Isomorphism between Frac($\mathbb Z[x]$) and Frac($\mathbb Q[x]$)

Question

I saw from another question that Frac($\mathbb Z[x]$) and Frac($\mathbb Q[x]$) are isomorphic. But by simply counting the elements in $\mathbb Z[x]$ and $\mathbb Q[x]$, I don't know why is it so. It is clear that the set $\mathbb Q[x]$ contains more elements than $\mathbb Z[x]$.

Answer 1

It is clear that $\text{Frac}(\Bbb Z[x]) \subseteq \text{Frac}(\Bbb Q[x])$.

On the other hand, suppose we have $\dfrac{p(x)}{q(x)} \in \text{Frac}(\Bbb Q[x])$.

Since the coefficients of the polynomial $q(x)$ are rational, and there are only finitely many of them, we can take the lcm of the denominators (when they are in "reduced form"), say it is $b$, and re-write:

$q(x) = \dfrac{1}{b} bq(x)$, where $bq(x) \in \Bbb Z[x]$.

We can do the same thing for $p(x)$, if $p(x) \neq 0$ (if $p(x) = 0$, how would you handle this special case?). So:

$p(x) = \dfrac{1}{a}ap(x)$ with $ap(x) \in \Bbb Z[x]$.

Thus: $\dfrac{p(x)}{q(x)} = \dfrac{\frac{1}{a}}{\frac{1}{b}}\dfrac{ap(x)}{bq(x)} = \dfrac{bap(x)}{abq(x)} \in \text{Frac}(\Bbb Z[x])$, since $bap(x),abq(x) \in \Bbb Z[x]$.

This shows that $\text{Frac}(\Bbb Z[x]) = \text{Frac}(\Bbb Q[x])$.

For example, take $p(x) = \frac{1}{3}x^2 - \frac{1}{4}$, and $q(x) = \frac{1}{5}x - \frac{1}{6}$. Then $a = 12$ and $b = 30$, and we have:

$\dfrac{\frac{1}{3}x^2 - \frac{1}{4}}{\frac{1}{5}x - \frac{1}{6}} = \dfrac{30(4x^2 - 3)}{12(6x - 5)}$