| Известия Российской академии наук. Серия математическая |
|
|
|
|
|
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 Cl(K) for the (finite commutative) ideal class group of K, cl(K) for the class number of K (i. e., the cardinality of Cl(K)) and exp(K) for the exponent of Cl(K). Clearly, exp(K) divides cl(K). (The equality holds if and only if Cl(K) is cyclic, which is not always the case, see [1; Tables].) In addition, exp(K) is odd if and only if cl(K) is odd. We write OK for the ring of integers in K and UK for the group of units, i. e., the multiplicative group of invertible elements in OK. As usual, an element of UK 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 UK. Let p be a prime and q a positive integer that is a power of p. We write Fp for the p-element finite field and Fq for its q-element overfield. As usual, ¯Fp stands for an algebraic closure of Fp, which is also an algebraic closure of Fq. We have If X is an abelian variety over ¯Fp, then we write End0(X) for its endomorphism algebra End(X)⊗Q, which is a finite-dimensional semisimple algebra over the field Q of rational numbers. If X is defined over k=Fq, then we write Endk(X) for its ring of k-endomorphisms and End0k(X) for the Q-algebra Endk(X)⊗Q; one may view End0k(X) as the Q-subalgebra of End0(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 Q. It is also known (Grothendieck–Tate) that End0(X) uniquely determines the dimension of X [2]. Namely, 2dim(X) is the maximal Q-dimension of a semisimple commutative Q-subalgebra of End0(X). However, it turns out that there are nonisogenous abelian varieties over ¯Fp 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]. 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=pexp(K). Then there exists an elliptic curve E that is defined with all its endomorphisms over Fq and such that End0(E)≅K. Remark 1.4. One may deduce from ([5; Satz 3], [6; Section 6, Corollary 1, p. 507]) that if we put q1=pcl(K), then there exists an elliptic curve E that is defined with all its endomorphisms over Fq1 and such that End(E)≅OK (and therefore End0(E)≅K). The next result is an analogue of Proposition 1.2 for abelian surfaces and quartic fields. Proposition 1.5. Let K be a CM quartic field that is a cyclic extension of Q.
Our main result is the following assertion. Theorem 1.6. Let n be a positive integer and K be a CM field that is a cyclic degree 2n extension of Q. Let K0 be the only degree 2n−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 ¯Fp such that End0(A) is isomorphic to K. Then p splits completely in K and A is an ordinary simple abelian variety of dimension 2n−1. (ii) Let p be a prime that splits completely in K. Let us put q=pc.
Remark 1.7. (a) If n=1, then K is an imaginary quadratic field and therefore K0=Q and UQ={±1}. The only (totally) positive unit in Q is 1, which is obviously a square in 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⩾3, then the number 22n−1−n of the corresponding isogeny classes is strictly greater than 1. This gives us examples of nonisogenous abelian varieties over ¯Fp, 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 1/2n). 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 isg(r) isogeny classes of simple ((r−1)/2)-dimensional ordinary abelian varieties A over ¯Fp, whose endomorphism algebra is isomorphic to the rth cyclotomic field Q(ζr). In addition, if c=exp(Q(ζr)) and q=pc, then each of these isogeny classes contains an abelian variety that is defined with all its endomorphisms over Fq.Remark 1.9. The congruence condition on p means that p splits completely in Q(ζr). 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, isg(r)=1 if and only if r⩽5. Let p be a prime that is congruent to 1mod. 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
\begin{equation*}
\operatorname{Gal}(\mathbb{Q}(\zeta_r)/\mathbb{Q})=(\mathbb{Z}/r\mathbb{Z})^*
\end{equation*}
\notag
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 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
\begin{equation*}
\pi \mathcal{O}_K=\prod_{\mathfrak{P}\in \Psi(\pi)}\mathfrak{P}^m=\mathfrak{B}^m, \quad \text{where } \mathfrak{B}:=\prod_{\mathfrak{P}\in \Psi(\pi)}\mathfrak{P}.
\end{equation*}
\notag
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
\begin{equation*}
p\cdot \mathcal{O}_K=\mathfrak{B}\cdot \rho(\mathfrak{B})=\mathfrak{B}^2,
\end{equation*}
\notag
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
\begin{equation*}
\frac{\#(H(p))}{\#(G)}=\frac{2^{2^{n-1}}}{2^n}=2^{2^{n-1}-n}.
\end{equation*}
\notag
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
\begin{equation}
\mathcal{Z}\colon H(p) \hookrightarrow \mathcal{O}_K \setminus\{0\}\subset \mathcal{O}_K\subset K
\end{equation}
\tag{26}
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
\begin{equation*}
\mathfrak{B}^c=\sigma(\mathfrak{B}_O)^c=\sigma(\mathfrak{B}_O^c)=\sigma(z_O)\mathcal{O}_K,
\end{equation*}
\notag
i. e., \sigma(z_O) is a generator of \mathfrak{B}^c. It remains to put Corollary is proved. \Box 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 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 KAs 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
\begin{equation*}
\operatorname{End}^0(\mathcal{A}_k)=\mathbb{Q}[\mathrm{Fr}_{\mathcal{A}_k}] =\mathbb{Q}[\mathrm{Fr}_{\mathcal{A}}]=\operatorname{End}_{\mathbb{F}_{q}}^{0}(\mathcal{A})
\end{equation*}
\notag
for all finite overfields k of \mathbb{F}_{q}. This implies that
\begin{equation*}
\operatorname{End}^0(\mathcal{A}_k)=\operatorname{End}_{\mathbb{F}_{q}}^{0}(\mathcal{A}),
\end{equation*}
\notag
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 rth 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^nth 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].
Цитирование в формате AMSBIB
\RBibitem{Zar23} |
|
|||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
|
Обратная связь: |
Пользовательское соглашение
|
Регистрация посетителей портала |
Логотипы |
|