Finding all $f \in \mathbb{C}[x]$ such that $\mathbb{C}(f,x^2)=\mathbb{C}(x)$ Call $f=f(x) \in \mathbb{C}[x]$ a friend of $x^2$ if the field generated by $f$ and $x^2$ is $\mathbb{C}(x)$, namely, $\mathbb{C}(f,x^2)=\mathbb{C}(x)$.

Question 1: Is it possible to find all friends of $x^2$, namely, a general form of a friend of $x^2$?

Examples: $f=x^3$, $f=x^4+x$.

Question 2: Is it possible to find all friends $f$ of $x^2$, with the additional property that $f(0)=0$?

We can assume that $\deg(f) \geq 3$; indeed:

*

*If $\deg(f)=0$, then $f \in \mathbb{C}$, hence $\mathbb{C}(f,x^2)=\mathbb{C}(x^2)$, so $f$ is not a friend of $x^2$.

*If $\deg(f)=1$, then $f$ is linear, hence $\mathbb{C}(f,x^2)=\mathbb{C}(x)$, so $f$ is a friend of $x^2$.

*If $\deg(f)=2$, then $f=ax^2+bx+c$, for some $0\neq a,b,c \in \mathbb{C}$.
If $b=0$, then $\mathbb{C}(f,x^2)=\mathbb{C}(x^2)$, so $f$ is not a friend.
If $b\neq 0$, then $\mathbb{C}(f,x^2)=\mathbb{C}(bx+c)=\mathbb{C}(x)$, so $f$ is a friend

Thank you very much!
 A: The extension $\mathbb{C}(x) \supseteq \mathbb{C}(x^2)$ is of degree $2$, so if $f \in \mathbb{C}(x) \setminus \mathbb{C}(x^2)$, then $\mathbb{C}(f,x^2) = \mathbb{C}(x)$. So any rational function $f$ that is not just a rational function in $x^2$, i.e., not of the form $f = g(x^2)$, will work. Thus, if an odd power of $x$ appears in either the numerator or denominator of $f$ (written in lowest terms), then $f$ is a friend of $x^2$.
A: You can prove this directly and constructively, without knowing about field extensions. Every polynomial can be written uniquely as:
$$f(x)=g_1(x^2)+xg_2(x^2).$$
Namely, with $f(x)=\sum_i a_ix^i,$ you use: $$g_1(x^2)=\sum a_{2i}x^{2i}, g_2(x)=\sum_i a_{2i+1}x^{2i}.$$
When $g_2\neq 0,$ then you get: $$x=\frac{f(x)-g_1(x^2)}{g_2(x)}$$
But $g_2=0$ precisely when $f\in \mathbb C[x^2].$

It is only a little harder to prove it for rational functions $f$ this way.
You essentially need to show that even elements of $\mathbb C(x)$ - the elements $g(x)\in \mathbb C(x)$ such that $g(x)=g(-x)$ - are exactly the elements of $\mathbb C(x^2),$ and the odd elements are the elements of $x\mathbb C(x^2).$
You show that by showing that $f(x)=\frac{p(x)}{q(x)}$ is even if and only if $p(x)q(-x)$ an even polynomial, and thus a polynomial in $x^2.$
But then $$f(x)=\frac{p(x)q(-x)}{q(x)q(-x)}.$$
And you just need to note $q(x)q(-x)$ is also an even polynomial.
Now given any $f\in \mathbb C(x),$ define $$g_1(x)=\frac12(f(x)+f(-x))\\g_2(x)=\frac2x(f(x)-f(-x))$$ and show $g_1,g_2$ are even, so in $\mathbb C(x^2),$ and $$x=\frac{f(x)-g_1(x)}{g_2(x)},$$ provided that $g_2\neq 0.$
