It is a fact that given $D>0$ there exist only finitely many isomorphism classes of elliptic curves over $\overline{\mathbb{Q}}$ with complex multiplication by $O_D=\mathbb{Z}[\frac{1}{2}(D+\sqrt{-D})]$ (where $D\equiv 0,3\operatorname{mod} 4$), whose $j$-invariants are all conjugate algebraic integers. Let $P_D(x)$ be the monic polynomial whose roots are the $j$-invariants. We can further consider $P_D(x)$ in characteristic $p$.
In Elkies' paper on the existence of infinitely many supersingular primes for rational every elliptic curve, he states "Since by Deuring's Lifting Lemma, complex multiplication in characteristic $p$ can be lifted to characteristic $0$, the roots of $P_D(x)$ are $j$-invariants of curves with an endomorphism $\frac{1}{2}(D+\sqrt{-D})$, that is, with complex multiplication by $O_{D'}$ for some factor $D'$ of $D$ such that $D/D'$ is a perfect square."
Deuring's lifting lemma states: If $E_0$ is an elliptic curve over $\mathbb{F}_p$ and $\alpha_0$ is a non-trivial endomorphism of $E_0$, then there exists an elliptic curve $E/\mathcal{O}_k$ for $K$ a number field, and an endomorphism $\alpha$ of $E$ and a prime $\mathfrak{p}$ of $K$ lying above $p$ with residue field $k$ such that $E_k\cong_{\overline{\mathbb{F}_p}}E_0$ and $\alpha_{\overline{\mathbb{F}_p}}=\alpha_0$.
My question is: how does the lifting lemma imply "the roots of $P_D(x)$ are $j$-invariants of curves with an endomorphism $\frac{1}{2}(D+\sqrt{-D})$, that is, with complex multiplication by $O_{D'}$ for some factor $D'$ of $D$ such that $D/D'$ is a perfect square"? How can I see this?
Best Answer
The exact quote is "the roots of $P_D(X)$ in characteristic $p$ are $j$-invariants of curves with an endomorphism $\frac{1}{2}(D+\sqrt{-D})$, that is, with complex multiplication by $\mathcal{O}_{D'}$ for some factor $D'$ of $D$ such that $D/D'$ is a perfect square." (Emphasis added; you left out the emphasized words.)
There are two parts to this claim:
"the roots of $P_D(X)$ in characteristic $p$ are $j$-invariants of curves with an endomorphism $\frac{1}{2}(D+\sqrt{-D})$": Let $\tilde{P}_D(X)$ denote $P_D(X) \bmod p$. Let $\tilde{j}_0$ denote a root of $\tilde{P}_D(X)$ over $\overline{\mathbb{F}}_p$. Let $\mathbb{F}_{p^k}$ be a finite field where $\tilde{P}_D(X)$ splits. Choose a number field $K \subset \overline{\mathbb{Q}}$ and a prime ideal $\mathfrak{p} \subset K$ lying over $p$ such that $\mathcal{O}_K/\mathfrak{p} \cong \mathbb{F}_{p^k}$ and the reduction map $\mathcal{O}_K \to \mathcal{O}_K/\mathfrak{p}$ sends $P_D(X)$ to $\tilde{P}_D(X)$. Choose a root $j_0$ of $P_D(X)$ such that the reduction map $\mathcal{O}_K \to \mathcal{O}_K/\mathfrak{p}$ sends $j_0$ to $\tilde{j}_0$. (If no such root $j_0$ exists, you may need to extend to larger fields $K$ and larger values of $k$ -- do so.) Let $E_0$ be any elliptic curve with $j(E_0) = j_0$. Then, by definition of $P_D(X)$, $E_0$ has complex multiplication by $\mathcal{O}_D$. Consider the reduction $\tilde{E}_0$ of $E_0$ to $\mathcal{O}_K/\mathfrak{p} \cong \mathbb{F}_{p^k}$, which satisfies $j(\tilde{E}_0) = \tilde{j}_0$. Observe that $\tilde{E}_0$ also has complex multiplication by (at least) $\mathcal{O}_D$ -- if $\frac{1}{2}(D+\sqrt{-D})$ acts on $E_0$ then it certainly acts on $\tilde{E}_0$, since reducing an endomorphism mod $p$ (or mod $\mathfrak{p}$) doesn't make it suddenly stop being an endomorphism. In other words, $\mathcal{O}_D \subset \operatorname{End}(\tilde{E}_0)$.
"that is, with complex multiplication by $\mathcal{O}_{D'}$ for some factor $D'$ of $D$ such that $D/D'$ is a perfect square": This second part of the claim comes from the observation that, even though $\mathcal{O}_D \subset \operatorname{End}(\tilde{E}_0)$ holds, this does not in any way guarantee that $\mathcal{O}_D$ is maximal in $\operatorname{End}(\tilde{E}_0)$. Maximal is defined in the same paper in the second sentence of the paper: "maximal in the sense that $\operatorname{Im}(\mathcal{O}_D)\otimes \mathbb{Q}$ contains only those endomorphisms already in $\operatorname{Im}(\mathcal{O}_D)$." If $\mathcal{O}_D$ is not maximal in $\operatorname{End}(\tilde{E}_0)$, then let $\mathcal{O}_{D'}$ be a maximal order in $\operatorname{End}(\tilde{E}_0)$ containing $\mathcal{O}_D$. By the theory of (non-maximal) imaginary quadratic orders, we must have $D' \mid D$, and $D/D'$ must be a perfect square in $\mathbb{Z}$.
You might ask, where is Deuring's lifting theorem used? It doesn't seem to be used above. The trick is in the wording of the sentence that you quoted, which I will quote again, with different emphasis: "the roots of $P_D(X)$ in characteristic $p$ are $j$-invariants of curves with an endomorphism $\frac{1}{2}(D+\sqrt{-D})$." The key word here is "are". The correct interpretation of this sentence is:
And now you see where Deuring's lifting theorem is used. In the above discussion, we only proved that the first set is contained inside the second set. But to show set equality, we also need to prove that the second set is contained in the first set. This last part requires Deuring's lifting theorem. Given a curve mod $p$ (or mod $\mathfrak{p}$) with an endomorphism $\frac{1}{2}(D+\sqrt{-D})$, Deuring's lifting theorem says exactly that you can lift it to characteristic $0$ while maintaining the endomorphism, so therefore its $j$-invariant lifts to a root of $P_D(X)$ in characteristic $0$.