Is there a nice description of the field of fractions of the ring of polynomials with integer coefficients?

Let $\mathbb{Z}[x]$ denote the ring of polynomials (in the formal variable $x$) with integer coefficients. Since $\mathbb{Z}[x]$ is an integral domain, we can form its field of quotients, call it $Q.$

Is there a nice description of $Q$? I don't think $Q = \mathbb{Q}((x)),$ or in other words I don't think $Q = $ the set of all formal Laurent polynomials with rational coefficients, because for example I think that $$\sum_{n=0}^\infty \frac{x^n}{n!} \notin Q.$$

Solution 1:

When you've found an injection of a domain $D$ into a field $F$, there's just one thing to check before concluding that $F$ is (isomorphic to) the field of fractions for $D$: you need to check that $D$ is "dense" in $F$ in the following sense.

For any $q\in F\setminus\{0\}$, there exists $d\in D$ such that $qd\in D\setminus\{0\}$.

If $D$ has field of fractions $Q$, and $D$ is also embedded in a field $F$, you are always going to have an injection $Q\to F$. The above condition, if it is satisfied, guarantees the injection is also a surjection, and in that case $Q$ and $F$ would be isomorphic fields.

Now $\Bbb Z[x]$ can be identified as a subset of $\Bbb Q(x)$ as elements of the form $\frac{p(x)}{1}$ where $p(x)$ is an integer polynomial. Given an arbitrary nonzero element $\frac{s(x)}{t(x)}\in \Bbb Q(x)$, find the least common multiple of denominators of coefficents of $s(x)$ and of $t(x)$ and call them $m$ and $m'$ respectively. Then $m\cdot m'\cdot t(x)\in \Bbb Z[x]$ and $(m\cdot m'\cdot t(x))\frac{s(x)}{t(x)}\in \Bbb Z[x]\setminus\{0\}$. So, $\Bbb Z[x]$ is "dense" in $\Bbb Q(x)$ and therefore $\Bbb Q(x)$ is the field of fractions for $\Bbb Z[x]$.

This also helps us see why $\Bbb Q((x))$ isn't the field of fractions for $\Bbb Z[x]$. Certainly $\Bbb Z[x]$ is contained in $\Bbb Q((x))$, but the field is too big, and $\Bbb Z[x]$ isn't dense in it! The field $\Bbb Q((x))$ contains power series $S$ such that $S$ does not lie in $\Bbb Q(x)$, and if it were to happen that $dS=d'$ for nonzero $d,d'$ in $\Bbb Z{x}$, then $\frac{d'}{d}=S$ lying in $\Bbb Q(x)$ would be a contradiction.