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"No" for both questions about CM elliptic curves and "I'm not even sure I dunnoknow 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 all $j$-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 $\mathbf{Q}(S) \hookrightarrow \mathbf{R}$ while mathbf{R}$. To see this, it's enough to note that for any two CM$j$-invariants$j_1$and$j_2$that there can exists an embedding$\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}$, which is absurd. Therefore, there is no such embedding for$K$.K\hookrightarrow \mathbf{Q}(S)$.

To show this, 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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"No" for both questions about CM elliptic curves and "I dunno" 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 all $j$-invariants of elliptic curves 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 $\mathbf{Q}(S) \hookrightarrow \mathbf{R}$ while there can be no such embedding for $K$.

To show this, 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 $\mathbf{Q}(j) \hookrightarrow \mathbf{R}$.