Showing that $\{1,X\}$ is a basis for $\mathbb{Q}(X)$ as a vector space over $\mathbb{Q}(X^2)$ The goal is to show that $[\mathbb{Q}(X) : \mathbb{Q}(X^2)] = 2$, by constructing a basis for $\mathbb{Q}(X)$ as a vector space over $\mathbb{Q}(X^2)$. So I don't want to use minimum polynomials or the theory of algebraic numbers.
The TA said that $\{1,X\}$ would be a basis, but I'm having trouble showing this. 
My idea is the following, let $\frac{f(x)}{g(x)} \in \mathbb{Q}(X)$, then we want to write $\frac{f(x)}{g(x)} = 1 * \frac{f_1(x)}{g_1(x)} + x* \frac{f_2(x)}{g_2(x)} = \frac{f_1(x) g_2(x)+ f_2(x) x g_1(x)}{g_1(x)g_2(x)}$, with $\frac{f_1(x)}{g_1(x)},\frac{f_2(x)}{g_2(x)} \in \mathbb{Q}(X^2)$. But it is not clear to me that such $f_1, f_2, g_1, g_2 \in \mathbb{Q}(X^2)$ would exists?
Regards,
Jens

Hint: Write $f(x) = F_1(x^2)x + F_0(x^2)$. Similarly for $g$. Then
$$
\frac{f}{g}
=\frac{F_1 x + F_0}{G_1 x + G_0}
=\frac{F_1 x + F_0}{G_1 x + G_0} \cdot \frac{G_1 x - G_0}{G_1 x - G_0}
=\frac{(F_1 x + F_0)(G_1 x - G_0)}{G_1^2 x^2 - G_0^2}
$$
Just works just as in $\mathbb Q(\sqrt 2)$.
