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Let $A$ be an abelian variety of dimension $g$ over $C$ with complex multiplication by a CM field $K$ where $[K:Q] =2g$. By this I mean that End($A$) $\cong \mathcal{O}_K$. Then, $A$ has a model over the Hilbert class field of the reflex field of $K$ which I denote by $H$. Let $\mathfrak{P}$ be a prime of $H$ above a rational prime which splits completely in the reflex field of $K$. Furthermore, assume that $A$ has good reduction at this prime.

My question is does the reduction of $A$ modulo $\mathfrak{P}$ have complex multiplication by $K$? If not, are there some splitting conditions one can put on the prime $\mathfrak{P}$ to ensure that the reduction has complex multiplication?

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  • $\begingroup$ Yes. The endomorphism ring injects under the reduction map, for any abelian variety having a good reduction. You may have more endomorphisms of $A_{\mathfrak{P}}$, but you definitely have at least the ones of $A$. $\endgroup$ – Vesselin Dimitrov Mar 27 '16 at 23:27
  • $\begingroup$ See the discussion for mathoverflow.net/questions/8887/… $\endgroup$ – ACL Mar 28 '16 at 12:24
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Let $B$ be a Dedekind scheme (e.g. $B$ is the spectrum of the ring of integers in a number field). Let $K$ be the function field of $B$. Let $\mathcal A$ be an abelian scheme over $B$.

It is a consequence of the theory of Neron models that $\mathcal A$ is a Neron model for its generic fibre $A := \mathcal A_K$. In particular, the canonical morphism $$\mathrm{End}_B(\mathcal A) = \mathrm{Hom}_B(A,A) \to \mathrm{Hom}_K(A,A)$$ is an isomorphism.

This shows that all endomorphisms of $A$ extend uniquely to endomorphisms of $\mathcal A$ over $B$, and thus induce endomorphisms of the fibres $\mathcal A_b$ over $k(b)$, where $b$ is a closed point of $B$, upon restricting an endomorphism of $\mathcal A$ to the fiber $\mathcal A_b$.

As Vesselin says in his comment above, for all $b$ in $B$, the reduction map $$\mathrm{End}_B(\mathcal A)\to \mathrm{End}_{k(b)}(\mathcal A_b)$$ is injective.

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