A is an abelian variety over number field K, with simple good reduction at a finite field $\kappa$, can we deduce that $A$ itself is simple?
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9$\begingroup$ Yes. Let $\mathcal{A}$ denote the Neron model over the dvr local ring at the chosen good place. The natural map ${\rm{End}}(\mathcal{A}) \rightarrow {\rm{End}}(A)$ is an equality and the natural specialization map ${\rm{End}}(\mathcal{A}) \rightarrow {\rm{End}}(\mathcal{A}_{\kappa})$ is injective (using $\ell$-adic Tate modules for $\ell \ne {\rm{char}}(\kappa)$), so we get an injective map ${\rm{End}}^0(A) \hookrightarrow {\rm{End}}^0(\mathcal{A}_{\kappa})$. The target here is a division algebra, so the source is too (as everything finite-dim'l over $\mathbf{Q}$). $\endgroup$– BCnrdCommented Nov 26, 2010 at 17:08
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$\begingroup$ Just want to make sure I understand the way to prove the injectivity correctly: Do I need to first go to endomorphism of $A[l^n]$ over formal scheme $Spf(R)$, then use formal GAGA and lifting property of etale scheme $A[l^n]$ over $R$? $\endgroup$– TJCMCommented Nov 26, 2010 at 21:23
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$\begingroup$ Formal GAGA is not needed; it is simpler. Each $\mathcal{A}[\ell^n]$ is a finite etale $R$-scheme ($R$ is base dvr), and any map between finite etale $R$-schemes is determined by its effect on a single geometric fiber. (This is true way more generally over any connected scheme, but let's not get into that.) Indeed, can assume by faithfully flat scalar extension to completion, then maximal unramified extension, then completion again, that $R$ is complete with sep. closed residue field. Then by Hensel every finite etale algebra is a product of copies of $R$. So it is "physically obvious". $\endgroup$– BCnrdCommented Nov 27, 2010 at 0:35
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Here is a slightly different proof. We have the following facts:
(1) If $B, C$ are abelian varieties over $K$, then the Néron model of $B\times C$ is the product of the Néron models (this is a simple onsequence of the universal property of Néron models). In particular, if $B\times C$ has good reduction, then $B, C$ also have good reduction, and the reduction of $B\times C$ is the product of the reductions of $B$ and $C$.
(2) If $A$ and $B$ are isogeneous abelian varieties over $K$ and if $A$ has good reduction, then $B$ has good reduction and any isogeny $A\to B$ induces an isogeny on the reductions.
Proof of (2): if $A\to B$ is an isogeny of degree $n$, then there exists an isogeny $B\to A$ such that the composition $A\to B\to A$ is the multiplication-by-$n$ map $[n]_A: A\to A$ on $A$. On the reductions we have the maps $A_k \to B_k^0 \to A_k$ whose composition is $[n]_{A_k}$. Consider a positive integer $\ell$ prime to $n$ and to $\mathrm{char}(k)$. As the restriction of $[n]_{A_k}$ to $A_k[\ell]$ is an isomorphism, $A_k[\ell]\to B^0_k[\ell]$ is injective, so $B^0_k[\ell]$ contains $(\mathbb Z/\ell \mathbb Z)^{2\dim B}$. Therefore $B_k^0$ is an abelian variety and $B$ has good reduction. Finally $A_k\to B_k=B_k^0$ is an isogeny because its kernel is contained in $A_k[n]$.
This is a special case of Néron-Ogg-Shafarevich criterion (Serre-Tate: Good reduction of abelian varieties, §1).
Now to answer the original quesiton, if $A$ was not simple, then it is isogeneous to a product of abelian varieties $B\times C$. Then $B\times C$, $B, C$ have good reduction by (2) and (1), and $A_k$ is isogeneous to $B_k\times C_k$. So $A_k$ would not be simple.
