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Isogeny classes and endomorphism algebras of abelian varieties over finite fields Yu. G. Zarhin Pennsylvania State University, Department of Mathematics, PA, USA
Англоязычная версия:
Izvestiya: Mathematics, 2023, Volume 87, Issue 3, Pages DOI: https://doi.org/10.4213/im9332e 
Реферативные базы данных:
    § 1. Introduction1.1. If $K$ is a number field, then we write $\mathrm{Cl}(K)$ for the (finite commutative) ideal class group of $K$, $\operatorname{cl}(K)$ for the class number of $K$ (i. e., the cardinality of $\mathrm{Cl}(K)$) and $\exp(K)$ for the exponent of $\mathrm{Cl}(K)$. Clearly, $\exp(K)$ divides $\operatorname{cl}(K)$. (The equality holds if and only if $\mathrm{Cl}(K)$ is cyclic, which is not always the case, see [1; Tables].) In addition, $\exp(K)$ is odd if and only if $\operatorname{cl}(K)$ is odd. We write $\mathcal{O}_K$ for the ring of integers in $K$ and $\operatorname{U}_K$ for the group of units, i. e., the multiplicative group of invertible elements in $\mathcal{O}_K$. As usual, an element of $\operatorname{U}_K$ is called a unit in $K$ or a $K$-unit. It is well known (and can be easily checked) that if a unit $u$ in $K$ is a square in $K$, then it is also a square in $\operatorname{U}_K$. Let $p$ be a prime and $q$ a positive integer that is a power of $p$. We write $\mathbb{F}_p$ for the $p$-element finite field and $\mathbb{F}_q$ for its $q$-element overfield. As usual, $\overline{\mathbb{F}}_p$ stands for an algebraic closure of $\mathbb{F}_p$, which is also an algebraic closure of $\mathbb{F}_q$. We have
$$
\begin{equation*}
\mathbb{F}_p\subset \mathbb{F}_q\subset \overline{\mathbb{F}}_p.
\end{equation*}
\notag
$$
If $X$ is an abelian variety over $\overline{\mathbb{F}}_p$, then we write $\operatorname{End}^{0}(X)$ for its endomorphism algebra $\operatorname{End}(X)\otimes\mathbb{Q}$, which is a finite-dimensional semisimple algebra over the field $\mathbb{Q}$ of rational numbers. If $X$ is defined over $k=\mathbb{F}_q$, then we write $\operatorname{End}_k(X)$ for its ring of $k$-endomorphisms and $\operatorname{End}_k^{0}(X)$ for the $\mathbb{Q}$-algebra $\operatorname{End}_k(X)\otimes\mathbb{Q}$; one may view $\operatorname{End}_k^{0}(X)$ as the $\mathbb{Q}$-subalgebra of $\operatorname{End}^{0}(X)$ with the same $1$. It is well known that isogenous abelian varieties have isomorphic endomorphism algebras and the same dimension (and $p$-adic Newton polygon). In addition, an abelian variety is simple if and only if its endomorphism algebra is a division algebra over $\mathbb{Q}$. It is also known (Grothendieck–Tate) that $\operatorname{End}^{0}(X)$ uniquely determines the dimension of $X$ [2]. Namely, $2\dim(X)$ is the maximal $\mathbb{Q}$-dimension of a semisimple commutative $\mathbb{Q}$-subalgebra of $\operatorname{End}^0(X)$. However, it turns out that there are nonisogenous abelian varieties over $\overline{\mathbb{F}}_p$ with isomorphic endomorphism algebras. The aim of this note is to provide explicit examples of such varieties. Let me start with a classical result of M. Deuring about elliptic curves [3], [4; Chapter 4]. Proposition 1.2. Let $K$ be an imaginary quadratic field. I did not find in the literature the following assertion that complements Proposition 1.2. Proposition 1.3. Let $K$ be an imaginary quadratic field and $p$ a prime that splits in $K$. Let us put $q=p^{\exp(K)}$. Then there exists an elliptic curve $E$ that is defined with all its endomorphisms over $\mathbb{F}_{q}$ and such that $\operatorname{End}^0(E) \cong K$. Remark 1.4. One may deduce from ([5; Satz 3], [6; Section 6, Corollary 1, p. 507]) that if we put $q_1=p^{\operatorname{cl}(K)}$, then there exists an elliptic curve $E$ that is defined with all its endomorphisms over $\mathbb{F}_{q_1}$ and such that $\operatorname{End}(E) \cong \mathcal{O}_K$ (and therefore $\operatorname{End}^0(E) \cong K$). The next result is an analogue of Proposition 1.2 for abelian surfaces and quartic fields. Proposition 1.5. Let $K$ be a $\mathrm{CM}$ quartic field that is a cyclic extension of $\mathbb{Q}$. Our main result is the following assertion. Theorem 1.6. Let $n$ be a positive integer and $K$ be a $\mathrm{CM}$ field that is a cyclic degree $2^n$ extension of $\mathbb{Q}$. Let $K_0$ be the only degree $2^{n-1}$ subfield of $K$, which is the maximal totally real subfield of $K$. Let us put $c=\exp(K)$. (i) Let $p$ be a prime and $A$ an abelian variety over $\overline{\mathbb{F}}_p$ such that $\operatorname{End}^0(A)$ is isomorphic to $K$. Then $p$ splits completely in $K$ and $A$ is an ordinary simple abelian variety of dimension $2^{n-1}$. (ii) Let $p$ be a prime that splits completely in $K$. Let us put $q=p^c$.
Remark 1.7. (a) If $n=1$, then $K$ is an imaginary quadratic field and therefore $K_0=\mathbb{Q}$ and $\operatorname{U}_{\mathbb{Q}}=\{\pm 1\}$. The only (totally) positive unit in $\mathbb{Q}$ is $1$, which is obviously a square in $\mathbb{Q}$. Hence, Propositions 1.2 and 1.3 are the special cases of Theorem 1.6 with $n=1$. On the other hand, Proposition 1.5 follows readily from the special case of Theorem 1.6 with $n=2$. (b) If $n \geqslant 3$, then the number $2^{2^{n-1}-n}$ of the corresponding isogeny classes is strictly greater than $1$. This gives us examples of nonisogenous abelian varieties over $\overline{\mathbb{F}}_p$, whose endomorphism algebras are isomorphic to $K$ and therefore are mutually isomorphic. (c) Now let $n$ be an arbitrary positive integer. By Chebotarev’s density theorem, the set of primes that split completely in $K$ is infinite (and even has a positive density $1/2^n$). Corollary 1.8. Let $r$ be a Fermat prime (e. g., $r=3,5,17,257, 65537$). Let $p$ be a prime that is congruent to $1$ modulo $r$. Let us put Then there are precisely $\operatorname{isg}(r)$ isogeny classes of simple ($(r-1)/2$)-dimensional ordinary abelian varieties $A$ over $\overline{\mathbb{F}}_p$, whose endomorphism algebra is isomorphic to the $r$th cyclotomic field $\mathbb{Q}(\zeta_r)$. In addition, if $c=\exp(\mathbb{Q}(\zeta_r))$ and $q=p^c$, then each of these isogeny classes contains an abelian variety that is defined with all its endomorphisms over $\mathbb{F}_{q}$. Remark 1.9. The congruence condition on $p$ means that $p$ splits completely in $\mathbb{Q}(\zeta_r)$. There are infinitely many such $p$ thanks to Dirichlet’s theorem about primes in an arithmetic progression. More precisely, the set of such primes has density $1/(r-1)$. Remark 1.10. It is well known that the property of being simple (respectively, ordinary) is invariant under isogenies. Remark 1.11. Let $r$ be a Fermat prime. Clearly, $\operatorname{isg}(r)=1$ if and only if $r \leqslant 5$. Let $p$ be a prime that is congruent to $1\bmod r$. It follows from Theorem 1.6 that $r \leqslant 5$ if and only if there is a precisely one isogeny class of simple ordinary ($(r-1)/2$)-dimensional abelelian varieties over $\overline{\mathbb{F}}_p$, whose endomorphiam algebra is isomorphic to $\mathbb{Q}(\zeta_r)$. In other words, all such abelian varieties are mutually isogenous over $\overline{\mathbb{F}}_p$, if and only if $r \in \{3,5\}$. Example 1.12. (i) Take $r=3$. We have $\operatorname{isg}(3)=1$. It follows from Remark 1.11 that if $p \equiv 1 \bmod 3$, then all ordinary elliptic curves over $\overline{\mathbb{F}}_p$ with endomorphism algebra $\mathbb{Q}(\zeta_3)$ are isogenous. (This assertion seems to be well known.) This implies that each such elliptic curve is isogenous over $\overline{\mathbb{F}}_p$ to $y^2=x^3-1$. (ii) Take $r=5$. We have $\operatorname{isg}(5)=1$. It follows from Remark 1.11 that if $p \equiv 1 \bmod 5$, then all abelian varieties over $\overline{\mathbb{F}}_p$ with endomorphism algebra $\mathbb{Q}(\zeta_5)$ are two-dimensional simple ordinary and mutually isogenous. This implies that each such abelian variety is isogenous to the jacobian of the genus $2$ curve $y^2=x^5-1$. Example 1.13. Let us take $r=17$. Then $\operatorname{cl}(\mathbb{Q}(\zeta_{17}))=1$ [7]. Let us choose a prime $p$ that is congruent to $1$ modulo $17$ (e. g., $p=103$). We have By Theorem 1.6, there are precisely $16$ isogeny classes of simple ordinary $\frac{16}{2}=8$-dimensional abelian varieties over $\overline{\mathbb{F}}_p$ with endomorphism algebras $\mathbb{Q}(\zeta_{17})$. In addition, each of these isogeny classes contains an abelian eightfold that is defined with all its endomorphisms over $\mathbb{F}_{p}$. This implies that there exist sixteen $8$-dimensional ordinary simple abelian varieties $A_1, \dots, A_{16}$ over $\overline{\mathbb{F}}_p$ that are mutually non-isogenous but each endomorphism algebra $\operatorname{End}^0(A_i)$ is isomorphic to $\mathbb{Q}(\zeta_{17})$ (for all $i$ with $1 \leqslant i \leqslant 16$). In particular,
$$
\begin{equation*}
\operatorname{End}^0(A_i) \cong \operatorname{End}^0(A_j) \quad \forall\, i,j,\qquad 1 \leqslant i<j \leqslant 16.
\end{equation*}
\notag
$$
In addition, each $A_i$ and all its endomorphisms are defined over $\mathbb{F}_{p}$. This gives an answer to a question of L. Watson [8]. The following assertion is a natural generalization of Corollary 1.8. Corollary 1.14. Let $r$ be an odd prime and $r-1=2^n \cdot m$ where $n$ is a positive integer and $m$ is a positive odd integer. Let $\mathbf{H}$ be the only order $m$ subgroup of the cyclic Galois group of order $r-1$. Let be the subfield of $\mathbf{H}$-invariants in $\mathbb{Q}(\zeta_r)$. Then
Remark 1.15. Let $K=K^{(r)}$. It is well known that $r$ is totally ramified in $\mathbb{Q}(\zeta_r)$ and therefore in its subfield $K$ as well. This implies that if $K_0$ is the only degree $2^{n-1}$ subfield of $K$, which is the maximal totally real subfield of $K$, then the quadratic extension $K/K_0$ is ramified. On the other hand, it is known ([9; Section 38], [10; Section 13, pp. 77, 78]) that $\operatorname{cl}(K^{(r)})$ is odd (and therefore $c=\exp(K^{(r)})$ is also odd). It follows from [9; Section 37, Satz 42] (see also [10; Corollary 13.10, p. 76]) that $K_0$ has units with independent signs (there are units of $K_0$ of every possible signature), which implies (thanks to [10; Lemma 12.2, p. 55]) that every totally positive unit in $K_0$ is a square in $K_0$ and therefore is a square in $\operatorname{U}_{K_0}$. Example 1.16. Let us fix an integer $n \geqslant 2$. Here is a construction of infinitely many mutually nonisomorphic $\mathrm{CM}$ fields that are cyclic degree $2^n$ extensions of $\mathbb{Q}$. Let us consider the infinite (thanks to Dirichlet’s theorem) set of primes $r$ that are congruent to $1+2^n$ modulo $2^{n+1}$. Then $r-1=2^n \cdot m$, where $m$ is an odd positive integer. In light of Corollary 1.14, the corresponding subfield $K^{(r)}$ of $\mathbb{Q}(\zeta_r)$ defined by (2) enjoys the desired properties. Since $K^{(r)}$ is a subfield of $\mathbb{Q}(\zeta_r)$, the field extension $K^{(r)}/\mathbb{Q}$ is ramified precisely at $r$. This implies that the fields $K^{(r)}$ are mutually nonisomorphic (and even linearly disjoint) for distinct $r$. The paper is organized as follows. In Section 2 we review basic results about maximal ideals of $\mathcal{O}_K$. In Section 3 we concentrate on the so called ordinary Weil’s $q$-numbers in $K$. In Section 4 we discuss simple abelian varieties over $\mathbb{F}_q$, whose Weil’s numbers lie in $K$. In Section 5 we discuss some basic facts of Honda–Tate theory [11]–[13]. Section 6 contains proofs of main results. In what follows, we will freely use the following elementary well-known observation. Any $\mathbb{Q}$-subalgebra with $1$ of a number field $K$ is actually a subfield of $K$; in particular, it is also a number field. E. g., if $u$ is an element of $L$, then the subfield $\mathbb{Q}(u)$ generated by $u$ coincides with the $\mathbb{Q}$-subalgebra $\mathbb{Q}[u]$ generated by $u$. § 2. Preliminaries2.1. We keep the notation and assumptions of Subsection 1.1 and Theorem 1.6. As usual, $\mathbb{Q}$, $\mathbb{R}$, $\mathbb{C}$ stand for the fields of rational, real and complex numbers and $\overline{\mathbb{Q}}$ for the (algebraically closed) subfield of all algebraic numbers in $\mathbb{C}$. We write $\mathbb{Z}$ (respectively, $\mathbb{Z}_+$) for the ring of integers (respectively, for the additive semigroup of nonnegative integers). If $T$ is a finite set, then we write $\#(T)$ for the number of its elements. Recall [12], [13] that an algebraic integer $\pi\in \overline{\mathbb{Q}}$ is called a Weil’s $q$-number if all its Galois-conjugates have the archimedean absolute value $\sqrt{q}$. Throughout this paper, $n$ is a positive integer and $K$ is a $\mathrm{CM}$ field that is a degree $2^n$ cyclic extension of $\mathbb{Q}$. We view $K$ as a subfield of $\mathbb{C}$; in particular, $K$ is a subfield of $\overline{\mathbb{Q}}$ that is stable under the complex conjugation. We denote by the restriction of the complex conjugation to $K$; one may view $\rho$ as an element of order $2$ in the Galois group where $G$ is a cyclic group of order $2^n$. Remark 2.2. Let $\pi \in K\subset \mathbb{C}$. We write $W(q,K)$ for the set of Weil’s $q$-numbers in $K$ and $\mu_K$ for the (finite cyclic) multiplicative group of roots of unity in $K$. Clearly, $W(q,K)$ is a finite $G$-stable subset of $\mathcal{O}_K$, which is also stable under multiplication by elements of $\mu_K$. The latter gives rise to the free action of $\mu_K$ on $W(q,K)$ defined by
$$
\begin{equation*}
\mu_K \times W(q,K) \to W(q,K), \qquad \zeta, \pi \mapsto \zeta \pi \quad \forall\, \zeta\in \mu_K,\quad \pi \in W(q,K).
\end{equation*}
\notag
$$
Remark 2.3. It is well known (and follows easily from a theorem of Kronecker [14; Chapter IV, Section 4, Theorem 8]) that $\pi_1,\pi_2 \in W(q,K)$ lie in the same $\mu_K$-orbit (i. e., $\pi_2/\pi_1$ is a root of unity) if and only if the ideals $\pi_1\mathcal{O}_K$ and $\pi_2\mathcal{O}_K$ of $\mathcal{O}_K$ coincide. Recall (Subsection 2.1) that $K$ is a subfield of the field $\mathbb{C}$ of complex numbers that is stable under the complex conjugation. Then is a (maximal) totally real number (sub)field, whose degree $[K_0:\mathbb{Q}]$ is2.4. Recall that the Galois group $G=\operatorname{Gal}(K/\mathbb{Q})$ is a cyclic group of order $2^n$. Hence, it has precisely one element of order $2$ and therefore this element must coincide with the complex conjugation The properties of $G$ imply that every nontrivial subgroup $H$ of $G$ contains $\rho$. It follows that every proper subfield of $K$ is totally real. Indeed, each such subfield is the subfield $K ^H$ of $H$-invariants for a certain nontrivial subgroup $H$ of $G$. Since $H$ contains $\rho$, the subfield $K^H$ consists of $\rho$-invaraiants and therefore is totally real; in particular, 2.5. Let $\ell$ be a prime and $S(\ell)$ be the set of maximal ideals $\mathfrak{P}$ of $\mathcal{O}_K$ that divide $\ell$. Since $K/\mathbb{Q}$ is a Galois extension, $G$ acts transitively on $S(\ell)$. In particular, $\#(S(\ell))$ divides $\#(G)=2^n$. This implies that if $\ell$ splits completely in $K$, i. e., then the action of $G$ on $S(\ell)$ is free. On the other hand, if a prime $\ell$ does not split completely in $K$, i. e., then the action of $G$ on $S(\ell)$ is not free. Let $H(\ell)$ be the stabilizer of any $\mathfrak{P}\in S(\ell)$, which does not depend on a choice of $\mathfrak{P}$, because $G$ is commutative. Then $H(\ell)$ is a nontrivial subgroup of $G$ and therefore contains $\rho$, i. e.,
$$
\begin{equation*}
\rho(\mathfrak{P})=\mathfrak{P} \quad \forall\, \mathfrak{P} \in S(\ell)
\end{equation*}
\notag
$$
if $\ell$ does not split completely in $K$. Let $e(\ell)$ be the ramification index in $K/\mathbb{Q}$ of $\mathfrak{P}\in S(\ell)$, which does not depend on $\mathfrak{P}$, because $K/\mathbb{Q}$ is Galois. We have the equality of ideals
$$
\begin{equation}
\ell \mathcal{O}_K=\prod_{\mathfrak{P}\in S(\ell)}\mathfrak{P}^{e(\ell)}.
\end{equation}
\tag{4}
$$
It follows that $K/\mathbb{Q}$ is unramified at $\ell$ if and only if $e(\ell)=1$. We write
$$
\begin{equation}
\operatorname{ord}_{\mathfrak{P}}\colon K^* \twoheadrightarrow \mathbb{Z}
\end{equation}
\tag{5}
$$
for the discrete valuation map attached to $\mathfrak{P}$. We have
$$
\begin{equation}
\operatorname{ord}_{\mathfrak{P}}(\ell)=e(\ell) \quad \forall\, \mathfrak{P}\in S(\ell),
\end{equation}
\tag{6}
$$
2.6. Let $p$ be a prime, $j$ a positive integer, and $q=p^j$. Let $\pi \in O_K$ be a Weil’s $q=p^j$-number. Let us consider the ideal $\pi\mathcal{O}_K$ in $\mathcal{O}_K$. Then there is a nonnegative integer-valued function
$$
\begin{equation}
d_{\pi}\colon S(p) \to \mathbb{Z}_+, \qquad \mathfrak{P} \mapsto d_{\pi}(\mathfrak{P}):=\operatorname{ord}_{\mathfrak{P}}(\pi)
\end{equation}
\tag{9}
$$
such that
$$
\begin{equation}
\pi\mathcal{O}_K=\prod_{\mathfrak{P}\in S(p)}\mathfrak{P}^{d_{\pi}(\mathfrak{P})}.
\end{equation}
\tag{10}
$$
It follows from (3) that Lemma 2.7. Let $\pi \in O_K$ be a Weil’s $q=p^j$-number. If $p$ does not split completely in $K$, then $\pi^2/q$ is a root of unity. Proof. Since $p$ does not split completely in $K$, it follows from arguments of Subsection 2.4 that
$$
\begin{equation*}
\rho(\mathfrak{P})=\mathfrak{P} \quad \forall\, \mathfrak{P}\in S(p).
\end{equation*}
\notag
$$
It follows from (11) that
$$
\begin{equation*}
d_{\pi}(\mathfrak{P})=\frac{j\cdot e(p)}{2} \quad \forall\, \mathfrak{P}\in S(p);
\end{equation*}
\notag
$$
in particular, $j$ is even if $e(p)=1$ (i. e., if $K/\mathbb{Q}$ is unramified at $p$). This implies that $\pi^2/q$ is a $\mathfrak{P}$-adic unit for all $\mathfrak{P}\in S(p)$. On the other hand, it follows from (3) that $\pi^2/q$ is an $\ell$-adic unit for all primes $\ell \ne p$. It follows from the very definition of Weil’s numbers that
$$
\begin{equation*}
\biggl|\sigma\biggl(\frac{\pi^2}{q}\biggr)\biggr|_{\infty}=1 \quad \forall\, \sigma \in G.
\end{equation*}
\notag
$$
(Here $|\,{\cdot}\,|_{\infty}\colon \mathbb{C} \to \mathbb{R}_+$ is the standard archimedean value on $\mathbb{C}$.) Now it follows from a classical theorem of Kronecker [14; Chapter IV, Section 4, Theorem 8] that $\pi^2/q$ is a root of unity. $\Box$ Lemma 2.8. Suppose that a prime $p$ completely splits in $K$. (In particular, $K/\mathbb{Q}$ is unramified at $p$.) Let $\pi \in O_K$ be a Weil’s $q=p^j$-number. Then either $\mathbb{Q}(\pi)=K$ or $j$ is even and $\pi=\pm p^{j/2}$ . Proof. So, $K/\mathbb{Q}$ is unramified at $p$, i. e., $e(p)=1$ and
$$
\begin{equation}
p\mathcal{O}_K=\prod_{\mathfrak{P}\in S(p)}\mathfrak{P}.
\end{equation}
\tag{12}
$$
This implies that
$$
\begin{equation}
q\mathcal{O}_K=\prod_{\mathfrak{P}\in S(p)}\mathfrak{P}^{j}.
\end{equation}
\tag{13}
$$
Since $p$ splits completely in $K$, the group $G$ acts freely on $S(p)$, in light of Subsection 2.5. In particular,
If the subfield $\mathbb{Q}(\pi)$ of $K$ does not coincide with $K$, then it is totally real, thanks to arguments of Subsection 2.4. This implies that $\rho(\pi)=\pi$. It follows from (3) that $\pi^2=q$, i. e., $\pi=\pm p^{j/2}$. This implies that the ideal $q\mathcal{O}_K$ is a square. It follows from (13) that $j$ is even. $\Box$ Definition 2.10. Let $\pi \in O_K$ be a Weil’s $q=p^j$-number. We say that $\pi$ is an ordinary Weil’s $q$-number if the “slope” $\operatorname{ord}_{\mathfrak{P}}(\alpha)/\operatorname{ord}_{\mathfrak{P}}(q)$ is an integer for all $\mathfrak{P}\in S(p)$. It (is well known and) follows from (3), (7) and (8) that if $\pi$ is an ordinary Weil’s $q$-number, then
$$
\begin{equation}
\frac{\operatorname{ord}_{\mathfrak{P}}(\pi)}{\operatorname{ord}_{\mathfrak{P}}(q)}=0 \quad \text{or}\quad 1.
\end{equation}
\tag{15}
$$
§ 3. Equivalence classes of ordinary Weil’s $q$-numbersLet $p$ be a prime that splits completely in $K$. Throughout this section, by Weil’s numbers we mean Weil’s $q$-numbers where $q$ is a power of $p$. We write $W(q,K)$ for the set of Weil’s $q$-numbers in $K$. We write $\mu_K$ for the (finite cyclic) multiplicative group of roots of unity in $K$. Definition 3.1. Let $q$ and $q'$ be integers $>1$ that are integral powers of $p$. Let $\pi \in K$ (respectively, $\pi'\in K$) be a Weil’s $q$-number (respectively, Weil’s $q'$-number). Following Honda [12], we say that $\pi$ and $\pi'$ are equivalent, if there are positive integers $a$ and $b$ such that $\pi^a$ is Galois-conjugate to ${\pi'}^b$. Clearly, if $\pi$ and $\pi'$ are equivalent, then $\pi$ is ordinary if and only if $\pi'$ is ordinary. In order to classify ordinary Weil’s numbers in $K$ up to equivalence, we introduce the following notion that is inspired by the notion of $\mathrm{CM}$ type for complex abelian varieties [15] (see also [12; Section 1, Theorem 2] and [13; Section 5]). Definition 3.2. We call a subset $\Phi\subset S(p)$ a $p$-type if $S(p)$ is a disjoint union of $\Phi$ and $\rho(\Phi)$. Clearly, $\Phi\subset S(p)$ is a $p$-type if and only if the following two conditions hold (recall that $[K:\mathbb{Q}]=2^n$). (i) $\#(\Phi)=2^{n-1}$. (ii) If $\mathfrak{P} \in \Phi$, then $\rho(\mathfrak{P}) \notin \Phi$. It is also clear that $\Phi\subset S(p)$ is a $p$-type if and only if $\rho(\Phi)$ is a $p$-type. Let $H(p)$ be the set of nonzero ideals $\mathfrak{B}$ of $\mathcal{O}_K$ such that
$$
\begin{equation*}
\mathfrak{B}\cdot \rho(\mathfrak{B})=p \cdot \mathcal{O}_K.
\end{equation*}
\notag
$$
In light of (12) and (14), an ideal $\mathfrak{B}$ of $\mathcal{O}_K$ lies in $H(p)$ if and only if there exists a $2^{n-1}$-element subset $\Phi=\Phi(\mathfrak{B})$ of $H(p)$ that meets every $\rho$-orbit of $S(p)$ at exactly one place and
$$
\begin{equation}
\mathfrak{B}=\prod_{\mathfrak{P}\in \Phi(\mathfrak{B})}\mathfrak{P}.
\end{equation}
\tag{16}
$$
Such a $\Phi(\mathfrak{B})$ is uniquely determined by $\mathfrak{B}\in H(p)$: namely, it coincides with the set of maximal ideals in $\mathcal{O}_K$ that contain $\mathfrak{B}$. This implies that Clearly,
$$
\begin{equation}
\Phi(\sigma(\mathfrak{B}))=\sigma(\Phi(\mathfrak{B})) \quad \forall \, \sigma \in G.
\end{equation}
\tag{18}
$$
Lemma 3.3. Let $m$ be a positive integer and $\pi$ be a Weil’s $q=p^m$-number in $K$. Then the following conditions are equivalent: If these equivalent conditions hold then such an ideal $\mathfrak{B}$ is unique and Proof. We have
$$
\begin{equation}
\pi\mathcal{O}_K=\prod_{\mathfrak{P}\in S(p)}\mathfrak{P}^{d(\mathfrak{P})},
\end{equation}
\tag{21}
$$
for some $d(\mathfrak{P}) \in \mathbb{Z}_+$ such that
$$
\begin{equation}
\frac{\operatorname{ord}_{\mathfrak{P}}(\pi)}{\operatorname{ord}_{\mathfrak{P}}(q)} =\frac{d(\mathfrak{P})}{m} \quad \forall \, \mathfrak{P}\in S(p).
\end{equation}
\tag{23}
$$
This implies that
$$
\begin{equation}
\Psi(\pi):=\{\mathfrak{P}\in S(p)\mid d(\mathfrak{P})=m\}\subset S(p).
\end{equation}
\tag{24}
$$
Combining (24) with (22), we obtain that
$$
\begin{equation}
\rho(\Psi(\pi)):=\{\mathfrak{P}\in S(p)\mid d(\mathfrak{P})=0\}=\biggl\{\mathfrak{P}\in S(p)\biggm| \frac{\operatorname{ord}_{\mathfrak{P}}(\pi)}{\operatorname{ord}_{\mathfrak{P}}(q)}=0 \biggr\}\subset S(p);
\end{equation}
\tag{25}
$$
in particular, the subsets $\Psi(\pi)$ and $\rho(\Psi(\pi))$ do not meet each other. In light of (20) and (25) combined with (15), $\pi$ is ordinary if and only if $S(p)$ is a disjoint union of $\Psi(\pi)$ and $\rho(\Psi(\pi))$, i. e., $\Psi(\pi)$ is a $p$-type. This proves the equivalence of (i) and (iii). If (i) and (iii) hold, then it follows from (21) that Since $\Psi(\pi)$ is a $p$-type, $\mathfrak{B}\in H(p)$ and, obviously, $\Phi(\mathfrak{B})=\Psi(\pi)$. This proves that equivalent (i) and (iii) imply (ii).
Let us assume that (ii) holds. This means that there is $\mathfrak{B} \in H(p)$ that satisfies (19). This implies that
$$
\begin{equation*}
\mathfrak{B} =\prod_{\mathfrak{P}\in \Phi(\mathfrak{B})}\mathfrak{P}, \qquad \pi\mathcal{O}_K=\mathfrak{B}^m=\prod_{\mathfrak{P}\in \Phi(\mathfrak{B})}\mathfrak{P}^m.
\end{equation*}
\notag
$$
It follows that
$$
\begin{equation*}
\begin{alignedat}{2} \frac{\operatorname{ord}_{\mathfrak{P}}(\pi)}{\operatorname{ord}_{\mathfrak{P}}(q)} &=1 &\quad \forall \, \mathfrak{P} &\in \Phi(\mathfrak{B}), \\ \frac{\operatorname{ord}_{\mathfrak{P}}(\pi)}{\operatorname{ord}_{\mathfrak{P}}(q)}&=0 &\quad \forall \, \mathfrak{P}&\notin \Phi(\mathfrak{B}). \end{alignedat}
\end{equation*}
\notag
$$
This implies that $\pi$ is ordinary and therefore (ii) implies (i). So, we have proved the equivalence of (i), (ii), (iii). The uniqueness of such $\mathfrak{B} $ is obvious. $\Box$ Lemma 3.4. The natural action of $G$ on $H(p)$ is free. In particular, $H(p)$ partitions into a disjoint union of $2^{2^{n-1}-n}$ orbits of $G$, each of which consists of $2^n$ elements. Proof. Suppose that there exists $\mathfrak{B}\in H(p)$ such that its stabilizer
$$
\begin{equation*}
G_{\mathfrak{B}}=\{\sigma \in G\mid \sigma(\mathfrak{B})=\mathfrak{B}\}
\end{equation*}
\notag
$$
is a nontrivial subgroup. Then $G_{\mathfrak{B}}$ must contain $\rho$, thanks to the arguments of Subsection 2.4. This means that $\rho(\mathfrak{B})=\mathfrak{B}$ and therefore which is not true, since $p$ is unramified in $K$. The obtained contradiction proves that the action of $G$ on $H(p)$ is free. Hence, every $G$-orbit in $H(p)$ consists of $\#(G)=2^n$ elements and the number of such orbits is Lemma is proved. $\Box$ In what follows we define (noncanonically) certain $G$-equivariant injective maps $\mathcal{Z}$, $\Pi$ and $\Pi_1$ from $H(p)$ to $K$; they will play a crucial role in the classification of ordinary Weil’s numbers in $K$ up to equivalence. Corollary 3.5. Let $c=\exp(K)$. Then there exists a $G$-equivariant map such that $\mathcal{Z}(\mathfrak{B})$ is a generator of $\mathfrak{B}^c$ for all $\mathfrak{B}\in H(p)$. Proof. We define $\mathcal{Z}$ separately for each $G$-orbit $O\subset H(p)$. Pick $\mathfrak{B}_O \in O$ and choose a generator $z_O$ of the principal ideal $\mathfrak{B}_O^c$. In light of Lemma 3.4, if $\mathfrak{B} \in O$, then there is precisely one $\sigma \in G$ such that $\mathfrak{B}=\sigma(\mathfrak{B}_O)$. This implies that i. e., $\sigma(z_O)$ is a generator of $\mathfrak{B}^c$. It remains to put Corollary is proved. $\Box$ Theorem 3.6. Let us put Let $K_0=K^{\rho}$ be the maximal totally real subfield of $K$. There exists an injective map
$$
\begin{equation}
\Pi\colon H(p) \hookrightarrow W(q^2,K), \qquad \mathfrak{B} \mapsto \Pi(\mathfrak{B})
\end{equation}
\tag{27}
$$
that enjoys the following properties. (0) $\Pi$ is $G$-equivariant, i. e.,
$$
\begin{equation*}
\Pi(\sigma(\mathfrak{B}))=\sigma(\Pi(\mathfrak{B})) \quad \forall \, \sigma\in G, \quad \mathfrak{B}\in H(p).
\end{equation*}
\notag
$$
(i) For all $\mathfrak{B} \in H(p)$ the ideal $\Pi(\mathfrak{B})\mathcal{O}_K$ coincides with $\mathfrak{B}^{2 c}$. (ii) The image $\Pi(H(p))$ consists of ordinary Weil’s $q^2$-numbers. (iii) If $\pi'$ is an ordinary Weil’s $p^m$-number in $K$, then there exists precisely one $ \mathfrak{B} \in H(p)$ such that the ratio $(\pi')^{2c}/\Pi(\mathfrak{B})^m$ is a root of unity. (iv) Let $\mathfrak{B}_1, \mathfrak{B}_2 \in H(p)$. Then Weil’s $q^2$-numbers $\Pi(\mathfrak{B}_1)$ and $\Pi(\mathfrak{B}_2)$ are equivalent if and only if $\mathfrak{B}_1$ and $\mathfrak{B}_2$ lie in the same $G$-orbit. (v) If $h$ is a positive integer, then the subfield $\mathbb{Q}(\Pi(\mathfrak{B})^h)$ of $K$ generated by $\Pi(\mathfrak{B})^h$ coincides with $K$. (vi) Suppose that every totally positive unit in $\operatorname{U}_{K_0}$ is a square in $K_0$ (and therefore in $\operatorname{U}_{K_0}$). Then there exists a map that enjoys the following properties:
Proof. Let us choose $\mathcal{Z}\colon H(p) \to \mathcal{O}_E \setminus\{0\}$ that enjoys the properties described in Corollary 3.5. Let $\mathfrak{B}\in H(p)$. In order to define $\Pi(\mathfrak{B})$, notice that
$$
\begin{equation*}
\mathfrak{B}\cdot \rho(\mathfrak{B})=p\mathcal{O}_K, \qquad \mathfrak{B}^c=z \mathcal{O}_K,
\end{equation*}
\notag
$$
where
$$
\begin{equation}
z=\mathcal{Z}(\mathfrak{B}) \in \mathcal{O}_K\setminus \{0\}.
\end{equation}
\tag{28}
$$
Then $z\rho(z)$ is a generator of the ideal
$$
\begin{equation*}
\mathfrak{B}^c\cdot \rho(\mathfrak{B}^c)=\left(\mathfrak{B}\cdot \rho(\mathfrak{B})\right)^c=p^c\cdot\mathcal{O}_K=q\mathcal{O}_K.
\end{equation*}
\notag
$$
Since $\rho$ is the complex conjugation, $z\rho(z)$ is a real (i. e., $\rho$-invariant) totally positive element of $\mathcal{O}_K$. Clearly, is an invertible element of $\mathcal{O}_K$ that is also $\rho$-invariant and totally positive unit in $\operatorname{U}_{K_0}$. Obviously, Now let us put
$$
\begin{equation}
\Pi(\mathfrak{B}):=q \cdot \frac{z}{\rho(z)}=\frac{z^2}{z\rho(z)/q}=\frac{z^2}{u} \in \mathcal{O}_K.
\end{equation}
\tag{29}
$$
If $u$ is a square in $K_0$, then there is a unit $u_0$ in $K_0$ such that $u=u_0^2$. If this is the case, then let us put
$$
\begin{equation}
\Pi_0(\mathfrak{B}):=\frac{z}{u_0}\in \mathcal{O}_K \quad \text{and get } \Pi_0(\mathfrak{B})^2=\biggl(\frac{z}{u_0}\biggr)^2=\frac{z^2}{u}=\Pi(\mathfrak{B}).
\end{equation}
\tag{30}
$$
Clearly,
$$
\begin{equation}
\Pi(\mathfrak{B})\cdot \mathcal{O}_K=z^2\cdot \mathcal{O}_K =(z\cdot\mathcal{O}_K)^2=(\mathfrak{B}^{c})^2=\mathfrak{B}^{2c},
\end{equation}
\tag{31}
$$
which proves (i). In order to check that $\Pi(\mathfrak{B})$ is a Weil’s $q^2$-number, notice that
$$
\begin{equation*}
\Pi(\mathfrak{B})\cdot \rho(\Pi(\mathfrak{B}))=q \cdot \frac{z}{\rho(z)}\cdot \rho\biggl(q \cdot \frac{z}{\rho(z)}\biggr)=q^2\cdot \frac{z}{\rho(z)} \cdot \frac{\rho(z)}{z}=q^2.
\end{equation*}
\notag
$$
In light of Remark 2.2, this proves that $\Pi(\mathfrak{B})$ is a Weil’s $q^2$-number. It follows from (30) that if $\Pi_0(\mathfrak{B})$ is defined, then it is a Weil’s $q$-number. By construction, which also implies that $\Pi(\mathfrak{B})$ is $p^{2c}=q^2$-ordinary Weil’s number. The $G$-invariance of $\mathcal{Z}$ (see Corollary 3.5) combined with (28) and (29) implies the $G$-equivariance of $\Pi$, which proves (0). The injectiveness of $\Pi$ follows from (31). This proves (i) and (ii).
In order to prove (v), notice that if $\mathbb{Q}(\Pi(\mathfrak{B})^h)$ does not coincide with $K$, then it consists of $\rho$-invariants (Subsection 2.4). In particular, the ideal $\Pi(\mathfrak{B})^h\mathcal{O}_K=\mathfrak{B}^{2ch}$ coincides with its complex-conjugate
$$
\begin{equation*}
\rho\bigl(\Pi(\mathfrak{B})^h\mathcal{O}_K\bigr)=\rho(\mathfrak{B}^{2ch}) =\rho(\mathfrak{B})^{2ch}.
\end{equation*}
\notag
$$
This implies that $\mathfrak{B}=\rho(\mathfrak{B})$, which is not the case, since $\mathfrak{B}\in H(p)$. The obtained contradiction proves (v).
In order to prove (iii), we need to check that if $\pi'$ is an ordinary Weil’s $p^m$-number in $K$, then it is equivalent to $\Pi(\mathfrak{B})$ for some $\mathfrak{B}\in H(p)$. In order to do that, let us consider the ideal $\mathfrak{M}:=\pi'\mathcal{O}_K$ in $\mathcal{O}_K$. Since $\pi'\cdot\rho(\pi')=p^m$, we get $\mathfrak{M}\cdot \rho(\mathfrak{M})=p^m \mathcal{O}_K$. It follows that
$$
\begin{equation*}
\mathfrak{M}=\prod_{\mathfrak{P}\in S(p)}\mathfrak{P}^{d(\mathfrak{P})}, \qquad d(\mathfrak{P})+d(\mathfrak{\rho(P)})=m \quad \forall \, \mathfrak{P}\in S(p).
\end{equation*}
\notag
$$
The ordinarity of $\mathfrak{M}$ implies that
$$
\begin{equation*}
d(\mathfrak{P})=0\text{ or } m \quad \forall \, \mathfrak{P}\in S(p).
\end{equation*}
\notag
$$
This implies that if we put
$$
\begin{equation*}
\Phi=\{\mathfrak{P}\in S(p)\mid d(\mathfrak{P})=m\} \subset S(p),
\end{equation*}
\notag
$$
then $\Phi$ is a $p$-type and
$$
\begin{equation*}
\mathfrak{M}=\prod_{\mathfrak{P}\in \Phi}\mathfrak{P}^m=\biggl(\prod_{\mathfrak{P}\in \Phi}\mathfrak{P}\biggr)^m.
\end{equation*}
\notag
$$
It is also clear that
$$
\begin{equation*}
\mathfrak{B}:=\prod_{\mathfrak{P}\in \Phi}\mathfrak{P}\in H(p),
\end{equation*}
\notag
$$
and
$$
\begin{equation*}
(\pi')^{2c}\mathcal{O}_K=\mathfrak{M}^{2c}=\mathfrak{B}^{2cm}=(\mathfrak{B}^{2c})^m =\bigl(\Pi(\mathfrak{B})\mathcal{O}_K\bigr)^m=\Pi(\mathfrak{B}^m)\mathcal{O}_K.
\end{equation*}
\notag
$$
It follows from Remark 2.3 that the ratio $\Pi(\mathfrak{B})^m /(\pi')^{2c}$ is a root of unity. The uniqueness follows from the already proved (i).
Let us prove (iv). The already proved (0) tells us that if $\mathfrak{B}_2 =\sigma(\mathfrak{B}_1)$ for $\sigma \in G$, then $\Pi(\mathfrak{B}_2)=\sigma(\Pi(\mathfrak{B}_1))$ and therefore Weil’s numbers $\Pi(\mathfrak{B}_1)$ and $\Pi(\mathfrak{B}_2)$ are equivalent. Conversely, suppose that $\Pi(\mathfrak{B}_1)$ and $\Pi(\mathfrak{B}_2)$ are equivalent. This means that there are positive integers $a,b$, a Galois automorphism $\sigma \in G$, and a root of unity $\zeta \in \mu_K$ such that
$$
\begin{equation*}
\Pi(\mathfrak{B}_2)^a=\zeta \cdot \sigma(\Pi(\mathfrak{B}_1))^b.
\end{equation*}
\notag
$$
This implies the equality of the corresponding ideals in $\mathcal{O}_K$:
$$
\begin{equation*}
\Pi(\mathfrak{B}_2)^a\mathcal{O}_K=\sigma(\Pi(\mathfrak{B}_1))^b\mathcal{O}_K =\Pi(\sigma(\mathfrak{B}_1))^b.
\end{equation*}
\notag
$$
This means (in light of already proved (i)) that
$$
\begin{equation*}
\mathfrak{B}_2^{2 ca}=(\sigma(\mathfrak{B}_1))^{2cb},
\end{equation*}
\notag
$$
which implies $\mathfrak{B}_2=\sigma(\mathfrak{B}_1)$. Hence $\mathfrak{B}_1$ and $\mathfrak{B}_2$ lie in the same $G$-orbit.
Let us prove (vi). Actually, we have already constructed the map $\Pi_0\colon H(p) \to \mathcal{O}_K$, checked that its image lies in $W(q,K)$; we have also proved property (vi-a). As for (vi-b), it follows readily from (30) combined with the $G$-equivariance of $\Pi$. As for (vi-c), it follows readily from (v) combined with (30). In order to prove (vi-d), it suffices to recall that $\Pi(\mathfrak{B})$ is an ordinary Weil’s $q^2$-number and notice that in light of (30), the integer
$$
\begin{equation*}
\frac{\operatorname{ord}_{\mathfrak{P}}(\Pi(\mathfrak{B}))}{\operatorname{ord}_{\mathfrak{P}}(q^2)} =\frac{2\operatorname{ord}_{\mathfrak{P}}(\Pi_0(\mathfrak{B}))} {2\operatorname{ord}_{\mathfrak{P}}(q)} =\frac{\operatorname{ord}_{\mathfrak{P}}(\Pi_0(\mathfrak{B}))} {\operatorname{ord}_{\mathfrak{P}}(q)}.
\end{equation*}
\notag
$$
Theorem 3.6 is proved. $\Box$ § 4. Abelian varieties with Weil’s numbers in $K$As above, $p$ is a prime, $m$ a positive integer, and $q=p^m$. Theorem 4.1. Let $A$ be a simple abelian variety over $k=\mathbb{F}_q$ such that the corresponding Weil’s $q$-number Let $\mathbb{Q}(\pi_A)$ be the subfield of $K$ generated by $\pi_A$. (i) Suppose that either $\mathbb{Q}(\pi_A)\ne K$ or $p$ does not split completely in $K$. Then $A$ is supersingular. (ii) If $p$ splits completely in $K$, $\mathbb{Q}(\pi_A)=K$, and $\pi_A$ is not ordinary, then the division $\mathbb{Q}$-algebra $\operatorname{End}_k^{0}(A)$ is not commutative. (iii) If $\pi_A$ is ordinary, then $K=\mathbb{Q}(\pi_A)$ and $\operatorname{End}_k^{0}(A) \cong K$; in particular, $\operatorname{End}_k^{0}(A)$ is commutative. Proof. (i) It follows from Lemmas 2.7 and 2.8 that $\pi_A^2/q$ is a root of unity. This means that $A$ is supersingular.
(ii), (iii) Recall [11], [13] that $E:=\operatorname{End}_k^{0}(A)$ is a central division algebra over the field $\mathbb{Q}(\pi_A)=K$. Since $p$ splits completely in $K$, the $\mathfrak{P}$-adic completion $K_{\mathfrak{P}}$ of $K$ coincides with $\mathbb{Q}_p$, i. e.,
$$
\begin{equation*}
[K_{\mathfrak{P}}:\mathbb{Q}_p]=1 \quad \forall \, \mathfrak{P} \in S(p).
\end{equation*}
\notag
$$
By [13; Theorem 1], the local $\mathfrak{P}$-adic invariant
$$
\begin{equation*}
\operatorname{inv}_{\mathfrak{P}}(E)\in \mathbb{Q}/\mathbb{Z}
\end{equation*}
\notag
$$
of the central division $K$-algebra $E$ is given by the formula
$$
\begin{equation}
\operatorname{inv}_{\mathfrak{P}}(E)= \frac{\operatorname{ord}_{\mathfrak{P}}(\pi_A)}{\operatorname{ord}_{\mathfrak{P}}(q)} [K_{\mathfrak{P}}:\mathbb{Q}_p]\bmod \mathbb{Z}= \frac{\operatorname{ord}_{\mathfrak{P}}(\pi_A)}{\operatorname{ord}_{\mathfrak{P}}(q)}\bmod \mathbb{Z}\in \mathbb{Q}/\mathbb{Z}.
\end{equation}
\tag{32}
$$
All other local invariants of $E$ (outside $S(p)$) are $0$ (ibid).
Suppose that $\pi_A$ is ordinary. Then $\mathbb{Q}(\pi_A)=K$, because otherwise $\mathbb{Q}(\pi_A)\subset \mathbb{R}$ and therefore $\pi_A$ is real, i. e., $A$ is supersingular [13; Examples], which is not the case. Since $\pi_A$ is ordinary, all the slopes $\operatorname{ord}_{\mathfrak{P}}(\pi_A)/\operatorname{ord}_{\mathfrak{P}}(q)$ are integers and therefore $\operatorname{inv}_{\mathfrak{P}}(E)=0$ for all $\mathfrak{P}\in S(p)$. This implies that the division algebra $E=\operatorname{End}_k^{0}(A)$ is actually a field, i. e., is isomorphic to $K$. This proves (iii). In order to prove (ii), assume that $\pi_A$ is not ordinary. Then there is a maximal ideal $\mathfrak{P} \in S(p)$ such that the ratio $\operatorname{ord}_{\mathfrak{P}}(\pi_A)/\operatorname{ord}_{\mathfrak{P}}(q)$ is not an integer, i. e.,
$$
\begin{equation}
\frac{\operatorname{ord}_{\mathfrak{P}}(\pi_A)}{\operatorname{ord}_{\mathfrak{P}}(q)}\bmod \mathbb{Z} \ne 0 \quad \text{in } \mathbb{Q}/\mathbb{Z}.
\end{equation}
\tag{33}
$$
Combining (33) with (32), we obtain that $\operatorname{inv}_{\mathfrak{P}}(E) \ne 0.$ It follows that $E=\operatorname{End}_k^{0}(A)$ does not coincide with its center, i. e., is noncommutative. This proves (ii). $\Box$ Remark 4.2. Let $A$ be a simple abelian variety over $\mathbb{F}_q$ such that $\pi_A \in K$. Obviously, $A$ is ordinary if and only if $\pi_A$ is ordinary. § 5. Honda–Tate theory for ordinary abelian varietiesAs above, $p$ is a prime that splits completely in $K$, $m$ a positive integer, and $q=p^m$. Let $\pi \in K$ be a Weil’s $q$-number. The Honda–Tate theory [11]–[13] attaches to $\pi$ a simple abelian variety $\mathcal{A}$ over $\mathbb{F}_{q}$ that is defined up to an $\mathbb{F}_{q}$-isogeny and enjoys the following properties. Let $\mathrm{Fr}_{\mathcal{A}}\colon \mathcal{A}\to \mathcal{A}$ be the Frobenius endomorphism of $\mathcal{A}$ and $F:=\mathbb{Q}[\mathrm{Fr}_{\mathcal{A}}]$ be the $\mathbb{Q}$-subalgebra of the division $\mathbb{Q}$-algebra $E:=\operatorname{End}_{\mathbb{F}_{q}}^{0}(\mathcal{A})$ (which is actually a subfield). Then $F$ is the center of $E$ and there is a field embedding
$$
\begin{equation*}
i\colon F \hookrightarrow \mathbb{C} \quad \text{such that} \quad i(\mathrm{Fr}_{\mathcal{A}})=\pi.
\end{equation*}
\notag
$$
Lemma 5.1. Suppose $\pi$ is ordinary and $\mathbb{Q}(\pi^h)=K$ for all positive integers $h$. Then $\mathcal{A}$ is an absolutely simple $2^{n-1}$-dimensional ordinary abelian variety, $\operatorname{End}^0(A) \cong K$, and all endomorphisms of $\mathcal{A}$ are defined over $\mathbb{F}_{q}$. Proof. Since $\mathbb{Q}(\pi)=K$, we get $i(F)=K$. In particular, number fields $K$ and $F$ are isomorphic. In light of Theorem 4.1, $\mathcal{A}$ is an ordinary abelian variety with commutative endomorphism algebra $E=F\cong K$. By Theorem 2 (c) of [11; Section 3],
$$
\begin{equation*}
\dim(\mathcal{A})=\frac{[E:\mathbb{Q}]}{2}=\frac{[K:\mathbb{Q}]}{2}=2^{n-1}.
\end{equation*}
\notag
$$
We are going to prove that $\mathcal{A}$ is absolutely simple and all its endomorphisms are defined over $\mathbb{F}_{q}$. Let $h$ be a positive integer and $k =\mathbb{F}_{q^{h}}$ a degree $h$ field extension of $\mathbb{F}_{q}$. Let $\mathcal{A}_k=\mathcal{A}\times_{\mathbb{F}_{q}}k$ be the abelian variety over $k$ obtained from $\mathcal{A}$ by the extension of scalars. There is the natural embedding (inclusion) of $\mathbb{Q}$-algebras
$$
\begin{equation*}
\operatorname{End}_{\mathbb{F}_{q}}^{0}(\mathcal{A})\subset \operatorname{End}_k^0(\mathcal{A}_k)
\end{equation*}
\notag
$$
such that the Frobenius endomorphism $\mathrm{Fr}_{\mathcal{A}_k}$ coincides with $\mathrm{Fr}_{\mathcal{A}}^h$. In particular,
$$
\begin{equation*}
\mathbb{Q}[\mathrm{Fr}_{\mathcal{A}_k}]\subset \mathbb{Q}[\mathrm{Fr}_{\mathcal{A}}]=F.
\end{equation*}
\notag
$$
In addition,
$$
\begin{equation*}
i(\mathrm{Fr}_{\mathcal{A}_k})=i(\mathrm{Fr}_{\mathcal{A}}^h)= i(\mathrm{Fr}_{\mathcal{A}})^h=\pi^h.
\end{equation*}
\notag
$$
Since $\mathbb{Q}[\pi^h]=K=\mathbb{Q}(\pi)$, we get
$$
\begin{equation*}
i(\mathbb{Q}[\mathrm{Fr}_{\mathcal{A}_k}])=K=i(\mathbb{Q}[\mathrm{Fr}_{\mathcal{A}}]).
\end{equation*}
\notag
$$
Hence, $\mathbb{Q}[\mathrm{Fr}_{\mathcal{A}_k}]=\mathbb{Q}[\mathrm{Fr}_{\mathcal{A}}]$ is a number field of degree $2\dim(\mathcal{A})=2\dim(\mathcal{A}_k)$. Applying again Theorem 2 (c) of [11; Section 3] to $\mathcal{A}_k$, we conclude that for all finite overfields $k$ of $\mathbb{F}_{q}$. This implies that i. e., all the endomorphisms of $\mathcal{A}$ are defined over $\mathbb{F}_{q}$. In particular, $\mathcal{A}$ is absolutely simple and $\operatorname{End}^0(\mathcal{A}) \cong K$. $\Box$ § 6. Proofs of main resultsAs above, $c=\exp(K)$, a prime $p$ splits completely in $K$ and $q=p^c$. Proof of Theorem 1.6. Let $\Pi\colon H(p) \to W(q^2,K)$ be as in Theorem 3.6. Let $\mathcal{B}\in H(p)$ and let $\Pi(\mathcal{B})$ be the corresponding ordinary Weil’s $q^2$-number in $K$. In light of Theorem 3.6 (v), $\mathbb{Q}[\Pi(\mathfrak{B})^h]=K$ for all positive integers $h$. In light of Lemma 5.1 applied to $q^2$ and $\Pi(\mathfrak{B})$, the Honda–Tate theory [11]–[13] attaches to $\Pi(\mathcal{B})$ an absolutely simple $2^{n-1}$-dimensional abelian variety $\mathcal{A}=A(\mathcal{B})$ over $\mathbb{F}_{q^2}$ (that is defined up to an $\mathbb{F}_{q^2}$-isogeny) such that $\operatorname{End}^0(A(\mathcal{B}))\cong K$, and all endomorphisms of $A(\mathcal{B})$ are defined over $\mathbb{F}_{q^2}$.
By Theorem 3.6 (iv), if $\mathfrak{B}_1, \mathfrak{B}_2 \in H(p)$, then the Weil numbers $\Pi(\mathfrak{B}_1)$ and $\Pi(\mathfrak{B}_2)$ are equivalent if and only if $\mathfrak{B}_1$, and $\mathfrak{B}_2$ belong to the same $G$-orbit. In light of [11; Theorem 1], [12; p. 84] combined with Lemma 3.4, all the $A(\mathfrak{B})$ lie in precisely $2^{2^{n-1}-n}$ isogeny classes of abelian varieties over $\overline{\mathbb{F}}_p$. We also know that each of these varieties is ordinary, has dimension $2^{n-1}$ and their endomorphism algebras are isomorphic to $K$. Now, let us prove that each abelian variety $\mathcal{B}$ over $\overline{\mathbb{F}}_p$, whose endomorphism algebra is isomorphic to $K$, is isogenous to one of $A(\mathfrak{B})$ over $\overline{\mathbb{F}}_p$. In order to do that, first, notice that since $K$ is a field, $\mathcal{B}$ is simple over $\overline{\mathbb{F}}_p$. Second, $\mathcal{B}$ is defined with all its endomorphisms over a certain finite field $k=\mathbb{F}_{q^{2h}}$ (where $h$ is a certain positive integer), i. e., there is a simple abelian variety $\mathcal{B}_k$ over $k$ such that
$$
\begin{equation*}
\mathcal{B}=\mathcal{B}_k\times_k \overline{\mathbb{F}}_p, \qquad \operatorname{End}_k^{0}(\mathcal{B}_k)=\operatorname{End}^0(\mathcal{B}) \cong K.
\end{equation*}
\notag
$$
Applying Theorem 2 (c) of [11; Section 3] to $\mathcal{B}_k$, we get
$$
\begin{equation*}
K \cong \operatorname{End}^0(\mathcal{B})= \operatorname{End}_k^0(\mathcal{B}_k) =\mathbb{Q}[\mathrm{Fr}_{\mathcal{B}_k}],
\end{equation*}
\notag
$$
where $\mathrm{Fr}_{\mathcal{B}_k}$ is the Frobenius endomorphism of $\mathcal{B}_k$. This gives us a field isomorphism $\mathbb{Q}[\mathrm{Fr}_{\mathcal{B}_k}] \to K$; let us denote by $\pi_{\mathcal{B}_k}$ the image of $\mathrm{Fr}_{\mathcal{B}_k}$ in $K$. Clearly, $\mathbb{Q}(\pi_{\mathcal{B}_k})\,{=}\,K$; according to a classical result of Weil [16], $\pi_{\mathcal{B}_k}$ is a Weil’s $q^{2h}$-number. By Theorem 4.1 (i) (applied to $q^{2h}$ instead of $q$), $\pi_{\mathcal{B}_k}$ is ordinary, since $\operatorname{End}_k^0(\mathcal{B}_k) \cong K$ is commutative. It follows from Theorem 3.6 (iii) that there is $\mathfrak{B} \in H(p)$ such that Weil’s numbers $\pi_{\mathcal{B}_k}$ and $\Pi(\mathfrak{B})$ are equivalent. This means (thanks to Theorem 1 of [11], see also [12; pp. 83, 84]) that absolutely simple abelian varieties $\mathcal{B}_k$ and $A(\mathfrak{B})$ become isogenous over $\overline{\mathbb{F}}_p$. It follows that absolutely simple abelian varieties $\mathcal{B}=\mathcal{B}_k\times_k \overline{\mathbb{F}}_p$ and $A(\mathfrak{B})$ are isogenous over $\overline{\mathbb{F}}_p$.
This proves (i), (ii) (1), and (ii) (2). It remains to prove (ii) (3). It suffices to check that for each $\mathcal{B}\in H(p)$ there exists an abelian variety $A_0$ that is defined over $\mathbb{F}_q$ with all its endomorphisms and such that $A(\mathcal{B})$ is isogenous to $A_0$ over $\overline{\mathbb{F}}_p$. Let $\Pi_0\colon H(p) \to W(q,K)$ be as in Theorem 3.6 (vi) and $\Pi_0(\mathcal{B})$ be the corresponding ordinary Weil’s $q$-number in $K$. In light of Theorem 3.6 (vi-c), $\mathbb{Q}[\Pi_0(\mathfrak{B})^h]=K$ for all positive integers $h$. In light of Lemma 5.1 applied to $q$ and $\Pi_0(\mathfrak{B})$, the Honda–Tate theory [11]–[13] attaches to Weil’s $q$-number $\Pi_0(\mathcal{B})$ an absolutely simple $2^{n-1}$-dimensional abelian variety $\mathcal{A}_0$ over $\mathbb{F}_{q}$ (that is defined up to an $\mathbb{F}_{q}$-isogeny) such that $\operatorname{End}^0(\mathcal{A}_0)\cong K$, and all endomorphisms of $\mathcal{A}_0$ are defined over $\mathbb{F}_{q}$. Since $\Pi_0(\mathcal{B})^2=\Pi(\mathcal{B})$, Weil’s numbers $\Pi_0(\mathcal{B})$ and $\Pi(\mathcal{B})$ are equivalent. As above, in light of Theorem 1 of [11] (see also [12; pp. 83, 84]), the corresponding absolutely simple abelian varieties $\mathcal{A}_0$ and $A(\mathcal{B})$ are isogenous over $\overline{\mathbb{F}}_p$. $\Box$ Proof of Corollary 1.14. Recall that $r$ is an odd prime and $\zeta_r$ is a primitive $r$th root of unity. Clearly, $\mathbb{Q}(\zeta_r)$ is a $\mathrm{CM}$ field. Hence, its subfield $K$ is either $\mathrm{CM}$ or totally real. Since $\mathbf{H}$ has odd order $m$, it does not contain the complex conjugation $\rho\colon \mathbb{Q}(\zeta_r) \to \mathbb{Q}(\zeta_r)$, because $\rho$ has order $2$. Hence, $\rho$ acts nontrivially on $K=\mathbb{Q}(\zeta_r)^{\mathbf{H}}=K^{(r)}$, which implies that $K$ is a $\mathrm{CM}$ field. (See also [10; p. 78].) Its degree
$$
\begin{equation*}
[K:\mathbb{Q}]=\frac{[\mathbb{Q}(\zeta_r):\mathbb{Q}]}{\#(\mathbf{H})}=\frac{m \cdot 2^n}{m}=2^n.
\end{equation*}
\notag
$$
We also know (Remark 1.15) that every totally positive unit in $K_0$ is a square in $K_0$.
Clearly, $K/\mathbb{Q}$ is ramified at $r$ and unramified at every prime $p \ne r$. Let us find which $p\ne r$ split completely in $K$. Let
$$
\begin{equation*}
f_p \in \operatorname{Gal}(\mathbb{Q}(\zeta_r)/\mathbb{Q})=(\mathbb{Z}/r\mathbb{Z})^*
\end{equation*}
\notag
$$
be the Frobenius element attached to $p$, which is characterized by the property In other words,
$$
\begin{equation*}
f_p=(p\ \operatorname{mod} r)\in (\mathbb{Z}/r\mathbb{Z})^* =\operatorname{Gal}(\mathbb{Q}(\zeta_r)/\mathbb{Q}).
\end{equation*}
\notag
$$
Clearly, $p$ splits completely in $K$ if and only if $f_p\in \mathbf{H}$. So, we need to find when $f_p$ lies in $\mathbf{H}$. In order to do it, notice that
$$
\begin{equation*}
\mathbf{H}=\{\sigma^{2^n}\mid \sigma \in \operatorname{Gal}(\mathbb{Q}(\zeta_r)/\mathbb{Q})= (\mathbb{Z}/r\mathbb{Z})^*\}.
\end{equation*}
\notag
$$
This implies that $f_p$ lies in $\mathbf{H}$ if and only if $p \bmod r$ is a $2^n$th power in $\mathbb{Z}/r\mathbb{Z}=\mathbb{F}_r$. This ends the proof of (0).
The remaining assertions (i) and (ii) follow from Theorem 1.6 combined with (0). $\Box$ Proof of Corollary 1.8. In the notation of Corollary 1.14, this is the case when $m=1$ and $2^n=r-1$. By little Fermat’s theorem, every nonzero $a\in \mathbb{Z}/r\mathbb{Z}$ satisfies Now the desired result follows readily from Corollary 1.14. $\Box$ AcknowledgementsI am grateful to Ley Watson for an interesting stimulating question [8].
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