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I'm studying complex multiplication at the moment, and I'm completely stuck on the proof of a classical result of reduction theory.


Let $\mathcal{O}_K$ be the maximal order of an imaginary quadratic number field $K$, and let $L$ be the Hilbert Class Field of $K$. Let $p \notin \{2, 3\}$ be a prime such that $p$ splits completely in $K$. We denote by $\mathfrak{P}$ a prime $\mathcal{O}_L$-ideal lying above $p$, and $n$ is its residue degree in $L/K$.

I'm trying to prove the following : "The reduction map modulo $\mathfrak{P}$ induces a one-to-one correspondence between $j$-invariants of elliptic curves over $L$ with complex multiplication by $\mathcal{O}_K$, and $j$-invariants of elliptic curves over $\mathbb{F}_{p^n}$ whose endomorphism ring is $\mathcal{O}_K$."

While proving surjectivity is easy by Deuring lifting theorem, proving injectivity seems hard. Serge Lang, in his book "Elliptic functions", proves the result as part of theorem 13 of chapter 13,ยง4. Sadly, the proof presented there is quite incomprehensible for me and seems to involve deeper results of general reduction theory.

Taking two invariants $j(\bar{A}) = \bar{j_A}, j(\bar{B})=\bar{j_B}$ of elliptic curves over $\mathbb{F}_{p^n}$, the first part of the proof consists in proving that two such curves are isomorphic by an isomorphism $\varepsilon$ defined over $\mathbb{F}_{p^n}$. The problem is to find a way to construct an isomorphism upstairs, so that finally we get $j_A = j_B$.

Does somebody know where I could find a proof of this result, which would be, if not easy, at least more self-contained than Lang's proof ?

Thank you a lot for your answers.

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There is a nice proof using "Serre tensor construction", but needs schemes to express ideas clearly. First handle alg. closed ground fields, then push down field of def'n. Lang's book uses alg. geometry methods poorly-suited to questions involving mixed characteristic (though Deuring managed to get results with the weaker tools available to him at the time). That's why Lang's version is unpleasant. A tiny merit of scheme approach without Weierstrass equations is that all $p$ are treated on equal footing (no weirdness for 2, 3). I do not know a reference; had to figure it out for myself. – BCnrd May 22 '10 at 7:07

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