---
address: 'Department of Mathematics, Pennsylvania State University, University Park, PA 16802, USA'
author:
- 'Yuri G. Zarhin'
title: Hodge classes on certain hyperelliptic prymians
---

Definitions and statements
==========================

Throughout this paper $K$ is a field, $\K$ its algebraic closure and $\Gal(K)=\Aut(\K/K)$ the absolute Galois group of $K$.

If $X$ is an abelian variety over $\K$ then we write $\End(X)$ for the ring of all its $\K$-endomorphisms ; the notation $1_X$ stands for the identity automorphism of $X$. If $Y$ is an abelian variety over $\K$ then we write $\Hom(X,Y)$ for the corresponding group of all $\K$-homomorphisms.

Let $f(x)\in K[x]$ be a polynomial of degree $n\ge 2$ with coefficients in $K$ and without multiple roots, $\RR_f\subset \K_a$ the ($n$-element) set of roots of $f$ and $K(\RR_f)\subset \K_a$ the splitting field of $f$. We write $\Gal(f)=\Gal(f/K)$ for the Galois group $\Gal(K(\RR_f)/K)$ and call it the Galois group of $f(x)$ over $K$; it permutes roots of $f$ and may be viewed as a certain permutation group of $\RR_f$, i.e., as a subgroup of the group $\Perm(\RR_f)\cong\Sn$ of permutation of $\RR_f$. (It is well known that $\Gal(f)$ is transitive if and only if $f$ is irreducible.) Let us put $$g=\left[\frac{n-1}{2}\right].$$ Clearly, $g$ is a nonnegative integer and either $n=2g+1$ or $n=2g+2$.

Let us assume that $\fchar(K)\ne 2$. We write $C_{f}$ for the genus $g$ hyperelliptic $K$-curve $y^2=f(x)$ and $J(C_{f})$ for its jacobian. Clearly, $J(C_{f})$ is a $g$-dimensional abelian variety that is defined over $K$. In particular, $J(C_{f})=\{0\}$ if and only if $n=2$. The abelian variety $J(C_{f})$ is an elliptic curve if and only if $n=4$.

Let us assume that $K$ is a subfield of the field $\C$ of complex numbers (and $\bar{K}$ is the algebraic closure of $K$ in $\C$). Then one may view $J(C_{f})$ as a complex abelian variety and consider its first rational homology group $\H_1(J(C_{f}),\Q)$ and the Hodge group $\Hdg(J(C_{f}))$ of $J(C_{f})$, which is a certain connected reductive algebraic $\Q$-subgroup of the general linear group $\GL(\H_1(J(C_{f}),\Q))$ [@MumfordSh; @Deligne; @Ribet; @ZarhinIzv; @MZ2]. The canonical principal polarization on $J(C_{f})$ gives rise to the nondegenerate alternating bilinear form $$\H_1(J(C_{f}),\Q) \times \H_1(J(C_{f}),\Q) \to \Q$$ and the corresponding symplectic group $\Sp( \H_1(J(C_{f}),\Q))$ contains $\Hdg(J(C_{f}))$ as a (closed) algebraic $\Q$-subgroup. In addition, $\End(J(C_{f}))$ coincides with the endomorphism ring of the complex abelian variety $J(C_{f})$ and $\End^{0}(J(C_{f}))$ coincides with the centralizer of $\Hdg(J(C_{f}))$ in $\End_{\Q}(\H_1(J(C_{f}),\Q))$ [@MumfordSh; @Ribet; @MZ2].

The following result was obtained by the author in [@ZarhinMRL Th. 2.1], [@ZarhinMMJ Sect. 10]. (See also [@ZarhinPLMS2], [@ZarhinBSMF; @ZarhinL].)

\[jacobian\] Suppose that $K\subset \C$, $n \ge 5$ (i.e., $g \ge
2$) and $\Gal(f)=\ST_n$ or the alternating group $\A_n$. Then $\End(J(C_{f}))=\Z$ and $\Hdg(J(C_{f}))=\Sp( \H_1(J(C_{f}),\Q))$. Every Hodge class on each self-product of $J(C_{f})$ can be presented as a linear combination of products of divisor classes. In particular, the Hodge conjecture is valid for each self-product of $J(C_{f})$.

The assertion that $\Hdg(J(C_{f}))=\Sp( \H_1(J(C_{f}),\Q))$ was not stated explicitly in [@ZarhinMMJ]. However, it follows immediately from the description of the Lie algebra $\mathrm{mt}$ of the corresponding Mumford-Tate group [@ZarhinMMJ p. 429] as the direct sum of the scalars $\Q \mathrm{Id}$ and the Lie algebra of the symplectic group, because the Lie algebra of the Hodge group coincides with the intersection of Lie algebras of the Mumford-Tate group and the symplectic group. (The same arguments prove the equality $\Hdg(J(C_{f}))=\Sp( \H_1(J(C_{f}),\Q))$ for all $f(x)$ that satisfy the conditions of Theorem 10.1 of [@ZarhinMMJ].)

Our next result that was obtained in [@ZarhinSh Th. 1.2 and Theorem 2.5] deals with homomorphisms of hyperelliptic jacobians.

\[homo\] Suppose that $\fchar(K)\ne 2$, $n\ge 3$ and $m\ge 3$ are integers and let $f(x)$ and $h(x)$ be irreducible polynomials over $K$ of degree $n$ and $m$ respectively. Suppose that $$\Gal(f)=\ST_n, \ \Gal(h)=\ST_m$$ and the corresponding splitting fields $K(\RR_f)$ and $K(\RR_h)$ are linearly disjoint over $K$. Then either $$\Hom(J(C_f),J(C_h))=\{0\}, \ \Hom(J(C_h),J(C_f))=\{0\}$$ or $\fchar(K)>0$ and both $J(C_f)$ and $J(C_h)$ are supersingular abelian varieties.

\[homoC\] If $K\subset \C$ then Theorem \[homo\] implies that (under its assumptions) there are no nonzero homomorphisms between complex abelian varieties $J(C_f)$ and $J(C_h)$.

The main results of the present paper are the following statements.

\[main\] Suppose that $n=2g+2=\deg(f)\ge 8$. Let $\tilde{C}_f \to C_{f}$ be an unramified double cover of complex smooth projective irreducible curves and let $P$ be the corresponding Prym variety, which is a $(g-1)$-dimensional (principally polarized) complex abelian variety.

If $\Gal(f)=\ST_n$ then:

-   $\End(P)=\Z$ or $\Z\oplus \Z$.

-   Every Hodge class on each self-product of $P$ can be presented as a linear combination of products of divisor classes. In particular, the Hodge conjecture holds true for each self-product of $P$.

\[main1\] Suppose that $n=2g+2\ge 10$, $K\subset\C$ and $f(x)=(x-a)h(x)$ where $a\in K$ and $h(x)\in K[x]$ is an irreducible degree $n-1$ polynomial with $\Gal(h)=\ST_{n-1}$. Let $\tilde{C}_{f} \to C_{f}$ be an unramified double cover of complex smooth projective irreducible curves and let $P$ be the corresponding Prym variety, which is a $(g-1)$-dimensional (principally polarized) complex abelian variety.

Then:

-   $\End(P)=\Z$ or $\Z\oplus \Z$.

-   every Hodge class on each self-product of $P$ can be presented as a linear combination of products of divisor classes. In particular, the Hodge conjecture holds true for each self-product of $P$.

Our proof is based on the explicit description of Prym varieties of hyperelliptic curves [@MumfordP; @Dal] and our results about Hodge groups of hyperelliptic jacobians mentioned above.

If $n=2g+2\le 10$ then $\dim(P)=g-1\le 3$. Notice that if $A$ is a complex abelian varietiy of dimension $\le 3$ then it is well known that every Hodge class on each self-product of $A$ can be presented as a linear combination of products of divisor classes [@MZ2 Th. 0.1(iv)]. In particular, the Hodge conjecture holds true for each self-product of $A$.

The paper is organized as follows. In Section \[Galois\] we discuss an elementary construction from Galois theory and apply it in Section \[homohyper\] to homomorphisms of hyperelliptic jacobians. Section \[hyperhodge\] deals with Hodge groups of hyperelliptic jacobians. In Section \[hyperprym\] we discuss hyperelliptic prymians and prove the main results.

Galois theory {#Galois}
=============

Throughout this Section, $K$ is an arbitrary field and $n\ge 3$ is an integer, $f(x)\in K[x]$ is a degree $n$ irreducible polynomial, whose Galois group $$\Gal(f)=\Gal(K(\RR_f)/K)$$ is the full symmetric group $\Perm(\RR_f)=\ST_n$. If $T\subset
\RR_f$ is a non-empty subset then we put $$f_T(x)=\prod_{\alpha\in T}(x-\alpha)\in K(\RR_f)[x].$$ By definition, $$\deg(f_T)=\#(T), \ \RR_{f_T}=T.$$ We view $\Perm(T)$ as a subgroup of $\Perm(\RR_f)=\Gal(f)$ that consists of all permutations that leave invariant every element outside $T$.

\[oneT\] Let us consider the subfield $E_0=K(\RR_f)^{\Perm(T)}$ of $\Perm(T)$-invariants. Since $\Perm(T)$ leaves invariant $T=\RR_{f_T}$, $$f_T(x)\in E_0[x].$$ Clearly, $$\Gal(K(\RR_f)/E_0)=\Perm(T)\subset \Perm(\RR_f)=\Gal(\RR_f/K).$$ Let us prove that the splitting field $E_0(\RR_{f_T})=E_0(T)$ of $f_T(x)$ over $E_0$ coincides with $K(\RR_f)$. Indeed, $\Gal(K(\RR_f))/E_0(T))$ consists of all elements of $\Perm(T)=\Gal(K(\RR_f)/E_0)$ that leave invariant every element of $T$. Since every element of $\Perm(T)$ leaves invariant every element of $\RR_f\setminus T$, $\Gal(K(\RR_f))/E_0(T))=\{1\}$, i.e., $K(\RR_f)=E_0(T)=E_0(\RR_{f_T})$. This implies that the Galois group $$\Gal(E_0(\RR_{f_T})/E_0)=\Gal(K(\RR_f)/E_0)=\Perm(T).$$

\[key\] Let $T$ and $S$ be two nonempty disjoint subsets of $\RR_f$. Then there exists a field subextension $E/K \subset
K(\RR_f)/K$ that enjoys the following properties.

-   The Galois group $\Gal(K(\RR_f)/E)$ coincides with the subgroup $$\Gal(T)\times \Gal(S)\subset \Perm(\RR_f)=\Gal(K(\RR_f)/K),$$ which consists of all permutations that leave invariant $T,S$ and every element outside $T \sqcup S$.

-   Both $f_T(x)$ and $f_S(x)$ lie in $E[x]$, i.e., all their coefficients belong to $E$.

-   Let $E(\RR_T)=E(T)$ and $E(\RR_S)=E(S)$ be the splitting fields over $E$ of $f_T(x)$ and $f_S(x)$ respectively. Then the natural injective homomorphisms $$\Gal(E(T)/E)\hookrightarrow \Perm(T), \ \Gal(E(S)/E)\hookrightarrow
    \Perm(S)$$ are group isomorphisms, i.e., $$\Gal(E(T)/E)=\Perm(T), \ \Gal(E(S)/E)=
    \Perm(S).$$

-   $E(T)$ and $E(S)$ are linearly disjoint over $E$.

-   The compositum $E(T) E(S)$ coincides with $K(\RR_f)$.

Recall that $$\RR_{f_T}=T, \ \RR_{f_S}=S.$$ We define $E$ as the subfield $K(\RR_f)^{\Perm(T)\times
\Perm(S)}$ of $\Perm(T)\times \Perm(S)$-invariants. Now Galois theory gives us (i). The subgroup $\Perm(T)\times \Perm(S)$ leaves invariant both sets $T=\RR_T$ and $S=\RR_S$. This implies that all the coefficients of $f_T(x)$ and $f_S(x)$ are $[\Perm(T)\times
\Perm(S)]$-invariant, i.e., lie in $E$. This proves (ii). Clearly, $$[K(\RR_f):E]=\#(\Perm(T)\times \Perm(S)).$$

Clearly, the subgroup of $\Perm(T)\times \Perm(S)$ that consists of all permutations that act identically on $S$ coincides with $\Perm(T)$. Similarly, the subgroup of $\Perm(T)\times \Perm(S)$ that consists of all permutations that act identically on $T$ coincides with $\Perm(S)$. This implies that $$\Gal(E(T)/E)=[\Perm(T)\times \Perm(S)]/\Perm(S)=\Perm(T),$$ $$\Gal(E(S)/E)=[\Perm(T)\times \Perm(S)]/\Perm(T)=\Perm(S),$$ which proves (iii). This implies that $$[E(T):E]=\#(\Perm(T)), \ [E(S):E]=\#(\Perm(S)).$$ Let $L$ be the compositum $E(T) E(S)$. Clearly, $L$ contains $T$, $S$ and $E$. Therefore $\Gal(K(\RR_f)/L)$ consists of elements of $\Perm(T)\times \Perm(S)=\Gal(K(\RR_f)/E)$ that act identically on $T$ and $T$. Since all elements of $\Perm(T)\times \Perm(S)$ act identically on the complement to $T \sqcup S$, we conclude that $\Gal(K(\RR_f)/L)=\{1\}$, i.e., $$K(\RR_f)=L=E(T) E(S).$$ This proves (v). We also obtain that $$[E(T) E(S):E]=[K(\RR_f):E]=\#(\Perm(T)\times \Perm(S))=[E(T):E]
[E(S):E],$$ i.e. $$[E(T) E(S):E]=[E(T):E]
[E(S):E],$$ which means that $E(T)/E$ and $E(S)/E$ are linearly disjoint. This proves (iv).

Homomorphisms of hyperelliptic jacobians {#homohyper}
========================================

We keep the notation and assumptions of Section \[Galois\]. Also we assume that $\fchar(K)\ne 2$.

\[homoG\] Let $T$ and $S$ be disjoint nonempty subsets of $\RR_f$ and consider the hyperelliptic curves $$C_{f_T}:y^2=f_T(x), \ C_{f_S}:y^2=f_S(x)$$ and their jacobians $J(C_{f_T})$ and $J(C_{f_S})$. Then either $$\Hom(J(C_{f_T}), J(C_{f_S}))=\{0\}, \ \Hom(J(C_{f_S}),
J(C_{f_T}))=\{0\}$$ or $\fchar(K)>0$ and both $J(C_{f_T})$ and $J(C_{f_S})$ are supersingular abelian varieties.

If $\#(T)<3$ (resp. $\#(S)<3$) then $C_{f_T}$ (resp. $C_{f_S}$) has genus zero and $J(C_{f_T})=0$ (resp. $J(C_{f_S})=0$), which implies that there are no nonzero homomorphisms between $J(C_{f_T})$ and $J(C_{f_S})$. So, further we assume that $$n_1:=\#(T)\ge 3, \ n_2:=\#(S)\ge 3.$$ By Lemma \[key\], there exists a field $E$ such that both $f_T(x)$ and $f_S(x)$ lie in $E[x]$, their Galois groups are $\Perm(T)\cong\ST_{n_1}$ and $\Perm(S)\cong\ST_{n_2}$ respectively. In addition, their splitting fields are linearly disjoint over $E$. Now the result follows from Theorem \[homo\] applied to $E,
f_T(x), f_S(x)$ instead of $K, f(x), h(x)$.

\[odd\] Suppose that $m:=\#(S)=2r+1$ is odd and let $b$ be an arbitrary element of $K$. Let us consider the hyperelliptic curve $C^{b}_{f_S}:y^2=(x-b)f_S(x)$. By Remark \[oneT\], there exists a field $E_0\subset K(\RR_f)$ such that $f_S(x)$ lies in $E_0[x]$ and $\Gal(E_0(\RR_{f_S})/E_0)=\Perm(S)=\ST_m$. Then the standard substitution [@ZarhinPLMS2 p. 25] $$x_1=\frac{1}{x-b}, \ y_1:=\frac{y}{(x-b)^{r+1}}$$ gives us a degree $m$ irreducible polynomial $h(x_1)\in E[x_1]$ such that $$E_0(\RR_h)=E_0(\RR_{f_S}), \
\Gal(E_0(\RR_h)/E)=\Gal(E_0(\RR_{f_S})/E)=\ST_m$$ and $C^{b}_{f_S}$ is $E_0$-birationally isomorphic to the hyperelliptic curve $C_{h}:y_1^2=h(x_1)$. (It is assumed in [@ZarhinPLMS2 p. 25] that $m \ge 5$ but the substitution works for any positive odd $m$.)

\[homoGt\] Let $T$ and $S$ be disjoint nonempty subsets of $\RR_f$ and assume that $\#(S)$ is odd. Let $b$ be an arbitrary element of $K$. Let us consider the hyperelliptic curves $$C_{f_T}:y^2=f_T(x), \ C^{b}_{f_S}:y^2=(x-b)f_S(x)$$ and their jacobians $J(C_{f_T})$ and $J(C^{b}_{f_S})$. Then either $$\Hom(J(C_{f_T}), J(C^{b}_{f_S}))=\{0\}, \ \Hom(J(C^{b}_{f_S}),
J(C_{f_T}))=\{0\}$$ or $\fchar(K)>0$ and both $J(C_{f_T})$ and $J(C^{b}_{f_S})$ are supersingular abelian varieties.

We may assume that both $m_1=\#(T)$ and $m_2=\#(S)$ are, at least, $3$. Let $E$ be as in Lemma \[key\]. In particular, the splitting fields $E(T)$ and $E(S)$ are linearly disjoint over $E$ and $$\Gal(E(T)/E)\cong \ST_{m_1}, \ \Gal(E(S)/E) \cong \ST_{m_2}.$$ Using Remark \[homoGt\] over $E$ (instead of $E_0$), we obtain that there is a degree $m$ irreducible polynomial $h(x)\in E[x]$ such that $$E(\RR_h)=E(\RR_{f_S}), \
\Gal(E(\RR_h)/E)=\Gal(E(\RR_{f_S})/E)=\ST_{m_2}$$ and $C^{b}_{f_S}$ is birationally $E$-isomorphic to the hyperelliptic curve $C_{h}:y^2=h(x)$. Clearly, the jacobians $J(C^{b}_{f_S})$ and $J(C_h)$ are isomorphic. Applying Theorem \[homo\] to $E,
f_T(x),h(x)$, we conclude that either $$\Hom(J(C_{f_T}), J(C_h))=\{0\}, \ \Hom(J(C_h)),
J(C_{f_T}))=\{0\}$$ or $\fchar(K)>0$ and both $J(C_{f_T})$ and $J(C_{f_S})$ are supersingular abelian varieties. Since $J(C^{b}_{f_S})$ and $J(C_h)$ are isomorphic, we are done.

\[homoGab\] Let $T$ and $S$ be disjoint nonempty subsets of $\RR_f$ and assume that both $\#(T)$ and $\#(S)$ are odd. Let $a$ and $b$ be arbitrary (not necessarily distinct) elements of $K$. Let us consider the hyperelliptic curves $$C^{a}_{f_T}:y^2=(x-a)f_T(x), \ C^{b}_{f_S}:y^2=(x-b)f_S(x)$$ and their jacobians $J(C^{a}_{f_T})$ and $J(C^{b}_{f_S})$. Then either $$\Hom(J(C^{a}_{f_T}), J(C^{b}_{f_S}))=\{0\}, \ \Hom(J(C^{b}_{f_S}),
J(C^{a}_{f_T}))=\{0\}$$ or $\fchar(K)>0$ and both $J(C^{b}_{f_T})$ and $J(C^{a}_{f_S})$ are supersingular abelian varieties.

We may assume that both $m_1:=\#(T)$ and $m_2:=\#(S)$ are, at least, $3$. Again, let $E$ be as in Lemma \[key\]. Applying Remark \[homoGt\] two times over $E$ (instead of $E_0$) to the polynomials $(x-a)f_T(x)$ and $(x-b)f_S(x)$, we conclude that there are degree $m$ irreducible polynomials $h_1(x)\in E[x]$ and $h_2 (x)\in E[x]$ such that $$E(\RR_{h_1})=E(T), \ E(\RR_{h_2})=E(S),$$ $$\Gal(E(\RR_{h_1})/E)=\ST_{m_1}, \ \Gal(E(\RR_{h_2})/E)=\ST_{m_2},$$ $C^{a}_{f_T}$ is $E$-birationally isomorphic to $C_{h_1}$ and $C^{b}_{f_T}$ is $E$-birationally isomorphic to $C_{h_2}$. Clearly, $J(C^{a}_{f_T})\cong J(C_{h_1})$ and $J(C^{b}_{f_S})\cong
J(C_{h_2})$. Applying Theorem \[homo\] to $E, h_1(x),h_2(x)$, we conclude that either $$\Hom(J(C_{h_1}), J(C_{h_2}))=\{0\}, \ \Hom(J(C_{h_2})),
J(C_{h_1}))=\{0\}$$ or $\fchar(K)>0$ and both $J(C_{h_1})$ and $J(C_{h_2})$ are supersingular abelian varieties. The rest is clear.

Let $K_2/K$ be the only quadratic subextension of $K(\RR_f)/K$. Clearly, $K_2(\RR_f)=K(\RR_f)$ and the Galois group $\Gal(K_2(\RR_f)/K_2)$ coincides with the alternating group $\A_n$.

\[ellipticZ\] Suppose that $\fchar(K)\ne 2$. Let $T \subset
\RR_f$ be a $4$-element subset. Let us consider the corresponding elliptic curve $$C_{f_T}:y^2=f_T(x)$$ and its jacobian $J(C_{f_T})$. If $n \ge 8$ then one of the following conditions holds:

-   $\End(J(C_{f_T}))=\Z$ for all $T$.

-   $\fchar(K)>0$ and all $J(C_{f_T})$’s are supersingular elliptic curves mutually isomorphic over $\K$.

Let $j_T$ be the $j$-invariant of the elliptic curve $J(C_{f_T})$ ([@Tate], [@Knapp Ch. III, Sect. 2]). Clearly, $$j_T \in K(T)\subset K(\RR_f)$$ and $$j_{\sigma T}=\sigma j_T \ \forall \sigma\in
\Gal(K(\RR_f)/K)=\Gal(f).$$ Suppose that $J(C_{f_T})$ admits complex multiplication. Then one of the following two conditions holds.

-   $p=\fchar(K)>0$. Then a classical result of M. Deuring asserts that $j_T$ is [*algebraic*]{}, i.e., lies in a finite field $\F_q$ where $q$ is a power of the prime $p$. (See [@Deuring], [@Oort Sect. 3.2], [@Lang Ch. 13, Sect. 5].) In particular, $K(j_T)/K$ is an abelian field extension.

-   $\fchar(K)=0$. Then there exists an imaginary quadratic field $k$ such that $\End^0(J(C_{f_T}))=k$. In addition, a classical result of the theory of complex multiplication asserts that $j_T$ is an algebraic number such that the field extension $k(j_T)/k$ is abelian. (See [@Shimura Sect. 5.4], [@Lang Ch. 10, Sect. 3].)

Let us consider the overfield $K^{\prime}$ of $K$ that is defined as follows. If $\fchar(K)>0$ then $K^{\prime}=K_2$. If $\fchar(K)=0$ then $K^{\prime}$ is the compositum $K_2 k$ of $K_2$ and the imaginary quadratic field $k$; in particular, $K^{\prime}$ contains $k$.

Since $\A_n=\Gal(K^{\prime}(\RR_f)/K^{\prime})$ is simple nonabelian, the field extension $K^{\prime}(\RR_f)/K^{\prime}$ does not contain nontrivial abelian subextensions. However, $j_T \in K^{\prime}(\RR_f)$ and the field (sub)extension $K^{\prime}(j_T)/K^{\prime}$ is abelian. This implies that this subextension is trivial, i.e., $j_T\in K^{\prime}$. This means that for all $\sigma \in \Gal(K^{\prime}(\RR_f)/K^{\prime})=\A_n$ $$j_T=\sigma j_T=j_{\sigma T}.$$ Since $n \ge 8$, the permutatation group $\A_n$ is $4$-transitive and therefore the jacobians $J(C_{f_T})$’s are mutually isomorphic over $\K$ for all $4$-element subsets $T\subset \RR_f$.

Let $T_1$ and $T_2$ be two [*disjoint*]{} $4$-element subsets of $\RR_f$. (Since $n\ge 8$, such $T_1$ and $T_2$ do exist.) Applying Theorem \[homoG\] to $T_1$ and $T_2$ (instead of $T$ and $S$) and taking into account that $J(C_{f_{T_1}})$ and $J(C_{f_{T_2}})$ are isomorphic over $\K$ (i.e., $\Hom(J(C_{f_{T_1}}),J(C_{f_{T_2}}))\ne\{0\}$), we conclude that $\fchar(K)>0$ and both $J(C_{f_{T_1}})$ and $J(C_{f_{T_2}}))$ are supersingular elliptic curves.

\[ellipticZ1\] Suppose that $\fchar(K)\ne 2$. Let $a$ be an arbitrary element of $K$. Let $T \subset \RR_f$ be a $3$-element subset. Let us consider the corresponding elliptic curve $$C^{a}_{f_T}:y^2=(x-a)f_T(x)$$ and its jacobian $J(C^{a}_{f_T})$. If $n \ge 6$ then one of the following conditions holds:

-   $\End(J(C^{a}_{f_T}))=\Z$ for all $T$.

-   $\fchar(K)>0$ and all $J(C^{a}_{f_T})$’s are supersingular elliptic curves mutually isomorphic over $\K$.

Let $j_{T,a}$ be the $j$-invariant of the elliptic curve $J(C^{a}_{f_T})$. Clearly, $$j_{T,a} \in K(T)\subset K(\RR_f)$$ and $$j_{\sigma T,a}=\sigma j_{T,a} \ \forall \sigma\in
\Gal(K(\RR_f)/K)=\Gal(f).$$ Suppose that $J(C^{a}_{f_T})$ admits complex multiplication. Then, as in the proof of Theorem \[ellipticZ\], there exists an overfield $K^{\prime}\supset K_2$ such that either $K^{\prime}=K_2$ or $K^{\prime}$ is a quadratic extension of $K_2$ and in both cases $K^{\prime}(j_{T,a})\subset
K^{\prime}(\RR_f)$ and the field (sub)extension $K^{\prime}(j_{T,a})/K^{\prime}$ is abelian. Again, $\A_n=\Gal(K^{\prime}(\RR_f)/K^{\prime})$ is simple nonabelian and therefore there are no nontrivial abelian subextensions of $K^{\prime}(\RR_f)/K^{\prime}$. This implies that $j_{T,a}\in K^{\prime}$, i.e., for all $\sigma \in
\Gal(K^{\prime}(\RR_f)/K^{\prime})=\A_n$ $$j_{T,a}=\sigma j_{T,a}=j_{\sigma T, a}.$$

Since $n \ge 6$, the permutatation group $\A_n$ is $3$-transitive and therefore the jacobians $J(C^{a}_{f_T})$’s are mutually isomorphic over $\K$ for all $3$-element subsets $T\subset \RR_f$. Let $T_1$ and $T_2$ be two [*disjoint*]{} $3$-element subsets of $\RR_f$. (Since $n\ge 6$, such $T_1$ and $T_2$ do exist.) Applying Theorem \[homoGab\] to $T_1,a$ and $T_2,a$ (instead of $T,a$ and $S,b$) and taking into account that $J(C^{a}_{f_{T_1}})$ and $J(C^{a}_{f_{T_2}})$ are isomorphic over $\K$ (i.e., $\Hom(J(C_{f_{T_1}}),J(C_{f_{T_2}}))\ne\{0\}$), we conclude that $\fchar(K)>0$ and both $J(C^{a}_{f_{T_1}})$ and $J(C^{a}_{f_{T_2}}))$ are supersingular elliptic curves. The rest is clear.

Hodge groups of hyperelliptic jacobians {#hyperhodge}
=======================================

We keep the notation and assumptions of Sections \[Galois\] and \[homohyper\]. Also we assume that $K\subset \C$.

We say (as in [@MZ2 Sect. 1.8]) that a complex abelian variety $X$ satisfies property (D) if every Hodge class on each self-product $X^r$ of $X$ can be presented as a linear combination of products of divisor classes. If this condition is satisfied then the Hodge conjecture is true for all $X^r$.

Abelian varieties that satisfy (D) are also called [*stably nondegenerate*]{} [@hazamaT]; see also [@murty].

\[DE\] If $Y$ is an elliptic curve over $\C$ with $\End(Y)=\Z$ then it is well known [@MZ2 Th. 0.1(iv)] that $Y$ satisfies (D) and $\Hdg(Y)=\Sp(\H_1(Y,\Q))$.

\[nonsimple\] Let $X_1$ and $X_2$ be complex abelian varieties of positive dimension and $X=X_1\times X_2$. Suppose that $$\End(X_1)=\Z, \ \End(X_2)=\Z, \ \Hom(X_1,X_2)=\{0\}.$$

Then:

1.  $\End(X_1\times X_2)=\Z\oplus \Z$.

2.  If both $X_1$ and $X_2$ satisfy (D) then $\Hdg(X)= \ \Hdg(X_1)\times
    \Hdg(X_2)$ and $X$ satisfies (D).

\(i) is obvious. (ii) follows from [@Hazama Th. 0.1 and Prop. 1.8] (see also Theorem 3.2(i) of [@MZ2]).

Since $\End(X_i)=\Z$ and $X_i$ satisfies (D), $$\Hdg(X_i)=\Sp(H_1(X_i,\Q)$$ [@hazamaT; @murty]. (See also [@MZ2 Sect. 1.8].)

\[jacprym\] Suppose that $T$ is a subset of $\RR_f$ with $\#(T)\ge 5$. Let us consider the hyperelliptic curve $C_{f_T}:y^2=f_T(x)$ and its jacobian $J(C_{f_T})$. Then $\End(J(C_{f_T}))=\Z$ and $\Hdg(J(C_{f_T}))=\Sp(
\H_1(J(C_{f_T}),\Q))$. In addition, $J(C_{f_T})$ satisfies (D).

Let us put $m=\#(T)$. We have $m \ge 5$ and $\deg(f_T)=m\ge 5$.

By Remark \[oneT\], there exists a (sub)field $$E_0\subset K(\RR_f)\subset\K\subset \C$$ such that $f_T(x)\in E_0(T)$ and the Galois group of $f_T(x)$ over $E_0$ is $\Perm(T)\cong \ST_m$. Now the result follows from Theorem \[jacobian\] applied to $m,E_0, f_T(x)$ instead of $n,K,f(x)$.

\[jacfam\] Suppose that $T$ is a subset of $\RR_f$ with $\#(T)\ge 5$. Suppose that $m$ is odd and let $a$ be an arbitrary element of $K$. Let us consider the hyperelliptic curve $C^{a}_{f_T}:y^2=(x-a)f_T(x)$ and its jacobian $J(C^{a}_{f_T})$. Then $\End(J(C^{a}_{f_T}))=\Z$ and $\Hdg(J(C^{a}_{f_T}))=\Sp( \H_1(J(C^{a}_{f_T}),\Q))$. In addition, $J(C^{a}_{f_T})$ satisfies (D).

By Remark \[odd\], there exists a field $$E_0\subset
K(\RR_f)\subset\K\subset\C$$ and a degree $m$ irreducible polynomial $h(x)\in E_0[x]$ such that $$E(\RR_h)=E(\RR_{f_T}), \
\Gal(E(\RR_h)/E)=\Gal(E(\RR_{f_T})/E)=\ST_m$$ and $C^{a}_{f_S}$ is birationally $E$-isomorphic to to the hyperelliptic curve $C_{h}:y^2=h(x)$. Clearly, the jacobians $J(C^{a}_{f_S})$ and $J(C_h)$ are isomorphic. It follows from Lemma \[jacprym\] applied to $m,E_0,h(x)$ (instead of $n,K,f(x)$) that $\End(J(C_{h}))=\Z$, $\Hdg(J(C_h)=\Sp( \H_1(J(C_h),\Q))$ and $J(C_{h})$ satisfies (D). Since $J(C^{a}_{f_S})$ and $J(C_h)$ are isomorphic, we are done.

Prym varieties {#hyperprym}
==============

Following [@MumfordP; @Dal], let us give an explicit description of hyperelliptic prymians $P$, assuming that $\fchar(K)\ne 2$. Suppose that $n=2g+2\ge 6$ and $$f(x)\in
K[x]\subset \K[x]$$ is a degree $n$ polynomial without multiple roots. Let us split the $n$-element set $\RR_f$ of roots of $f(x)$ into a [*disjoint*]{} union $$\RR_f =\RR_1 \sqcup \RR_2$$ of [*non-empty*]{} sets $\RR_1$ and $\RR_2$ of [*even*]{} cardinalities $n_1$ and $n_2$ respectively. Further we assume that $$n_1 \ge n_2\ge 2.$$ and put $$f_1(x)=\prod_{\alpha\in\RR_1}(x-\alpha), \
f_2(x)=\prod_{\alpha\in\RR_2}(x-\alpha).$$ We have $n_1+n_2=n$ and define nonnegative integers $g_1$ and $g_2$ by $$n_1=2g_1+2, \ n_2=2g_2+2.$$ Clearly, $$g_1+g_2=g-1.$$ Let us consider the hyperelliptic curves $C_{f_1}:y^2=f_1(x)$ and $C_{f_2}:y^2=f_2(x)$ of genus $g_1$ and $g_2$ respectively and corresponding hyperellptic jacobians $J(C_{f_1})$ and $J(C_{f_2})$ of dimension $g_1$ and $g_2$ respectively. Then the prymians $P$ of $C_f: y^2=f(x)$ are just the products $J(C_{f_1})\times J(C_{f_2})$ for all the partitions $\RR_f =\RR_1 \sqcup \RR_2$.

Now Theorem \[main\] becomes an immediate corollary of the following statement.

\[help\] Suppose that $n=2g+2\ge 8$, $K\subset\C$ and $\Gal(f)=\ST_n$. Let us put $P=J(C_{f_1})\times J(C_{f_2})$. Then:

-   $\Hom(J(C_{f_1}),J(C_{f_2}))=\{0\}, \
    \Hom(J(C_{f_2}),J(C_{f_1}))=\{0\}$.

-   Suppose that $g_i \ge 1$, i.e., $n_i \ge 4$. Then $$\End(J(C_{f_i}))=\Z, \ \Hdg(J(C_{f_i}))=\Sp(
    \H_1(J(C_{f_i}),\Q))$$ and $J(C_{f_i})$ satisfies (D).

-   If $n_2=2$ then $P=J(C_{f_1})$. In particular, $\End(P)=\Z$, $\Hdg(P)=\Sp( \H_1(P,\Q))$ and $P$ satisfies (D).

-   If $n_2 \ge 4$ then $\End(P)=\Z\oplus\Z$, $$\Hdg(P)=\Hdg(J(C_{f_1}))\times \Hdg(J(C_{f_2}))=\Sp(
    \H_1(J(C_{f_1}),\Q))\times \Sp( \H_1(J(C_{f_2}),\Q))$$ and $P$ satisfies (D).

The assertion (i) follows from Theorem \[homoG\].

If $n_i\ge 6$ then (ii) follows from Lemma \[jacprym\]. Suppose that $n_i=4$. Then $J(C_{f_i})$ is an elliptic curve. It follows from Theorem \[ellipticZ\] that $\End(J(C_{f_i}))=\Z$. Now the assertion about its Hodge group and property (D) follows from Example \[DE\]. This completes the proof of (ii).

Let us prove (iii). If $n_2=2$ then $J(C_{f_2})=0$ and therefore $P=J(C_{f_1})$. Now the assertion follows from (ii).

Let us prove (iv). We assume that $$n_1 \ge n_2 \ge 4.$$ By already proven (i) and (ii), $$\End(J(C_{f_1}))=\Z, \ \End(J(C_{f_2}))=\Z,  \
 \Hom(J(C_{f_1}),J(C_{f_2}))=\{0\}.$$ Now (iv) follows from Theorem \[nonsimple\] applied to $X_1=J(C_{f_1})$, $X_2=J(C_{f_2})$ and $X=P$.

Theorem \[main1\] is an immediate corollary of the following statement.

\[help1\] Suppose that $n=2g+2\ge 10$, $K\subset\C$ and $f(x)=(x-a)h(x)$ where $a\in K$ and $h(x)\in K[x]$ is an irreducible degree $(n-1)$ polynomial with $\Gal(h)=\ST_{n-1}$. Let us put $P=J(C_{f_1})\times J(C_{f_2})$. Then:

-   $\Hom(J(C_{f_1}),J(C_{f_2}))=\{0\}, \
    \Hom(J(C_{f_2}),J(C_{f_1}))=\{0\}$.

-   Suppose that $g_i \ge 1$, i.e., $n_i \ge 4$. Then $$\End(J(C_{f_i}))=\Z, \ \Hdg(J(C_{f_i}))=\Sp(
    \H_1(J(C_{f_i}),\Q))$$ and $J(C_{f_i})$ satisfies (D).

-   If $n_2=2$ then $P=J(C_{f_1})$. In particular, $\End(P)=\Z$, $\Hdg(P)=\Sp( \H_1(P,\Q))$ and $P$ satisfies (D).

-   If $n_2 \ge 4$ then $\End(P)=\Z\oplus\Z$, $$\Hdg(P)=\Hdg(J(C_{f_1}))\times \Hdg(J(C_{f_2}))=\Sp(
    \H_1(J(C_{f_1}),\Q))\times \Sp( \H_1(J(C_{f_2}),\Q))$$ and $P$ satisfies (D).

Clearly, $\RR_f=\RR_h \sqcup\{a\}$; in particular, $a$ belongs to precisely one of $\RR_1$ and $\RR_2$. Suppose that $a$ lies in $\RR_j$ and does [*not*]{} belong to $\RR_k$ and put $$T=\RR_k\subset\RR_h, \ S=\RR_j\setminus \{a\}\subset\RR_h.$$ Now the assertion (i) follows from Corollary \[homoGt\] applied to $h(x)$ (instead of $f(x)$).

If $n_k\ge 6$ then the assertion (ii) for $J(C_{f_k})$ follows from Lemma \[jacprym\]. Suppose that $n_k=4$, i.e., $J(C_{f_k})$ is an elliptic curve. Then it follows from Theorem \[ellipticZ\] applied to $m=n-1\ge 9$ and $h(x)$ (instead of $f(x)$) that $\End(J(C_{f_k}))=\Z$. Now the assertion about its Hodge group and property (D) follows from Example \[DE\].

If $n_j\ge 6$ then the assertion (ii) for $J(C_{f_j})$ follows from Lemma \[jacfam\]. Suppose that $n_j=4$, i.e., $J(C_{f_j})$ is an elliptic curve. Then it follows from Theorem \[ellipticZ1\] applied to $m=n-1$ and $h(x)$ (instead of $f(x)$) that $\End(J(C_{f_j}))=\Z$. Now the assertion about its Hodge group and property (D) follows from Example \[DE\]. This ends the proof of (ii).

The proof of the remaining assertions (iii) and (iv) goes literally as the proof of the corresponding assertions of Theorem \[help\].

Let us take $K=\Q$ and $f_n(x)=x^n-x-1$. It is known [@SerreGalois p. 42] that $\Gal(f_n)=\ST_n$. Let $a$ be a rational number. Suppose that $n=2g+2$ and let us consider the hyperelliptic genus $g$ curves $C_{f_n}: y^2=f_n(x)$ and $C^{a}_{f_{n-1}}:
y^2=(x-a) f_{n-1}(x)$. Then:

-   If $n=2g+2\ge 8$ then all $(2^{2g}-1)$ Prym varieties $P$ of $C_{f_{n}}$ satisfy (D). Among them there are exactly $n(n-1)/2$ complex abelian varieties with $\End(P)=\Z$; for all others $\End(P)=\Z\oplus \Z$.

-   If $n=2g+2\ge 10$ then all $(2^{2g}-1)$ Prym varieties $P$ of $C^{a}_{f_{n-1}}$ satisfy (D). Among them there are exactly $n(n-1)/2$ complex abelian varieties with $\End(P)=\Z$; for all others $\End(P)=\Z\oplus \Z$.

Let $z_1, \dots , z_n$ be algebraically independent (transcendental) complex numbers and $L=\Q(z_1, \dots , z_n)\subset \C$ the corresponding subfield of $\C$, which is isomorphic to the field of rational functions in $n$ variables over $\Q$. Let $K\subset L$ be the (sub)field of symmetric rational functions. Then $$f(x)=\prod_{i=1}^n (x-z_i)\in K[x], \ \RR_f=\{z_1, \dots , z_n\}, \
 \Gal(f)=\ST_n.$$ Suppose that $n=2g+2$ and let us consider the hyperelliptic genus $g$ curve $C_f:y^2=f(x)$.

If $g\ge 3$ (i.e., $n\ge 8$) then all $(2^{2g}-1)$ Prym varieties $P$ of $C_{f}$ satisfy (D). Among them there are exactly $n(n-1)/2$ complex abelian varieties with $\End(P)=\Z$; for all others $\End(P)=\Z\oplus \Z$.

If $n=6$ (i.e., $g=2$) then all fifteen Prym varieties $P$ are elliptic curves $y^2=\prod_{z\in T}(x-z)$ where $T$ is a $4$-element subset of $\{z_1, \dots , z_6\}$. The algebraic independence of $z_1, \dots , z_6$ implies that the $j$-invariants of these elliptic curves are transcendental numbers and therefore all $P$ have no complex multiplication, i.e., $\End(P)=\Z$.

The property (D) and equality $\End(P)=\Z$ for [*general*]{} (not necessarily unramified) Prym varieties $P$ of arbitrary smooth projective curves were proven in [@Biswas].

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