Frac(S)$).

Note that $Frac(Frac(D)[x])=Frac(D)(x)$, and this is the same as $Frac(D[x])$ (The latter has quotients of polynomials in $D[x]$, the former has quotient of polynomials in $Frac(D)[x]$ than can be turned into a quotient of polynomials in $D[x]$).

• That I know , but my question was , whether the inclusion is proper or not .... – user228168 Apr 6 '15 at 15:55
• I didn't answer your question properly. I'm sorry. Note that $Frac(Frac(D)[x])=Frac(D)(x)$, this is exactly the same (via isomoprhisms) as $Frac(D[x])$ (take quotients of polynomials in $D[x]$). I've edited the answer. – Daniel Apr 6 '15 at 16:04

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.

• Add $d \neq 0$. – Martin Brandenburg Apr 6 '15 at 17:04
• @Martin $\$ By defintion, $0$ is not a denominator, but I'll add it anyway for emphasis. – Bill Dubuque Apr 6 '15 at 17:04
• Yes, but in the offset formula it's missing. – Martin Brandenburg Apr 6 '15 at 17:53