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 enter image description here

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?

  • $\begingroup$ A (field) automorphism preserves $0$ and $1$, hence all positive integers, hence all integers. So if $p(x)=0$ for some integer polynomial $p$ then $p(\sigma(x))=0$. An automorphism can’t send algebraic numbers to transcendental ones. $\endgroup$ – Erick Wong Jul 27 at 5:08
  • $\begingroup$ @ErickWong isn't the only automorphism of C complex conjugation? $\endgroup$ – uhhhhidk Jul 27 at 5:13
  • $\begingroup$ @ErickWong oh so for example for some $\sigma \in Aut(\mathbb{C})$, $\sigma(2^{1/3})=w*2^{1/3}$? (w is a cube root of unity) Basically the automorphism group of C is like all the elements of the galios groups of all algebraic numbers? (And more automorphisms for transcendental numbers however those are defined) $\endgroup$ – uhhhhidk Jul 27 at 5:19
  • $\begingroup$ Yeah, that seems like a reasonable way to visualize automorphisms of $\mathbb C$. Apparently the cardinality of the automorphism group is $2$ to the power of continuum, so I suspect most of the freedom is in moving transcendentals around rather than the Galois actions among the algebraics. $\endgroup$ – Erick Wong Jul 27 at 5:42
  • $\begingroup$ @ErickWong I just have one more question: is each automorphism able to change more than one element? For example, if $\sigma \in Aut(\mathbb{C})$ satisfied $\sigma(\sqrt2)=-\sqrt2$, would it be inert for all other algebraic numbers, or could it also satisfy, say, $\sigma(2^{\frac{1}{3}})=w*2^{\frac{1}{3}}$? I don't know why but I'm thinking that if it changes more than one element it wouldn't hold as an automorphism. $\endgroup$ – uhhhhidk Jul 27 at 14:16

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.

  • $\begingroup$ Okay I think I understand a bit better. If we apply this automorphism to the coefficients $g_2(\Lambda)$ and $g_3(\Lambda)$, we get two new values which can be be represented by $g_2(\Lambda')$ and $g_3(\Lambda')$ of a new lattice $\Lambda'$ we can then prove that $\Lambda'$ has CM by elements $\sigma(\alpha)$ if the original lattice has CM by $\lambda$ since $\alpha$ is imaginary quadratic, the automorphism sends it to its conjugate, so the action of the automorphism sends the endomorphism ring to itself, this shows that the endomorphism rings of $E$ and $E^\sigma$ are equal. Is this close? $\endgroup$ – uhhhhidk Jul 27 at 6:38
  • $\begingroup$ There's some errors I made while typing that I can't edit for some reason, I hope you can see through it $\endgroup$ – uhhhhidk Jul 27 at 6:46

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