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Let $\tau$ in the upper half plane lie in an imaginary quadratic field $K$.

Then is $K \subset \mathbb{Q}(\{j(g \tau) \ | \ g \in GL_2^+(\mathbb{Q}) \})$?

(here $j$ is the modular $j$-function, and $GL_2^+(\mathbb{Q})$ means positive determinant - i.e. we adjoined all $j$ values of elliptic curves with CM by an order in $K$).

If the above isn't true, then is $K$ contained in $\mathbb{Q}(S)$ where $S$ is the set of $j$ values of all CM elliptic curves?

I'm also interested in whether the appropriate statement is true for a general Shimura variety.

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"No" for both questions about CM elliptic curves and "I'm not even sure I know what the question would be" about general Shimura varieties.

Basic idea: If $j$ is the $j$-invariant of a CM elliptic curve, then there is some imaginary quadratic discriminant $\Delta \equiv 0$ or $1 \bmod 4$ such that $\mathbf{Q}(j) \cong \mathbf{Q}[X]/H_\Delta(X)$ where $H_\Delta(X) \in \mathbf{Z}[X]$ is the Hilbert Class Polynomial of discriminant $\Delta$, whose roots are the $j$-invariants of elliptic curves over the complex numbers (actually $\overline{\mathbf{Q}}$ is enough) with CM by $\mathbf{Z}\left[\dfrac{ \Delta + \sqrt \Delta}{2}\right]$.

The point is that now we can see that there is an embedding $\mathbf{Q}(j) \hookrightarrow \mathbf{R}$, for all possible $j$. Therefore there is an embedding $\mathbf{Q}(S) \hookrightarrow \mathbf{R}$. To see this, it's enough to note that for any two CM $j$-invariants $j_1$ and $j_2$ that there exists an embedding $\mathbf{Q}(j_1,j_2)\hookrightarrow \mathbf{R}$. Let $J_1$ and $J_2$ be the canonical image of $j_1$ and $j_2$ in the real numbers. Then $\mathbf{Q}(j_1)$ embeds into the real numbers as $\mathbf{Q} + \mathbf{Q}J_1 + \dots + \mathbf{Q}J_1^{h_1 -1}$ and $\mathbf{Q}(j_2)$ embeds into the real numbers as $\mathbf{Q} + \mathbf{Q}J_2 + \dots + \mathbf{Q}J_2^{h_2 -1}$. Therefore $\mathbf{Q} + \mathbf{Q}J_1 + \mathbf{Q}J_2 + \dots + \mathbf{Q}J_1^{h_1 -1}J_2^{h_2 -1}$ is a copy of $\mathbf{Q}(j_1,j_2)$ inside of $\mathbf{R}$. Notice that I didn't use direct sums because $\mathbf{Q}(j_1)$ and $\mathbf{Q}(j_2)$ might not be linearly disjoint over $\mathbf{Q}$! This is also the reason I didn't use a tensor product argument. In any case, this inductive step allows us to work with direct limits and embed $\mathbf{Q}(S)$ into $\mathbf{R}$.

Therefore if we assume that there is an embedding $K\hookrightarrow \mathbf{Q}(S)$ then there must be an embedding $K\hookrightarrow\mathbf{Q}(S) \hookrightarrow \mathbf{R}$, which is absurd. Therefore, there is no embedding $K\hookrightarrow \mathbf{Q}(S)$.

To show that we have an embedding $\mathbf{Q}(j)\hookrightarrow\mathbf{R}$, consider that there is some $\tau \in \mathcal{H}$ of the form $\dfrac{1}{2}\sqrt\Delta$ or $\dfrac{ 1 + \sqrt \Delta}{2}$ such that $j(\tau)$ is a root of $H_\Delta(X)$. But then $j(\tau)$ is real, because the inverse image of the reals under the $j$-function contains the lines $\lbrace iy : y \ge 1\rbrace$ and $\lbrace 1/2 + iy : y \ge (1/2)\sqrt 3\rbrace$. Therefore we have our embedding $\mathbf{Q}(j) \hookrightarrow \mathbf{R}$.

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  • $\begingroup$ @stankewicz:Fair enough, you have an embedding $\mathbb{Q}(j)$ into $\mathbb{R}$. But could you please give more details on how you embed $\mathbb{Q}(S)$ into $\mathbb{R}$ (surely $H_{\Delta}(X)$ could have some complex roots)?. $\endgroup$
    – Adam Harris
    Commented Oct 8, 2012 at 13:25
  • $\begingroup$ @stankewicz: Isn't it true that if the class of $\mathfrak{a}$ in the ideal class group has order greater than 2 then $j(\mathfrak{a})$ isn't real? $\endgroup$
    – Adam Harris
    Commented Oct 8, 2012 at 13:33
  • $\begingroup$ I edited the answer to be more explicit. It is very possible for Hilbert Class Polynomials to have complex roots and it's easy to find examples by quickly playing around with MAGMA or sage. It just doesn't matter, because as soon as there is ONE real embedding, any subfield has to also have a real embedding. $\endgroup$
    – stankewicz
    Commented Oct 8, 2012 at 15:13
  • $\begingroup$ @stankewicz: Thanks, but I'm still confused: What is this `canonical image' of $j_i$ in the reals? If it is $j(\mathcal{O})$ where $\mathcal{O}$ is your order then what if the canonical images are equal i.e. $J_1=J_2$? i.e. how do you embed the splitting field of $H_D(X)$ in the reals if it has a complex root? $\endgroup$
    – Adam Harris
    Commented Oct 8, 2012 at 20:08
  • $\begingroup$ I don't embed the splitting field of $H_D(X)$ into the reals because you can't. Is your Hecke orbit field the union of $\mathbf{Q}(j(E))$ over the set of elliptic curves with CM by an order inside $K$, or is it the union of the Galois closures? If you mean Galois closures, the answer becomes yes and in fact the Galois closure for $E$ with CM by ANY order in $K$ will work. $\endgroup$
    – stankewicz
    Commented Oct 8, 2012 at 21:25

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