If $E$ has CM by an imaginary quadratic ring $\mathcal{O}_K$ such that $h(\mathcal{O}_K)=1$, how would we show that $j(E)$ is an integer? (or, equivalently, that $j(\frac{1+\sqrt -t}{2})\in\mathbb{Z}$ if $\bf Z$$[\frac{1+\sqrt -t}{2}]$ has class number 1)
The shortest proofs I've seen are based on using $\sigma \in Aut(\mathbb{C})$, for example
but I cant understand how the automorphisms of the complex numbers allow one to prove an element is algebraic. Since we're considering all of $\mathbb{C}$, why couldn't some automorphism send an algebraic number to a transcendental one? In other words, how would one even show that an algebraic number has a finite orbit under the automorphisms of $\mathbb{C}$? I'm able to understand the essence of the proof but the first part makes no sense to me. Is there another short proof which is more clear?
If $\alpha$ is algebraic, it is a zero of a polynomial $f$ over $\Bbb Q$. But $\sigma$ preserves the coefficients of $f$, so $0=f(\alpha)^\sigma=f(\alpha^\sigma)$, therefore $\alpha^\sigma$ is a zero of $f$, so one of the finitely many conjugates of $\alpha$.
If $\alpha$ is transcendental, then $\Bbb Q(\alpha)\cong\Bbb Q(X)$, the rational function field. Then $\Bbb Q(X)$ has automorphisms sending $X$ to $X+c$ for any $c\in\Bbb Q$. So $\Bbb Q(\alpha)$ has an automorphism $\sigma$ sending $\alpha\to\alpha+c$, and a Zorn's lemma argument extends this to an automorphism of $\Bbb C$. So $\alpha$ has infinitely many images under $\text{Aut}(\Bbb C)$.
The only other proof I know for the theorem on CM elliptic curves is the "honest" one using the modular equation.