Elements of $\mathbb{Q}(x) - \mathbb{Q}(x^{2})$

Question

Given the tower of field extensions $$ \mathbb{Q} \subseteq \mathbb{Q}(x^{2}) \subseteq \mathbb{Q}(x) $$ What does an arbitrary element of $\mathbb{Q}(x) - \mathbb{Q}(x^{2})$, that is $y \in \mathbb{Q}(x)$ but $y \not\in \mathbb{Q}(x^{2})$, look like? Is every element of the form $q_{1} + q_{2}x$ where $q_{1}$ and $q_{2}$ are in $\mathbb{Q}$ or is it possible to have $x^{k}$ terms where $k$ is odd?

Answer 1

$\mathbb{Q}(x)/\mathbb{Q}(x^2)$ is a degree $2$ algebraic extension so $\mathbb{Q}(x)=\mathbb{Q}(x^2)\oplus \mathbb{Q}(x^2)x$, i.e. every rational function of $x$ is uniquely expressible as $A(x^2)+B(x^2)x$ where $A$ and $B$ are rational functions.

Thus, $\mathbb{Q}(x)-\mathbb{Q}(x^2)$ consists of all elements expressible as $A(x^2)+B(x^2)x$ with $B\ne0$, or in other words we can say the set difference is $\mathbb{Q}(x^2)+\mathbb{Q}(x^2)^\times x$.

Given an arbitrary rational function $R(x)$, we can solve for $A(x)$ and $B(x)$ by symmetry:

$$ A(x)=\frac{R(x)+R(-x)}{2}, \quad B(x)=\frac{R(x)-R(-x)}{2x}. $$

(Compare with expressing $\sin\theta$ and $\cos\theta$ in terms of $e^{\pm i\theta}$.)

Since $\mathbb{Q}(x^2)$ is the subfield of even rational functions, we can say the elements of $\mathbb{Q}(x)-\mathbb{Q}(x^2)$ are the noneven rational functions. This is not equivalent to being odd, it just means the odd part is nonzero ($A(x)$ and $B(x)x$ are the even and odd parts of $R(x)$, respectively).

Every element of $\mathbb{Q}(x^2)$ written in simplest terms has only even powers of $x$ in the numerator and denominator, and conversely any rational function expressible this way is in $\mathbb{Q}(x^2)$. Thus, we have yet another characterization: the elements of the set difference are those rational functions which, when written in simplest form, contain an odd power of $x$ in the numerator or denominator (or both). It is necessary to say "simplest form," as examples like $x/x=1$ show.