When is $\mathbb{Q}(x)$ a finite extension of $\mathbb{Q}$?

Question

Let $x\in\mathbb{C}$, and consider the field $\mathbb{Q}(x)$. If this field is a finite extension of $\mathbb{Q}$, then $x$ is algebraic over $\mathbb{Q}$, so satisfies some polynomial $P$. Any field automorphism of $\mathbb{C}$ will send $x$ to a root of $P$, so $x$ has a finite orbit under $Aut(\mathbb{C})$.

How do I prove the converse? That is, if $x$ has a finite orbit under $Aut(\mathbb{C})$, how do we know that $\mathbb{Q}(x)/\mathbb{Q}$ is finite?

Answer 1

Let $x_1, \dots, x_n$ be the orbit of $x$. Then $p(t)=\Pi_{i=1}^n (t-x_i)$ is a polynomial with coefficients in $\mathbb Q$ since it is fixed by all automomorpishms of $\mathbb C$. But then $x$ is algebraic because $p(x)=0$.