Abelian varieties with Real multiplication :
classification and isogeny classes over finite fields

We show that $\operatorname{End}^{0}(A)=K$. Since the endomorphisms by ${\mathcal{O}}_{L}$ are defined over ${\mathbb{F}}_{q}$, we get a splitting of the $p$-divisible group and hence of its Dieuodonné module over ${\mathbb{F}}_{q}$ into rank two modules. By Lemma 3.1, we get a decomposition into ordinary and supersingular factors $\mathbb{D}_{\operatorname{ord}}\oplus\mathbb{D}_{\operatorname{ss}}$ where $\mathbb{D}_{\operatorname{ss}}$ is a product of $a$ supersingular rank. Since $p$ is completely split in ${\mathcal{O}}_{L}$, we write ${\mathcal{O}}_{L}\otimes{\mathbb{Z}}_{p}={\mathbb{Z}}_{p}^{g-a}\times{\mathbb{Z}}_{p}^{a}$, where the ${\mathbb{Z}}_{p}^{a}$ factor acts on $\mathbb{D_{\operatorname{ss}}}$. Choose a basis of idempotents $e_{1},e_{2},\dots,e_{a}$ of ${\mathbb{Z}}_{p}^{a}$. Since $e_{1},e_{2},\dots,e_{a}$ are idempotent elements, we see that ${\mathbb{D}}_{\operatorname{ss}}$ is isomorphic to the direct sum $e_{1}\cdot{\mathbb{D}}_{\operatorname{ss}}\oplus\cdots\oplus e_{a}\cdot{\mathbb{D}}_{\operatorname{ss}}$ such that the action of ${\mathbb{Z}}_{p}^{a}$ respects the direct sum. Let $f(x)$ denote the characteristic polynomial of $\pi$ over ${\mathbb{Z}}$. Since $A$ is geometrically simple, we have that $f(x)=r(x)^{m}$ where $r(x)$ is an irreducible polynomial over ${\mathbb{Q}}$. We have a decomposition of $f(x)=f_{\text{ord}}(x)f_{\operatorname{ss}}(x)$ over ${\mathbb{Z}}_{p}$ corresponding to the decomposition of the Dieudonné module. We note that $f_{\operatorname{ss}}$ is product of factors $f^{i}_{\operatorname{ss}}$ for $1\leq i\leq a$. We have a similar decomposition of $r(x)$ as well. We know that the exponent $m$ is either $1$ or $2$ since each $r^{i}_{\operatorname{ss}}$ is degree $2.$ Suppose that $f(x)=r(x)^{2}.$ Then each $r^{i}_{\operatorname{ss}}$ must be a degree $1$ polynomial. Let $\alpha$ be the root of say, $r^{1}_{\operatorname{ss}}(x).$ Then, $f^{1}_{\operatorname{ss}}(x)=(x-\alpha)^{2}$. However, this implies that the minimal polynomial is reducible over ${\mathbb{Q}}$, that is, $A_{{\mathbb{F}}_{q^{2}}}$ is not simple, which is a contradiction. This proves that $\operatorname{End}^{0}(A)$ equals its center. Therefore, $L$ is the degree $g$ totally real field of $K.$ Let $\iota$ denote the complex conjugation of $K$ over $L.$ Consider the $g-a$ ordinary primes in ${\mathcal{O}}_{L}$ over $p.$ These further split into two different primes ${\mathfrak{p}}$ and ${\mathfrak{p}}^{\prime}$ in $K$ such that ${\mathfrak{p}}^{\prime}=\iota{{\mathfrak{p}}}$. On the other hand, the supersingular primes either stay inert or ramified as they correspond to a degree $2$ extension of ${\mathbb{Q}}_{p}.$ This describes the Newton polygon of $A$ and hence gives a decomposition of the $p$-divisible group over ${\mathbb{F}}_{q}$ into ${\mathscr{G}}_{\text{\'{e}t}}\times{\mathscr{G}}_{\operatorname{tor}}\times{\mathscr{G}}_{\operatorname{ss}}$. Therefore, we get the corresponding decomposition of $S$ as well. Finally, in order to show that $S^{i}_{\operatorname{ss}}$ is maximal, we follow a similar argument given at the end of [OS20]. Without loss of any generality, we do this for $i=1.$ Suppose ${\mathscr{G}}^{\prime}$ is a $p$-divisible group isogenous to ${\mathscr{G}}^{1}_{\operatorname{ss}}$ such that $\operatorname{End}({\mathscr{G}}^{\prime})$ is maximal. Let $j:{\mathscr{G}}_{0}\rightarrow{\mathscr{G}}^{1}_{\operatorname{ss}}$ be any isogeny. We will show that $\operatorname{End}({\mathscr{G}}^{\prime})$ preserves $\ker j,$ proving that $\operatorname{End}({\mathscr{G}}^{\prime})\subset\operatorname{End}({\mathscr{G}}^{1}_{\operatorname{ss}}).$ Note that ${\mathscr{G}}_{0}$ is a one dimensional $p$-divisible group and so for each $r$, it has a unique subgroup of order $p^{r}$ corresponding to the unique quotient of order $p^{r}$ of its Dieudonné module. Therefore, $\operatorname{End}({\mathscr{G}}^{1}_{\operatorname{ss}})$ preserves the kernel of $j$ which is a unique subgroup of order $p^{r}$ in ${\mathscr{G}}_{0}$. ∎

We show that if $A=B_{1}\times B_{2}$, a product of two geometrically simple RM abelian varieties then $B_{1}$ is isogenous to $B_{2}$. Let $g_{1}$ be the dimension of $B_{1}$ and $g_{2}$ be the dimension of $B_{2}$. Without any loss of generality, we assume $g_{1}\geq g_{2}.$ Then by the definition of RM abelian varieties, $B_{1}$ has RM by a totally real field, say $L_{1}$, of degree $g_{1}$, while $B_{2}$ has RM by a totally real field $L_{2}$ of degree $g_{2}.$ Now, if $B_{1}$ were not isogenous to $B_{2}$, then $\operatorname{End}^{0}A=\operatorname{End}^{0}(B_{1}\times B_{2})=\operatorname{End}^{0}(B_{1})\times\operatorname{End}^{0}(B_{2})$. We note that $L$ is a degree $g$ field in $\operatorname{End}^{0}A$. However, $\operatorname{End}^{0}(B_{1})\times\operatorname{End}^{0}(B_{2})$, a direct sum of CM fields can at most have dimension $g_{1}$ totally real field embedded in it, which is a contradiction. ∎

We work over the algebraic closure $k$. By the previous lemma, we know that $A$ is isogenous to $B^{m}$ for some simple abelian variety $B$ that is RM by $\mathcal{O}_{L^{\prime}}$ over $k$. Let $\varphi:B^{m}\rightarrow A$ be an isogeny and let $\varphi[p^{\infty}]:B[p^{\infty}]^{m}\rightarrow A[p^{\infty}]$ be the corresponding isogeny on the $p$-divisible groups. Note that since $p$ is split in ${\mathcal{O}}_{L}$, it is completely split in ${\mathcal{O}}_{L^{\prime}}.$ Hence, we write $B[p^{\infty}]_{\operatorname{ss}}^{m}=\prod_{1\leq i\leq a}{\mathscr{G}}^{i}_{\operatorname{ss}}\times\cdots\times{\mathscr{G}}^{i}_{\operatorname{ss}}$, the $m$-fold product of ${\mathscr{G}}^{i}_{\operatorname{ss}}$, that is, the factor of the super-singular part of $B[p^{\infty}]$. We know that ${\mathbb{Z}}_{p}^{m}$ acts on $({\mathscr{G}}^{i}_{\operatorname{ss}})^{m}$ for each $1\leq i\leq a.$ As in Proposition 3.2, we can choose a basis of idempotents $e_{1},e_{2},\dots,e_{m}$ of ${\mathbb{Z}}_{p}^{m}$. Without loss of any generality consider $i=1$ and let $({\mathbb{D}}^{1})^{m}$ be the corresponding Dieudonné module of the $p$-divisible group $({\mathscr{G}}^{1}_{\operatorname{ss}})^{m}$. Since $e_{1},e_{2},\dots,e_{m}$ are idempotent elements, we see that $({\mathbb{D}}^{1})^{m}$ is isomorphic to the direct sum $e_{1}\cdot({\mathbb{D}}^{1})^{m}\oplus\cdots\oplus e_{m}\cdot({\mathbb{D}}^{1})^{m}$. This gives us an isomorphism of $({\mathscr{G}}^{1}_{\operatorname{ss}})^{m}$ with $\mathscr{H}_{1}\times\cdots\times\mathscr{H}_{m}$ such that the action of ${\mathbb{Z}}_{p}^{,}$ respects the direct sum. Here $\mathscr{H}_{j}$ is the $p$-divisible groups corresponding to the Dieudonné module $e_{j}\cdot({\mathbb{D}}^{1})$ for $1\leq j\leq m$. Now, let $G$ be the kernel of $\varphi$ such that the order of $G$ is a power of $p.$ Since the action of ${\mathbb{Z}}_{p}^{m}$ respects the direct sum, therefore, $G\simeq G_{1}\times\cdots\times G_{m}$ where $G_{j}\subset\mathscr{H}_{j}$ is of $p$-power order for each $1\leq j\leq m$. Hence, $A[p^{\infty}]\simeq\mathscr{H}_{1}/G_{1}\oplus\cdots\oplus\mathscr{H}_{m}/G_{m}.$ Since, each $\mathscr{H}_{j}/G_{j}$ is isogenous to $\mathscr{H}_{j}$ their endomorphism algebra is commutative, and therefore they lift to $W^{\prime}$, along with all their endomorphisms by Lemma 3.3.

∎

We know that for ordinary abelian varieties, there exists a Serre-Tate “canonical lift” of their $p$-divisible group to $W(k)$. We denote the corresponding lift of ${\mathscr{G}}_{\text{\'{e}t}}\times{\mathscr{G}}_{\operatorname{tor}}$ by $\widetilde{{\mathscr{G}}_{\text{\'{e}t}}}\times\widetilde{{\mathscr{G}}_{\operatorname{tor}}}.$ Since each of the super-singular factors of $A[p^{\infty}]$ has at most $2$ lifts, we see that ${\mathscr{G}}_{\operatorname{ss}}$ has at most $2^{a}$ lifts. Moreover, the Serre-Tate canonical lift is characterized by the fact that every endomorphism also lifts. This is guaranteed by the previous Lemma 3.3 for the lift of the supersingular part as well. ∎

Since the isogenies in $\mathcal{H}_{L,\mathfrak{a}}$ are compatible with the action of ${\mathcal{O}}_{L},$ we see that $G$ remains fixed by its action and therefore, it splits as $G_{\operatorname{et}}\times G_{\operatorname{tor}}\times\prod_{i=1}^{a}G_{\operatorname{ss}}^{i}$. It suffices to prove this result for each component. The result is certainly true for prime-to-$p$ subgroups of $A$. Therefore we assume that the order of $G$ is a power of $p$. We check this on the level of $p$-divisible groups. We know that the canonical lift of $A[p^{\infty}]={\mathscr{G}}_{\operatorname{\text{\'{e}t}}}\times{\mathscr{G}}_{\operatorname{tor}}\times\prod_{i=1}^{a}{\mathscr{G}}_{\operatorname{\operatorname{ss}}}$ to be the product of the canonical lifts of its étale, multiplicative and the local-local part. The étale and multiplicative subgroups lift uniquely by the definition of the canonical lift, so we are left to check that every local-local finite flat subgroup of $A$ lifts uniquely to $\widetilde{A}$. Since $G_{\operatorname{\operatorname{ss}}}\subset{\mathscr{G}}_{\operatorname{\operatorname{ss}}}$, it suffices to check that ${\mathscr{G}}_{\operatorname{\operatorname{ss}}}$ lifts to a subgroup $\widetilde{G}_{\operatorname{\operatorname{ss}}}\subset\widetilde{{\mathscr{G}}}_{\operatorname{\operatorname{ss}}}$. For each $1\leq i\leq a$, the $p$-divisible group ${\mathscr{G}}_{\operatorname{ss}}^{i}$ is a connected and $1$-dimensional group. Therefore it has a unique order $p^{r}$ subgroup for each positive integer $r$. Since each ${\mathscr{G}}_{\operatorname{ss}}$ is ramified, we see that this has to be the kernel of the endomorphism $\overline{\omega}^{r}$ where $\overline{\omega}$ is the uniformizer of ${\mathcal{O}}_{\operatorname{ss}}^{i}$. The canonical lift $\widetilde{G}_{\operatorname{ss}}^{i}$ has the property that every endomorphism of ${\mathscr{G}}_{\operatorname{ss}}^{i}$ lifts and hence it follows that $\widetilde{G}=\widetilde{G}_{\operatorname{ss}}^{i}[\overline{\omega}^{r}]$ is the required lift of $G$. ∎

We check this on the level of $p$-divisible groups. We have $\widetilde{B}[p^{\infty}]=\widetilde{{\mathscr{G}}}_{B,\operatorname{et}}\times\widetilde{{\mathscr{G}}}_{B,\operatorname{tor}}\times\prod_{i=1}^{2^{a}}\widetilde{{\mathscr{G}}}^{i}_{B,\operatorname{\operatorname{ss}}}$ where $\tilde{{\mathscr{G}}}_{B,\heartsuit}$ is the canonical lift of ${\mathscr{G}}_{B,\heartsuit}$ for $\heartsuit$ either $\operatorname{et},\operatorname{tor}$ or $\operatorname{ss}$. By Proposition 4.4, the subgroup $\widetilde{G}$ is the canonical lift of $G$ which is preserved under the action of ${\mathcal{O}}_{\operatorname{ss}}$, which implies that it acts on $\widetilde{B}[p^{\infty}]$ as well. Therefore $\widetilde{B}$ is a canonical lift of $B$.

Let $\psi:A\to B$ be an ${\mathbb{F}}_{q}$-isogeny. Denote by $G$ and $H$, the kernel of $\ker(\psi)$ and $\ker(\varphi\circ\psi)$ respectively. By Proposition 4.4, there are unique lifts of $B$ and $C$ respectively, namely $\widetilde{B}=\widetilde{A}/\widetilde{G}$ and $C=\widetilde{A}/\widetilde{H}$. Therefore we take $\widetilde{\varphi}:\widetilde{A}/\widetilde{G}\to\widetilde{A}/\widetilde{H}.$ ∎

By Proposition 4.3, the functor $\mathcal{F}$ is essentially surjective. The proof of [OS20, Proposition 3.4] shows that $\mathcal{F}$ is fully faithful as well. ∎