# Prove j(E) is an integer for an elliptic curve with CM by a quadratic field of class number 1

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?

• 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. – Erick Wong Jul 27 at 5:08
• @ErickWong isn't the only automorphism of C complex conjugation? – uhhhhidk Jul 27 at 5:13
• @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) – uhhhhidk Jul 27 at 5:19
• 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. – Erick Wong Jul 27 at 5:42
• @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. – 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 $$\text{Aut}(\Bbb C)$$.
• 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? – uhhhhidk Jul 27 at 6:38