Linear Algebra – Basis for [Q(?):Q]

Question

I'm trying to figure out whether the basis of $\mathbb{Q}(x)$ over $\mathbb{Q}$ is countable when $x$ is transcendental. I know that the elements in $\mathbb{Q}(x)$ will be rational functions in $x$ and so they are countable like algebraic numbers.

Let the rank be the sum of coefficients and degrees of polynomials in denominator and numerator.

So it passes the necessary condition for countable basis outlined by Asaf Karagila in Countable/uncountable basis of vector space. I have no more ideas to go with.

Edit:

Obviously a basis will have to be countable if the space is countable. So how do I go about finding a basis?

Answer 1

I'll answer my own question using the suggestion by GEdgar in the comments. $\mathbb Q[x]$ is a Euclidean Domain and by the theorem of Partial Fractions

Let $f$ and $g$ be nonzero polynomials over a field $K$. Write $g$ as a product of powers of distinct irreducible polynomials :

$$g=\prod_{i=1}^k p_i^{n_i}$$

There are (unique) polynomials $b$ and $a_{ij}$ with deg$(a_{ij})<$deg$(p_i)$ such that

$$\frac{f}{g}=b+\sum_{i=1}^k\sum_{j=1}^{n_i}\frac{a_{ij}}{p_i^j}$$

If deg$(f)<$deg$(g)$, then $b=0$.

Let $\mathcal{B_1}=\{x^n : n\in\mathbb{N}\}$ and $\mathcal{B_2}=\left\{\dfrac{x^n}{p^m} : p \text{ is a monic irreducible and } 0\le n < \text{deg}(p)\right\}$.

The theorem implies $\cal B_1 \cup B_2$ is a basis for $\mathbb Q(x)$ over $\mathbb Q$