Let $R$ be an integral domain and let $S={\rm Frac}(R)$. Do we always have $R(x)=S(x)$?

Question

I was reading this post which shows that the fraction field of $\mathbb Z[x]$ is $\mathbb Q(x)$.

I think a simpler proof than the one given there is the following:

We know $\mathbb Z(x) \subset \mathbb Q(x)$. Since $\mathbb Q \subset \mathbb Z(x)$, then $\mathbb Q[x] \subset \mathbb Z(x)$, hence $\mathbb Q(x) \subset \mathbb Z(x)$.

Is this correct?

If so, it seems we can replace $\mathbb Z$ by any integral domain $R$ and $\mathbb Q$ by $S={\rm Frac}(R)$ and say:

We know $R(x) \subset S(x)$. Since $S \subset R(x)$, then $S[x] \subset R(x)$, hence $S(x) \subset R(x)$.

Is this correct?

Answer 1

It is sound but not really simpler. To paraphrase and use your notation, the second paragraph says “if $q\in S(x)$ and $m$ is the product of denominators of coefficients appearing in the numerator and denominator of $q$, then $q=\frac mm q\in R(x)$ , so $S(x)\subseteq R(x)$.

If the rest made it sound complicated, it was because it is an explanation for the OP about why their guess was not the right candidate.