Proof. The statement on the essential image is obvious. The functor is fully faithful, because both $\mathrm{Hom}(k^n,k^m) $ and $\mathrm{Hom}(\mathscr {O}_X^{\oplus n},\mathscr {O}_X^{\oplus m})$ are given by $m\times n$ -matrices $(\varphi _{ij})$ with entries from $\mathrm{Hom}(k,k)=k=\Gamma (X,\mathscr {O}_X)=\mathrm{Hom}(\mathscr {O}_X,\mathscr {O}_X)$ . The functor is exact since the structure map $X\rightarrow \mathrm{Spec}(k)$ is flat.

Proof. We have $h^0(\mathscr {E}) = \dim _{k'}H^0(X',\mathscr {E}')=\dim _{k'} H^0(X',\mathscr {O}_{X'})\cdot r=r$ . Choose a basis $s_1,\ldots ,s_r\in H^0(X,\mathscr {E})$ , and consider the resulting homomorphism of quasicoherent sheaves

$$ \begin{align*}k^r\otimes_k\mathscr{O}_X\stackrel{s_1,\ldots,s_r}{\longrightarrow} H^0(X,\mathscr{E})\otimes_k\mathscr{O}_X\stackrel{\mathrm{can}}{\longrightarrow} \mathscr{E}. \end{align*} $$

The map on the left is bijective by construction, and the map on the right becomes bijective after base-change to $k'$ . It follows that both maps are bijective, hence $\mathscr {E}$ is free of rank r.

Proof. Set $r=\mathrm{rank}(\mathscr {E})$ and $V=H^0(X,\mathscr {E})$ . The canonical map $ V\otimes _k\mathscr {O}_X\rightarrow \mathscr {E}$ is surjective, and the same holds for its pullback to $X'$ . By Lemma 2.2, there is a vector subspace $U\subset V$ such that the pullback of $\varphi :U\otimes _k\mathscr {O}_X\rightarrow \mathscr {E}$ is bijective. We may regard $s=\det (\varphi )$ as a section for the invertible sheaf $\mathscr {L}=\det (\mathscr {E})$ , and the task is to verify that it has no zeros. Let $a\in X$ be a point and $a'\in X'$ be some preimage. By construction, $s(a')=s(a)\otimes \kappa (a')$ does not vanish, hence $s(a)\neq 0$ .

Proof. First note that the Kummer sequence $0\rightarrow \mu _{\ell ^\nu }\rightarrow \mathbb {G}_m\stackrel {\ell ^\nu }{\rightarrow } \mathbb {G}_m\rightarrow 0$ yields an exact sequence

(3) $$ \begin{align} 0\longrightarrow k^\times/k^{\times \ell^\nu} \longrightarrow H^1(X,\mathbb{Z}/\ell^\nu\mathbb{Z}(1))\longrightarrow \mathrm{Pic}(X)[\ell^\nu]\longrightarrow 0. \end{align} $$

Next recall that the connected component $P=\mathrm{Pic}^\tau _{X/k}$ is a group scheme of finite type. It is actually proper (see, e.g., [Reference Laurent and Schröer40, Proposition 2.3]) hence an extension of a finite group scheme G by an abelian variety A. We thus have exact sequences $0\rightarrow A[\ell ^\nu ]\rightarrow P[\ell ^\nu ]\rightarrow G[\ell ^\nu ]$ of group schemes and obtain exact sequences

$$ \begin{align*}0\longrightarrow \varprojlim_\nu A[\ell^\nu](k) \longrightarrow \varprojlim_\nu P[\ell^\nu](k)\longrightarrow \varprojlim_\nu G[\ell^\nu](k) \end{align*} $$

of groups, where the terms on the right are finite. To proceed, we may assume that k is algebraically closed. Then the term on the left is a free $\mathbb {Z}_\ell $ module of rank $2\dim (A)$ . The term in the middle can be identified with $H^i(X, \mathbb {Z}_\ell (1))$ by (3), and the assertion follows.

Proof. Both schemes are étale over X, hence the morphism is étale by [Reference Grothendieck27, Proposition 17.3.4]. Let $N\subset \Pi $ be the open normal subgroup corresponding to the composite map $X"\rightarrow X$ , and $\Gamma "\subset \Gamma $ be its image. Then $X\otimes k"$ corresponds to the subgroup $\Pi "=\Pi \times _\Gamma \Gamma "$ of $\Pi $ , and $X"\rightarrow X\otimes _kk"$ corresponds to $N\subset \Pi "$ . The first statement follows. The statement on $H^0(X",\mathscr {O}_{X"})$ is trivial, and the statement on $H^0(X\otimes k",\mathscr {O}_{X\otimes k"})$ follows from $h^0(\mathscr {O}_X)=1$ .

Proof. The implication (i) $\Rightarrow $ (ii) is trivial. For the converse, let $k\subset k'$ be any field extension, set $A'=A\otimes k'$ , and consider the ordered set of connected subgroup schemes $N^{\prime }_\lambda \subset A'$ , $\lambda \in L$ such that $A'/N^{\prime }_\lambda $ is supersingular. For any two members $N^{\prime }_\lambda $ and $N^{\prime }_\mu $ , the intersection $(N^{\prime }_\lambda \cap N^{\prime }_\mu )^0$ belongs to the family, because the class of supersingular abelian varieties is stable under products and subvarieties. Note that the same goes for $(N^{\prime }_\lambda \cap N^{\prime }_\mu )^0_{\mathrm{red}}$ , provided that this is a subgroup scheme. In any case, we see that our collection contains a smallest member $N^{\prime }_0\subset A'$ . Obviously, this is stable under the action of $G=\mathrm{Aut}(k'/k)$ .

For $k'=k^{\mathrm{alg}}$ , we have by assumption an abelian subvariety $B'\subset A'$ that admits a non-zero supersingular quotient $B'/M'$ . According to the Poincaré complete reducibility theorem, there is a finite subgroup scheme $H'\subset B'$ such that $B'/H'\subset A'/H'$ admits a retraction. Using the latter, we obtain a surjection $A'\rightarrow B'/(M'+H')$ and see that $A'/N^{\prime }_0$ is non-zero. Via Galois descent, we see that the base-change to the perfect closure $k^{\mathrm{perf}}$ admits a non-zero supersingular quotient.

Changing notation, we find a finite purely inseparable field extension $k\subset k'$ with a non-zero supersingular quotient $f':A\otimes k'\rightarrow B'$ . The corresponding homomorphism $f:A\rightarrow \mathrm{Res}_{k'/k}(B')$ factors over the abelian variety $\bar {A}=\mathrm {Im}(f)$ . Pulling back, we obtain

$$ \begin{align*}A\otimes k'\longrightarrow \bar{A}\otimes k'\longrightarrow \mathrm{Res}_{k'/k}(B)\otimes k'\longrightarrow B. \end{align*} $$

The map from $\bar {A}\otimes k'$ to B is surjective, because $A\otimes k'\rightarrow B$ is surjective, whereas the map to $\mathrm{Res}_{k'/k}(B)\otimes k'$ is a closed embedding, because this holds for $\bar {A}\rightarrow \mathrm{Res}_{k'/k}(B)$ . Using that the kernel for the adjunction $\mathrm{Res}_{k'/k}(B)\otimes k'\rightarrow B$ is affine [Reference Conrad, Gabber and Prasad17, Appendix A, Proposition 5.11], we conclude that $\bar {A}\otimes k'\rightarrow B$ is an isogeny and infer that $\bar {A}$ is supersingular.

For the remaining statement, suppose that $B=A/N$ is a supersingular quotient. Passing to a further quotient, we may assume that B is supersingular elliptic curve, so the canonical map $H^1(B^{(p)},\mathscr {O}_{B^{(p)}})\rightarrow H^1(B,\mathscr {O}_B)$ induced by the relative Frobenius, which is identical to the Verschiebung on $\mathrm{Pic}^0_{B/k}=B$ , vanishes. The cup product $\Lambda ^g H^1(A,\mathscr {O}_A)\rightarrow H^g(A,\mathscr {O}_A)$ is bijective (see, e.g., [Reference Rössler and Schröer53, Proposition 2.3]). Passing to the Stein factorization for the projection $A\rightarrow B$ , we may furthermore assume that the canonical map $H^1(B,\mathscr {O}_B)\rightarrow H^1(A,\mathscr {O}_A)$ is injective, and the same for Frobenius pullbacks. Choose a non-zero $\epsilon _1\in H^1(B^{(p)},\mathscr {O}_{B^{(p)}})$ and extend it to a basis $\epsilon _1,\ldots ,\epsilon _g\in H^1(A^{(p)},\mathscr {O}_{A^{(p)}})$ . The relative Frobenius vanishes on the generator $\epsilon _1\cup \dots \cup \epsilon _g$ of $H^g(A^{(p)},\mathscr {O}_{A^{(p)}})$ , because it vanishes the first factor.

Proof. Assertion (ii) is easy: If the order of h is not a p-power, we may pass to some multiple of h and assume that the order is prime to p. Then the h-action on $\mathscr {O}_{A,e}/\mathfrak {m}^{n+1}$ is semisimple, but trivial on the graded pieces $\mathfrak {m}^i/\mathfrak {m}^{i+1}=\mathrm {Sym}^i(\mathfrak {m}/\mathfrak {m}^2)$ . In turn, the action on $\mathscr {O}_{A,a}^\wedge $ and hence on X is trivial, contradiction.

The key idea for the other statements to understand the case where the ground field is a prime field, and then reduce to this situation via standard arguments. We proceed in three steps.

Step 1: Suppose that $k=\mathbb {F}_p$ is the prime field. First note that $A^{(p)}=A$ and that the relative and absolute Frobenius maps for A coincide. We thus have a short exact sequence $0\rightarrow A[F]\rightarrow A\stackrel {F}{\rightarrow }A\rightarrow 0$ . Also note that this sequence is functorial. Thus, F belongs to the center of the associative algebra $\mathrm {End}(A)$ , and every endomorphism stabilizes the Frobenius kernel $A[F]$ .

Replacing our group element $h\in \mathrm{Aut}(A)$ of finite order by a suitable power, we may assume $\mathrm{ord}(h)=p$ . The difference $ {\mathrm{id}}-h$ induces the zero map on $\mathrm{Lie}(A)$ . In light of the Demazure–Gabriel correspondence [Reference Demazure and Gabriel20, Chapter II, Section 7, Theorem 3.5], this actually means $A[F]\subset \mathrm{Ker}({\mathrm{id}}-h)$ . The isomorphism theorem gives an endomorphism $f:A\rightarrow A$ with ${\mathrm{id}}-h =f\circ F$ , or equivalently,

$$ \begin{align*}h={\mathrm{id}} -f\circ F. \end{align*} $$

Fix a complex embedding $\mathbb {Q}_\ell ^{\mathrm{alg}}\subset \mathbb {C}$ and consider the effect of f and F on the vector space $V=H^1(A\otimes k^{\mathrm{alg}},\mathbb {Q}_\ell )\otimes \mathbb {C}$ , which has dimension $2g$ . Since f and F commute, there is a basis $e_1,\ldots ,e_{2g}\in V$ such that the resulting matrices for $f^*$ and $F^*$ are both lower triangular [Reference Bourbaki9, Chapter VII, Section 5, No. 9, Proposition 19]. Let $\alpha _1,\ldots ,\alpha _{2g}$ and $\beta _1,\ldots ,\beta _{2g}$ be the matrix entries on the diagonal for $f^*$ and $F^*$ , respectively. In turn, the matrix for $h^*$ is lower triangular as well, and $1-\alpha _i\beta _i$ are its diagonal entries. These are the eigenvalues for $h^*$ , all of which are pth roots of unity. If all of them are trivial, then $h^*$ is the identity, because it is diagonalizable, and hence h is the identity [Reference Mumford46, Section 18, Theorem 3], contradiction.

Fix an index $1\leq d\leq 2g$ for which the pth root of unity $\zeta = 1-\alpha _d\beta _d$ is primitive. Since the characteristic polynomial for $h^*$ belongs to $\mathbb {Z}[T]$ , all conjugates of $\zeta $ appear among the $1-\alpha _i\beta _i$ . Choose indices $i_1,\ldots ,i_{p-1}$ so that

$$ \begin{align*}\{1-\alpha_{i_1}\beta_{i_1}, \ldots, 1-\alpha_{i_{p-1}}\beta_{i_{r-1}}\}=\{\zeta^1,\ldots,\zeta^{p-1}\}. \end{align*} $$

Using this and the cyclotomic polynomial $P(T)=T^{p-1}+\cdots + 1 = \prod _{j=1}^{p-1} (T-\zeta ^j)$ for substitutions, we obtain

$$ \begin{align*}p = P(1) = \prod_{j=1}^{p-1}(1-\zeta^j) = \prod_{j=1}^{p-1}\alpha_{i_j}\cdot \prod_{j=1}^{p-1}\beta_{i_j}. \end{align*} $$

Now recall that the $\beta _{i_j}$ are Weil numbers, having $|\beta _{i_j}|=p^{1/2}$ . Taking absolute values from the above equation gives $p=s p^{(p-1)/2}$ , with the factor $s= \prod _{j=1}^{r-1}|\alpha _{i_j}|$ . Each $\alpha =\alpha _{i_j}$ is an algebraic integer, so the same holds for the conjugate $\bar {\alpha }$ and the absolute value $|\alpha |=(\alpha \cdot \bar {\alpha })^{1/2}$ . Suppose now $p\neq 2$ . Then $3-p$ is even, hence $s=p^{(3-p)/2}$ is both a rational number and an algebraic integer, and thus an integer. Uniqueness of prime factorization gives $3-p\geq 0$ , so the only possibility is $p=3$ . This establishes (i).

It remains to verify (iii). Set $N=\mathrm{Ker}(h-{\mathrm{id}})$ and $A'=A/N$ . Recall that by [Reference Yu59, Theorem 2.9], we have to verify that, with respect to all complex embeddings $\mathbb {Q}_\ell \subset \mathbb {C}$ , the eigenvalues of Frobenius on $H^1(A'\otimes k^{\mathrm{alg}},\mathbb {Q}_\ell )\otimes \mathbb {C}$ take the form $\xi \cdot p^{1/2}$ for some root of unity $\xi \in \mathbb {C}^\times $ , and thus are supersingular Weil numbers.

Let us first reduce to the case that N is finite. Write $q:A\to A/N=A'$ for the quotient map. Since the commutative group schemes of finite type form an abelian category, there is a unique monomorphism $i:A'\to A$ such that $i\circ q=h-{\mathrm{id}}$ . Our $h\in \mathrm{Aut}(A)$ induces by construction an automorphism $h'\in \mathrm{Aut}(A')$ of finite order, which satisfies the equation $q\circ (h-{\mathrm{id}})=(h'-{\mathrm{id}})\circ q$ . We claim that $h'-{\mathrm{id}}$ has a finite kernel. Suppose this is not the case. To produce a contradiction, we may replace k by its algebraic closure. Using the surjectivity of $q:A\rightarrow A'$ , we find a closed point $a\in A$ such that $q(a)$ has infinite order and belongs to $\mathrm{Ker}(h'-{\mathrm{id}})$ . Then $(h-{\mathrm{id}})(a)\in N$ , and thus $(h-{\mathrm{id}})^2(a)=0$ . On the other hand, the gcd of $(T-1)^2$ and $T^r-1$ in $\mathbb {Q}[T]$ is $T-1$ , so there is an integer $m\geq 1$ and polynomials $Q(T), R(T)\in \mathbb {Z}[T]$ such that $Q(T)\cdot (T-1)^2+R(T)\cdot (T^r-1)=m(T-1)$ . We conclude

$$ \begin{align*}m\cdot(h-{\mathrm{id}})(a)=(h-{\mathrm{id}})(m\cdot a)=(i\circ q)(m\cdot a)=i(m\cdot q(a))=0. \end{align*} $$

Since i is a closed immersion, we see that $q(a)\in A$ is a torsion point, contradiction. We may thus replace A (resp. h) by $A'$ (resp. $h'$ ) and suppose that N is finite.

From $0=(h-{\mathrm{id}})\circ (h^{p-1}+h^{p-2}+\cdots +{\mathrm{id}}),$ we infer that all eigenvalues $\zeta _i=1-\alpha _i\beta _i$ for $h^*$ are primitive pth roots of unity. Now we can exploit $p=3$ : For every complex embedding $\mathbb {Q}_\ell \subset \mathbb {C,}$ we have $\zeta _i=e^{\pm 2\pi i/3}$ . Set $\omega = e^{2\pi i/6}$ . Using $- e^{2\pi i/3}=\omega ^{-1}$ and $-e^{-2\pi i/3}=\omega $ and $\omega ^{-1}+\omega =1,$ we compute

$$ \begin{align*}|1-\zeta_i|^2=(1-e^{2\pi i/3})(1-e^{-2\pi i/3}) =3, \end{align*} $$

and see $1-\zeta _i = \omega ^{\pm 1}\cdot p^{1/2}$ . In particular, $|1-\zeta _i|=p^{1/2}$ , which coincides with $|\beta _i|=p^{1/2}$ . From $\alpha _i\beta _i = 1-\zeta _i $ , we get $|\alpha _i|=1$ . Since this applies to all complex embeddings, the $\alpha _i\in \mathbb {C}$ are algebraic integers all whose conjugates lie on the unit circle, so they actually must be roots of unity [Reference Washington58, Chapter I, Lemma 1.6]. Setting $\xi _i= \alpha _i^{-1}\omega ^{\pm 1}$ , we get $\beta _i=(1-\zeta _i)/\alpha _i=\xi _i\cdot p^{1/2}$ , as desired.

Step 2: Suppose that the ground field k is finite. Let $k_0=\mathbb {F}_p$ be the prime field. Then $k_0\subset k$ is a finite Galois extension, with cyclic $G=\mathrm{Gal}(k/k_0)$ , say of order $m\geq 1$ . We now form the Weil restriction $A_0=\mathrm{Res}_{k/k_0}(A)$ , which comes with an induced automorphism $h_0=\mathrm{Res}_{k/k_0}(h)$ . By Lemma 2.1, we have

(5) $$ \begin{align} A_0\otimes_{k_0} k = \prod_{\sigma\in G} A_\sigma\quad\mathrm{and}\quad h_0\otimes_{k_0} k=\prod_{\sigma\in G} h_\sigma, \end{align} $$

where $A_\sigma $ denotes the abelian variety $A\otimes _kk$ , with base-change via $\sigma :k\rightarrow k$ . This shows that $A_0$ is para-abelian of dimension $g_0=mg$ . Using the image $e_0\in A_0$ of the origin $e\in A$ , it becomes an abelian variety, with $h_0\in \mathrm{Aut}(A_0)$ . Then $h_0\neq {\mathrm{id}}$ , while the induced action on $\mathrm{Lie}(A_0)$ is trivial, because this holds after base-change to k. Using step 1 with $h_0\in \mathrm{Aut}(A_0)$ , we immediately see $p\leq 3$ . Furthermore, the order of $h_0$ must be a p-power, so, by (5), this also holds for h. Finally, $A^{\prime }_0=A_0/\mathrm{Ker}(h_0-{\mathrm{id}})$ is supersingular, which carries over to

$$ \begin{align*}A^{\prime}_0\otimes_{k_0} k= \prod_{\sigma\in G} A_\sigma/\mathrm{Ker}(h-{\mathrm{id}}). \end{align*} $$

Then each factor, and in particular $A/\mathrm{Ker}(h-{\mathrm{id}})$ , is supersingular.

Step 3: Now the ground field k is general. Let $R_\lambda \subset k$ and $\lambda \in L$ be the ordered set of all subrings that are finitely generated over the prime field $\mathbb {F}_p$ . By [Reference Grothendieck26, Theorem 8.8.2], we find some member $R=R_\lambda $ such that A and h arise from a family of abelian varieties $\mathfrak {A}\rightarrow \mathrm{Spec}(R)$ and some relative automorphism $h:\mathfrak {A}\rightarrow \mathfrak {A}$ , which we denote by the same letter. Set $r=\mathrm{ord}(h_\eta )$ . Since the generic fiber $\mathfrak {A}_\eta $ is schematically dense, we actually have $h^r={\mathrm{id}}_{\mathfrak {A}}$ . Localizing further, we may assume that $\mathrm{Ker}(h^i-{\mathrm{id}}_{\mathfrak {A}})$ , $0\leq i\leq r-1$ are flat. This ensures that, for all $s\in S,$ the element $h_s\in \mathrm{Aut}(\mathfrak {A}_s)$ also has order r. Moreover, the quotient $\mathfrak {B}=\mathfrak {A}/\mathrm{Ker}(h-{\mathrm{id}}_{\mathfrak {A}})$ exists as an algebraic space (see, e.g., [Reference Laurent and Schröer40, Lemma 1.1]), which is actually a family of abelian varieties.

The closed points $s\in S$ are Zariski dense, because S is a Jacobson space, and the residue fields $\kappa (s)$ must by finite, by Hilbert’s Nullstellensatz. Applying step 2 with $h_s\in \mathrm{Aut}(\mathfrak {A}_s)$ , we see that the characteristic satisfies $p\leq 3$ , that the common order r is a p-power, and that the closed fiber $\mathfrak {B}_s$ is supersingular. In turn, for the finite flat group scheme $\mathfrak {B}[p],$ the closed fibers are geometrically connected. By [Reference Grothendieck26, Theorem 9.7.7], the generic fiber is geometrically connected as well, which implies that $\mathfrak {B}_\eta $ is supersingular.

Proof. By Lemma 4.2, for each element $g=(a,h)$ from the set (6), the entry a belongs to the kernel of $1+h+\cdots +h^{r-1}$ , and thus gives a cohomology class $[a]\in H^1(G, A(k))$ . Moreover, each cohomology class arises in this way. The conjugacy action in question is given by $(b,{\mathrm{id}})\cdot (a,h)\cdot (b,{\mathrm{id}})^{-1} = (b + a -h(b), h)$ , which shows that the map $(a,h)\mapsto [a]$ becomes injective when passing to $A(k)$ -conjugacy classes. The statement on the fixed points also follows from the lemma.

Proof. Let $M=A(k).$ Let $M'=M[r^\infty ]$ be the submodule of elements annihilated by some r-powers. The resulting short exact sequence $0\rightarrow M'\rightarrow M\rightarrow M"\rightarrow 0$ of G-modules induces an exact sequence

$$ \begin{align*}H^1(G,M')\longrightarrow H^1(G,M)\longrightarrow H^1(G,M"). \end{align*} $$

The term on the right vanishes, because it is annihilated by $r=|G|$ , while the homothety $M"\stackrel {r}{\rightarrow } M"$ is bijective. Thus, the map on the left is surjective, and the statement follows from the proposition.

Proof. Consider the polynomials $P=T^{r-1}+T^{r-2}+\cdots +1$ and $Q=T-1$ , and write $fP+gQ=r$ with some $f,g\in \mathbb {Z}[T]$ . The addition map in question takes the form $F:\mathrm{Ker} P(h)\times \mathrm{Ker} Q(h)\rightarrow A$ . Using $PQ=T^r-1$ , we also obtain a map

$$ \begin{align*}G:A\longrightarrow\mathrm{Ker} P(h)\times\mathrm{Ker} Q(h), \quad a\longmapsto \left( (gQ)(h)(a), (fP)(h)(a)\right) \end{align*} $$

in the other direction. Then $G\circ F=r=F\circ G$ , and the latter equation ensures surjectivity of F. If $F(a_1,a_2)=0$ , the former equation gives $r(a_1,a_2)=0$ . Identifying the kernel of the addition map with $\mathrm{Ker} P(h)\cap \mathrm{Ker} Q(h)$ , we see that it is isomorphic to a subgroup scheme of $A[r]$ .

Proof. We have $r\geq 2$ because g has no fixed points. Thus, $h\neq {\mathrm{id}}$ , and furthermore the abelian variety $A'=\mathrm {Im}({\mathrm{id}}-h)$ does not contain a by Lemma 4.2. Thus, $0\subsetneqq A'\subsetneqq A$ , hence A is not simple.

Seeking a contradiction, we assume that $A[r]$ is connected. It is then a local Artin scheme with residue field k, so we may also assume that k is algebraically closed. Being successive extensions, the kernels $A[r^\nu ]$ , $\nu \geq 0$ are connected as well. By Proposition 4.4, we may assume that some of these kernels contain a. Thus, $a=0$ , hence $g=(0,h)$ fixes the origin $0\in P$ , contradiction.

Proof. The first statement follows from $\Omega ^1_{(X\times Y)/k}=\mathrm {pr}_1^*(\Omega ^1_{X/k})\otimes \mathrm {pr}_2^*(\Omega ^1_{Y/k})$ . For (ii), it suffices to treat the case that X is irreducible. Since X is smooth over k, the finite extension $k\subset k'$ must be separable. The structure maps $X\rightarrow \mathrm{Spec}(k')\rightarrow \mathrm{Spec}(k)$ induce an exact sequence $\Omega ^1_{k'/k}\otimes _k\mathscr {O}_X\rightarrow \Omega ^1_{X/k}\rightarrow \Omega ^1_{X/k'}\rightarrow 0$ , where the term on the left vanishes. In the exact sequence ${f^*(\Omega ^1_{Y/k})\rightarrow \Omega ^1_{X/k}\rightarrow \Omega ^1_{X/Y}\rightarrow 0}$ , the term on the right vanishes, and the other terms are locally free of the same rank. Thus, the arrow on the left is bijective, so $\Omega ^1_{X/k'}$ is free, and (ii) follows.

Statement (iii) follows from Lemma 2.3. It remains to check (iv). In light of (iii), it suffices to verify that $X_0\otimes _{k_0}k$ is T-trivial. This indeed follows from Lemma 2.1, together with (i).

Proof. Set $m=\dim (X)$ and $n=\dim (Z)$ , and consider the exact sequence

$$ \begin{align*}h^*(\Omega^1_{Z/k})\longrightarrow \Omega^1_{X/k}\longrightarrow \Omega^1_{X/Z}\longrightarrow 0. \end{align*} $$

By assumption, the two terms on the left are free, and the map on the left is injective at the generic point of X. According to Lemma 2.2, the term on the right is free, the map on the left is injective, and the short exact sequence splits, which already gives (iv). Furthermore, $\mathrm{rank}(\Omega ^1_{X/Z})=m-n$ . It then follows from [Reference Hartshorne32, Chapter III, Proposition 10.4] that $h:X\rightarrow Z$ is smooth. If X and Z have the same dimension, the generic fiber is zero-dimensional, hence h is finite, and thus étale, giving (iii).

The smoothness of h ensures that the Stein factorization $f:Y\rightarrow Z$ is étale [Reference Grothendieck25, Remark 7.8.10]. So in the exact sequence $ f^*(\Omega ^1_{Z/k})\rightarrow \Omega ^1_{Y/k}\rightarrow \Omega ^1_{Y/Z}\rightarrow 0$ , the term on the right vanishes, and the map on the left is injective [Reference Grothendieck27, Theorem 17.11.1]. Thus, $\Omega ^1_{Y/k}$ is free, and it follows that Y is a T-trivial variety. Using that $\Omega ^1_{X/Y}$ is free, we infer that the $g^{-1}(y)$ are T-trivial varieties, which gives (ii). Applying the preceding paragraph with $g:X\rightarrow Y$ instead of $h:X\rightarrow Z$ , we see that the former is smooth, which establishes (i).

Proof. For the sake of completeness, we provide an argument: By Proposition 5.3, $h:X\rightarrow P$ must be smooth, and its Stein factorization $f:Q\rightarrow P$ is étale. Using Proposition 5.4, we infer that Q is para-abelian. From the universal property of the Albanese variety, one sees that f admits a section $s:P\rightarrow Q$ . Thus, the field extension $k(P)\subset k(Q)$ is an equality, and it follows that $f:Q\rightarrow P$ is an isomorphism.

Proof. The Albanese map $h:X\rightarrow P$ is smooth, according to Proposition 5.5, hence $\dim (X)\geq \dim (P)$ . The group scheme $\mathrm{Pic}^0_{X/k}$ is proper [Reference Laurent and Schröer40, Proposition 2.3], and thus $\dim (P)=\dim (\mathrm{Pic}_{X/k})$ . Finally, we have $2\dim (\mathrm{Pic}_{X/k}) = b_1(X)$ by Lemma 2.5. Thus,

$$ \begin{align*}b_1(X) = 2\dim(P) \leq 2\dim(X). \end{align*} $$

The outer terms are equal if and only if X and P have the same dimension. In this case, the Albanese map $h:X\rightarrow P$ is étale, by Proposition 5.3, and thus an isomorphism, in light of Proposition 5.5.

Proof. Hodge theory gives $H^1(X(\mathbb {C}),\mathbb {Z})\otimes \mathbb {C}=H^{1,0}(X)\oplus H^{0,1}(X)$ , where the summands on the right have the same dimension. Using $h^0(\Omega ^1_{X/k})=\dim (X),$ one gets $b_1(X)=2\dim (X)$ , and the assertion follows from Proposition 5.6.

Proof. The base-change $X'=X\otimes _kk'$ is a T-trivial variety over $k'$ , and the induced morphism ${f':P\rightarrow X'}$ is finite and surjective and respects the $k'$ -structures. By Lemma 2.6, we may replace X by $X\otimes k'$ and k by $k'$ , and thus may assume P is a para-abelian variety over k. In light of Lemma 3.1, it now suffices to treat the case that k is algebraically closed.

Choose a finite étale covering $Q\rightarrow P$ with connected total space such that the composite map $Q\rightarrow X$ is Galois. Then Q is para-abelian, by the Serre–Lang theorem. Fix a rational point $e_Q\in Q$ , and write $e_P\in P$ for the image point. One may regard the pairs $(Q,e_Q)$ and $(P,e_P)$ as abelian varieties, and the morphism $Q\rightarrow P$ as an isogeny. According to Lemma 3.1, we may replace P by Q, and thus may assume that $P\rightarrow X$ is Galois. Write $G=\mathrm{Aut}(P/X)$ for the Galois group, such that $X=P/G$ . Using Grothendieck’s spectral sequence for equivariant cohomology [Reference Grothendieck22, Theorem 5.2.1], we see that the canonical inclusion $H^0(X,\Omega ^1_{X/k})\subset H^0(P,\Omega ^1_{P/k})^G $ is an equality. From $h^0(\Omega ^1_{X/k})=h^0(\Omega ^1_{P/k}),$ we infer that the induced G-action on $H^0(P,\Omega ^1_{P/k})$ is trivial.

Fix a rational point $e\in P$ . The resulting map $A\rightarrow P$ is an isomorphism, so one may regard A as the pair $(P,e)$ and obtains $\mathrm{Aut}(P)=A(k)\rtimes \mathrm{Aut}(A)$ . The translational part $N=G\cap A(k)$ is normal in G, and the quotient $P/N$ remains para-abelian. So without restriction, we may assume that the projection $G\rightarrow \mathrm{Aut}(A)$ is injective. Note that the group G is non-trivial, because X is not para-abelian.

Fix a non-trivial element $g\in G$ , and write $g=(a,h)$ with $a\in A(k)$ and $h\in \mathrm{Aut}(A)$ . Let $r\geq 2$ be the common order for g and h. Since g and the translation a act trivially on $H^0(A,\Omega ^1_{A/k})$ , the same holds for h. In turn, it also acts trivially on the dual vector space $H^0(A,\Theta _{A/k})$ , and thus belongs to the kernel of the canonical map $\mathrm{Aut}(A)\rightarrow \mathrm{GL}(\mathrm{Lie}(A))$ . By Theorem 4.1, we must have $p\leq 3$ , and $r=\mathrm{ord}(h)$ is some p-power. Using Proposition 4.6, we see that A is not simple and contains a point of order p. For $p=3,$ we also have a supersingular quotient, again by Theorem 4.1.

Proof. We may assume that k is algebraically closed. According to the results of Mehta and Srinivas [Reference Mehta and Srinivas43,Theorem 1], there is a finite étale covering $A\rightarrow X$ by some abelian variety A, which must be ordinary [Reference Mehta and Srinivas43, Theorem 1 and Lemma 1.2]. Seeking a contradiction, we assume that X is not para-abelian. The theorem ensures that the characteristic is $p=3$ , and the abelian variety A fails to be ordinary, contradiction.

Proof. The algebraic space $\mathrm{Pic}^\tau _{X/S}$ is proper by [Reference Laurent and Schröer40, Proposition 2.3], and the same holds for the closed subspace A. For the remaining statements, it suffices to treat the case that S is the spectrum of a complete local Noetherian ring R, with closed point $s\in S$ and separably closed residue field.

To see (i) and (ii), we consider the group scheme $G=\mathrm{Pic}^\tau _{X_s/\kappa (s)}$ , which is of finite type over the residue field $k=R/\mathfrak {m}_R$ . By [Reference Demazure and Gabriel20, Section 3, Theorem 8.2], it sits in a short exact sequence $0\rightarrow N\rightarrow G\rightarrow G^{\mathrm {aff}}\rightarrow 0,$ where $G^{\mathrm {aff}}=\mathrm{Spec}\Gamma (G,\mathscr {O}_G)$ is the affinization, and the kernel N is anti-affine, which means $h^0(\mathscr {O}_N)=1$ . The latter is an extension of an abelian variety $N/H$ by some smooth connected affine group scheme H, according to [Reference Brion and Can10, Lemma 3.1.4]. In our situation, G is proper, whence $G^{\mathrm {aff}}$ is finite and H is trivial. This gives (i) and (ii).

Since A is irreducible, the closed fiber $A_s$ must be connected, according to Hensel’s lemma [Reference Grothendieck27, Theorem 18.5.11]. Since A contains the zero section of P, we thus get $(A_s)_{\mathrm{red}}\subset \mathrm{Pic}_{X_s/\kappa (s)}^\alpha $ , by (i) and (ii). Set $P=\mathrm{Pic}^\tau _{X/S}$ , suppose for the moment that the function $s\mapsto \dim (P_s)$ is constant, and write $g\geq 0$ for the common value. With Chevalley’s semicontinuity theorem [Reference Grothendieck26, Corollary 13.1.5], we get $g\geq \dim (A_s)\geq \dim (A_\eta )=g$ , so A is equi-dimensional. Moreover, the inclusion $(A_s)_{\mathrm{red}}\subset \mathrm{Pic}_{X_s/\kappa (s)}^\alpha $ is an equality, because both schemes are g-dimensional, and the right-hand side is irreducible. This yields (iii). To see (iv), suppose that $P^{\mathrm {aff}}_s$ is reduced. Then $P_s^\alpha $ is a connected component of $P_s$ , and thus the inclusion $P_s^\alpha \subset A_s$ must be an equality. This gives (iv).

It remains to check that $s\mapsto \dim (P_s)$ is constant, which by Chevalley’s semicontinuity theorem boils down to $\dim (P_s)\leq \dim (P_\eta )$ . Fix a prime $\ell $ that does not divide the order of the finite group scheme $P_s^{\mathrm {aff}}$ and $P_\eta ^{\mathrm {aff}}$ , and also differs from the characteristic $p\geq 0$ of the residue field $ R/\mathfrak {m}_R$ . Consider the infinitesimal neighborhoods $X_n=X\otimes R/\mathfrak {m}_R^{n+1}$ . The short exact sequence $0\rightarrow \mathscr {O}_{X_0}\rightarrow \mathscr {O}^\times _{X_{n+1}}\rightarrow \mathscr {O}^\times _{X_n}\rightarrow 1$ induces an exact sequence

$$ \begin{align*}H^1(X_0,\mathscr{O}_{X_0})\longrightarrow \mathrm{Pic}(X_{n+1})\longrightarrow \mathrm{Pic}(X_n)\longrightarrow H^2(X_0,\mathscr{O}_{X_0}). \end{align*} $$

Since $\ell $ is prime to the characteristic of the residue field $R/\mathfrak {m}_R$ , the map in the middle induces a bijection on $\ell $ -torsion elements. It then follows from Grothendieck’s existence theorem [Reference Grothendieck24, Theorem 5.4.1] that the restriction map $\mathrm{Pic}(X)\rightarrow \mathrm{Pic}(X_s)$ indeed induces a bijection on $\ell $ -torsion. So $\dim _{\mathbb {F}_\ell } \mathrm{Pic}(X_s)[\ell ] \leq \dim _{\mathbb {F}_\ell }\mathrm{Pic}(X_\eta )[\ell ]$ . In turn, we have $\dim (P_s)\leq \dim (P_\eta )$ .

Proof. We first check that $A\rightarrow S$ is a family of para-abelian varieties. Being the closure of the regular scheme $A_\eta $ , the total space A contains no embedded components. By Lemma 6.1, the structure morphism $A\rightarrow S$ is proper and equi-dimensional, and the fiber-wise inclusions $\mathrm{Pic}^\alpha _{X_s/\kappa (s)}\subset A_s$ are equalities. According to Kollár’s generalization of Hironaka’s flatness lemma [Reference Kollár38, Corollary 11], the morphism $f:A\rightarrow S$ must be flat, and thus is a family of para-abelian varieties.

Obviously, the zero section $e:S\rightarrow \mathrm{Pic}^\tau _{X/S}$ factors over A. By [Reference Laurent and Schröer40, Proposition 4.3], there is a unique group law that turns A into a family of abelian varieties. The inclusion $A\subset \mathrm{Pic}^\tau _{X/S}$ respects the zero section. Using [Reference Mumford, Fogarty and Kirwan47, Corollary 6.4], one infers that it actually respects the group laws. We already observed that each $A_s$ is the maximal abelian subvariety inside $\mathrm{Pic}^\tau _{X_s/\kappa (s)}$ , so by [Reference Laurent and Schröer40, Theorem 10.2], the relative Albanese map $g:X\rightarrow \mathrm{Alb}_{X/S}$ exists.

Proof. By Cartier’s theorem [Reference Demazure and Gabriel20, Chapter II, Theorem 1.1], group schemes of finite type over the points $s\in S$ are automatically reduced. Lemma 6.1 ensures that the fiber $A_s$ is reduced, so the theorem applies.

Proof. Since $A_\eta $ extends over some dense open subset, it suffices to treat the case that S is the spectrum of a discrete valuation ring R, and $s\in S$ is the closed point. The task is to check that $A_\eta $ has good reduction. For this, we further may assume that R is complete, with separably closed residue field [Reference Bosch, Lütkebohmert and Raynaud8, Section 7.2, Theorem 1].

Fix a prime $\ell>0$ that does not divide the order of the finite group scheme $\mathrm{Pic}^{\mathrm {aff}}_{X_\eta /\kappa (\eta )}$ , and also differs from the characteristic $p\geq 0$ of the residue field $ R/\mathfrak {m}_R$ . The former ensures that the inclusion $A_\eta [\ell ]\subset \mathrm{Pic}^\tau _{X_\eta /\kappa (\eta )}[\ell ]$ is an equality. In light of the Néron–Ogg–Shafarevich criterion [Reference Serre and Tate55, Theorem 1], our task is to verify that the finite étale group scheme $\mathrm{Pic}^\tau _{X_\eta /\kappa (\eta )}[\ell ]$ is constant. Equivalently, the $\mathbb {F}_\ell $ -vector space $\mathrm{Pic}(X_\eta )[\ell ]$ has dimension $2g$ , where $g\geq 0$ is the relative dimension of $\mathrm{Pic}^\tau _{X/S}$ . It thus suffices to verify that the restriction map $\mathrm{Pic}(X)[\ell ]\rightarrow \mathrm{Pic}(X_0)[\ell ]$ is surjective, where $X_0$ is the closed fiber. This indeed follows as in the proof of Lemma 6.1.

Proof. For the statement on U, it suffices to treat the case that S is the spectrum of a discrete valuation ring R, with closed point $s\in S$ . Our task is to verify that the inclusion $X_\eta \subset \mathrm {Dom}(g)$ is strict. For this, we argue as in [Reference Bosch, Lütkebohmert and Raynaud8, Section 2.5, Proposition 5]: Let $\Gamma \subset X\times P$ be the closure of the graph for the Albanese map $g_\eta :X_\eta \rightarrow P_\eta $ . We have to check that the projection $\mathrm {pr}_1:\Gamma \rightarrow X$ is an isomorphism. By fppf descent, we may replace S by X and assume that the structure morphism $f:X\rightarrow S$ has a section. The induced section for P endows it with the structure of a family of abelian varieties. This is actually the Néron model of $P_\eta $ . Since $X\rightarrow S$ is smooth, the Albanese map $g_\eta :X_\eta \rightarrow P_\eta $ extends to a morphism over $S=\mathrm{Spec}(R)$ .

For the remaining statement, we may assume that $\mathrm {Dom}(g)$ surjects onto the normal scheme S, such that $g:X\dashrightarrow P$ is an S-rational map. Each codimension-one point $\zeta \in X$ either belongs to the generic fiber $X_\eta $ , or maps to a codimension-one point $s\in S$ . By the previous paragraph, the rational map $g:X\dashrightarrow P$ is defined at all such $\zeta \in X$ . By the Weil extension theorem [Reference Bosch, Lütkebohmert and Raynaud8, Section 4.4, Theorem 1], the S-rational map g is defined everywhere.

Proof. Let $G_\eta =\mathrm{Pic}^\tau _{A_\eta /\kappa (\eta )}$ be the dual for the abelian variety $A_\eta =\mathrm{Pic}^\alpha _{X_\eta /\kappa (\eta )}$ . By Lemma 6.4, we may assume that it extends to a family of abelian varieties $G\rightarrow S$ . Fix a rational point on $X_\eta $ . The resulting rational point on the Albanese variety yields an identification $P_\eta =\mathrm{Alb}_{X_\eta /\kappa (\eta )}$ with $G_\eta $ . Thus, $P_\eta $ extends to a family $P\rightarrow S$ . By Proposition 6.5, the Albanese map $g_\eta :X_\eta \rightarrow P_\eta $ extends to a morphism $g:X\rightarrow P$ , at least after removing a closed subset $Z\subset S $ of codimension at least two.

Proof. It suffices to treat the case that k is separably closed. To start with, we also assume that k is algebraically closed. Consider the invertible sheaf $\omega _{G/C}=e^*(\Omega ^g_{G/C})$ on the smooth curve C, where $e:C\rightarrow G$ denotes the zero section and $g=\dim (G/C)$ . According to [Reference Faltings and Chai21, Chapter V, Proposition 2.2], it suffices to verify $\deg (\omega _{G/C})\leq 0$ . For this, we may replace C by any finite covering and assume $C=C'$ .

As observed above, the Albanese map $g_\eta :X_\eta \rightarrow G_\eta $ extends to a morphism $g:X\rightarrow G$ . This yields an exact sequence

(8) $$ \begin{align} g^*(\Omega^1_{G/C})\longrightarrow \Omega^1_{X_C/C}\longrightarrow \Omega^1_{X_C/G}\longrightarrow 0. \end{align} $$

The term on the left is locally free, and the term in the middle is free, because $\Omega ^1_{X/B}$ is free. Let $d\geq 0 $ be its rank. The generic fiber $X_\eta $ is a T-trivial variety, hence its Albanese map $g_\eta $ is smooth. So in (8) the map on the left is locally a direct summand over $X_\eta $ , and in particular injective. Pulling back along the section e yields an inclusion $e^*(\Omega ^1_{G/C})\subset \mathscr {O}_C^{\oplus d}$ , and thus $\deg (\omega _{G/C})\leq 0$ .

We still have to consider the situation where k is only supposed to be separably closed. Fix a prime $\ell \geq 3$ different from the characteristic $p\geq 0$ . Let D be the normalization of the reduction for $C^{\mathrm{alg}}=C\otimes k^{\mathrm{alg}}$ . The canonical map $D\rightarrow C^{\mathrm{alg}}$ is a finite universal homeomorphism. Using [Reference Grothendieck30, Expos\’e IX, Theorem 4.10], we find a symplectic level structure $(\mathbb {Z}/\ell \mathbb {Z})^{2g}_C\rightarrow G$ . According to [Reference Faltings and Chai21, Corollary 2 for Theorem 6.7], the stack $\mathscr {A}_{g,\ell }$ of g-dimensional abelian varieties with such a level structure is an algebraic space. Let $C\rightarrow \mathscr {A}_{g,\ell }$ be the classifying map for G. By the previous paragraph, it factors over a singleton after base-change to $k^{\mathrm{alg}}$ , so the same holds over k.

Proof. It suffices to treat the case that the ground field k is algebraically closed. Let $d\geq 0$ be the rank of the local system $R^1f_*(\mathbb {Q}_\ell )$ . Our task is to show that the corresponding representation $\pi _1(B,a)\rightarrow \mathrm{GL}_d(\mathbb {Q}_\ell )$ has finite image. This is trivial for $n=0$ , because then $d=0$ , so we assume $n\geq 1$ . We now introduce particular curves C and $C'$ to form the diagram (7).

By Bertini’s theorem together with the Lefschetz theorem for algebraic fundamental groups ([Reference Grothendieck28, Expos\’e XII, Corollary 3.5] and [Reference Jouanolou36, Chapter I, Theorem 6.3]), there is a smooth curve $C\subset B$ containing the image of $a\in X$ such that the induced map $\pi _1(C,a)\rightarrow \pi _1(B,a)$ is surjective. Choose an algebraic closure of the function field $\kappa (\eta )=k(C)$ . Working with the resulting geometric point $\bar {\eta }$ rather than a as base point, we have to show that the composite map $\rho :\pi _1(C,\bar {\eta })\rightarrow \mathrm{GL}_d(\mathbb {Q}_\ell )$ has finite image.

By Proposition 7.2, the family $g:G\rightarrow C$ is isotrivial. Moreover, the projection $f_C:X_C\rightarrow C$ is smooth. We thus find a finite branched covering $C'\rightarrow C$ such that $G_{C'}=G_0\times C'$ for some abelian variety $G_0$ over k, and that $f_{C'}:X_{C'}\rightarrow C'$ has a section. Choose a section to get the diagram (7). Now the Albanese map $X_{\eta '}\rightarrow G_{\eta '}$ extends to a morphism $X_{C'}\rightarrow G_{C'}$ , and $g_{C'}:G_{C'}\rightarrow C'$ is given by the projection $G_0\times C'\rightarrow C'$ .

To restate our findings in terms of fundamental groups, we lift the geometric point $\bar {\eta } $ to $C'$ . The Proper Base-Change Theorem [Reference Artin, Grothendieck and Verdier2, Expos\’e XII, Theorem 5.1] gives $R^1(f_C)_*(\mathbb {Q}_\ell )=R^1f_*(\mathbb {Q}_l)|_{C'}$ , so $\rho :\pi _1(C,\bar {\eta })\rightarrow \mathrm{GL}_d(\mathbb {Q}_\ell )$ is the representation corresponding to the local system $R^1(f_C)_*(\mathbb {Q}_\ell )$ . By Lemma 7.6, the local systems $R^1(f_{C'})_*(\mathbb {Q}_\ell )$ and $R^1(g_{C'})_*(\mathbb {Q}_\ell )$ are isomorphic at the generic point $\eta '$ , and by construction, $G\otimes _{\kappa (\eta )}\kappa (\eta ')=G_0\otimes _k\kappa (\eta ')$ . Thus, the diagram

(9)

is commutative, where $S=\mathrm{Spec}(k)$ . Of course, S is simply-connected and the lower horizontal map is trivial. In the upper row, the map on the left is injective with image of finite index, and the map on the right is surjective by [Reference Grothendieck30, Expos\’e V, Proposition 8.2]. It follows that the representation $\rho :\pi _1(B,\bar {\eta })\rightarrow \mathrm{GL}_d(\mathbb {Q}_\ell )$ vanishes on $\pi _1(\eta ',\bar {\eta })$ , and thus has finite image.

Proof. It suffices to treat the case that k is algebraically closed. The Leray–Serre spectral sequence for the Albanese map gives an exact sequence

$$ \begin{align*}H^1(B,\mathbb{Q}_\ell)\longrightarrow H^1(X,\mathbb{Q}_\ell)\longrightarrow H^0(B,R^1f_*(\mathbb{Q}_\ell))\longrightarrow H^2(B,\mathbb{Q}_\ell)\longrightarrow H^2(X,\mathbb{Q}_\ell), \end{align*} $$

where $B=\mathrm{Alb}_{X/k}$ . By assumption, the term in the middle is a $\mathbb {Q}_\ell $ -vector space of dimension ${d=\mathrm{rank}(R^1f_*(\mathbb {Q}_\ell ))}$ . The outer maps are injective by Lemma 7.5 below. The map on the left is actually bijective, which follows from Lemma 2.5. Thus, $d=0$ . According to the Proper Base-Change Theorem, all geometric fibers of $f:X\rightarrow B$ have $b_1=0$ .

Proof of Theorem 7.1

Let X be an n-dimensional T-trivial variety, $B=\mathrm{Alb}_{X/k}$ its Albanese variety, and $f:X\rightarrow B$ be the Albanese map. In light of Lemma 2.6, it suffices to treat the case that k is algebraically closed.

The case $n=0$ is trivial, so we assume $n\geq 1$ . Consider the collection of all finite étale cover ${X_\lambda \rightarrow X}$ , $\lambda \in L$ with connected total space. Each $X_\lambda $ is an n-dimensional T-trivial variety, and by Proposition 5.6, we have $b_1(X_\lambda )\leq 2n$ . Replacing X by some $X_\mu $ whose first Betti number attains the largest value, we may assume $b_1(X)=b_1(X_\lambda )$ for all $\lambda \in L$ .

Set $B=\mathrm{Alb}_{X/k}$ , and consider the Albanese map $f:X\rightarrow B$ and the ensuing local system $R^1f_*(\mathbb {Q}_\ell )$ . The latter is isotrivial, according to Proposition 7.3. Choose some finite étale cover $B'\rightarrow B$ on which it becomes constant. Without loss of generality, we may assume that $B'$ is connected. Then $B'$ can be seen as an abelian variety, and $X'=X\times _BB'$ is an n-dimensional T-trivial variety. Let $\mathrm {pr}_2:X'\rightarrow B'$ be the induced projection, and $f':X'\rightarrow \mathrm{Alb}_{X'/k}$ be the Albanese map. Then $\mathrm {pr}_2=g\circ f'$ for some unique $g:\mathrm{Alb}_{X'/k}\rightarrow B'$ . In light of Proposition 5.5, both $\mathrm {pr}_2$ and $f'$ are smooth, hence the same holds for g. By construction,

$$ \begin{align*}\dim(\mathrm{Alb}_{X'/k})=2b_1(X')=2b_1(X)=\dim(B)=\dim(B'). \end{align*} $$

Thus, the smooth proper morphism $g:\mathrm{Alb}_{X'/k}\rightarrow B'$ is finite. Since $\mathrm {pr}_2$ is in Stein factorization, the same holds for g. Thus, g yields an identification $\mathrm{Alb}_{X'/k}=B'$ . By construction, $R^1f^{\prime }_*(\mathbb {Q}_\ell )=R^1f_*(\mathbb {Q}_l)|B'$ is constant. The assertion now follows from Proposition 7.4.

Proof. Without loss of generality, we may assume $k=k^{\mathrm{alg}}$ and $h^0(\mathscr {O}_Z)=1$ . Set $n=\dim (Z)$ . Choose a closed point $\zeta $ in the generic fiber for $g:Y \rightarrow Z$ . By functoriality of cohomology, we may replace Y by the closure of $\zeta $ , and thus may assume that $Y $ is integral and n-dimensional. Let $\alpha \in H^i(Z,\mathbb {Q}_\ell )$ be a non-zero class. By Poincaré duality, there is a class $\beta \in H^{2n-i}(Z,\mathbb {Q}_\ell )$ such that $\alpha \cup \beta \neq 0$ . It thus suffices to treat the case $i=2n$ .

Set $\Lambda =\mathbb {Z}/\ell ^\nu \mathbb {Z}$ . Given an open set U in Y, we can consider cohomology with compact support $H^{2n}_c(U,\Lambda )$ . The inclusion map $i:U\rightarrow Y$ yields a short exact sequence $0\rightarrow i_!(\Lambda _U)\rightarrow \Lambda _Y\rightarrow \Lambda _D\rightarrow 0$ , where $D=Y\smallsetminus U$ . In the ensuing long exact sequence,

$$ \begin{align*}H^{2n-1}(D,\Lambda)\rightarrow H^{2n}(Y,i_!\Lambda)\longrightarrow H^{2n}(Y,\Lambda)\longrightarrow H^{2n}(D,\Lambda), \end{align*} $$

the outer terms vanish for dimension reasons, and the second terms coincides with $H^{2n}_c(U,\Lambda )$ . This gives identification $H^{2n}(Y,\Lambda )=H^{2n}_c(U,\Lambda )$ , compatible with respect to inclusions of open sets.

We now use the trace maps $ {\mathrm{Tr}}_g: R^{2d}g_!\Lambda _U(d)\rightarrow \Lambda _V $ constructed in [Reference Artin, Grothendieck and Verdier2, Expos\’e XVIII, Theorem 2.9]. These satisfy various naturality conditions and are defined for certain morphisms $g:U\rightarrow V$ between separated schemes of finite type. The condition is that $g|U_0$ is flat of relative dimension d on some open set $U_0$ , and the fibers for $g|U\smallsetminus U_0$ have dimension $<d$ .

For some dense open set $V\subset Z,$ the preimage $U=f^{-1}(V)$ is smooth and $g=f|U$ is finite and flat. This yields a commutative diagram

of trace maps. Note that by [Reference Artin, Grothendieck and Verdier2, Expos\’e XVIII, Theorem 2.14] the diagonal arrows are bijective, so the vertical map is bijective as well. The composition ${\mathrm{Tr}}_g \circ g^* $ with the canonical map $g^*:H^{2n}_c(V,\Lambda (n))\rightarrow H^{2n}_c(U,\Lambda (n))$ is multiplication by $\deg (U/V)$ , according to [Reference Artin, Grothendieck and Verdier2, Theorem 2.9].

Passing to the limit with respect to $\nu $ in $\Lambda =\mathbb {Z}/\ell ^\nu \mathbb {Z}$ and tensoring with $\mathbb {Q}_\ell $ , we infer that $g^*: H^{2n}(Z,\mathbb {Q}_\ell (n))\rightarrow H^{2n}(Y,\mathbb {Q}_\ell (n))$ is a bijection of one-dimensional vector spaces over $\mathbb {Q}_\ell $ .

Proof. It suffices to treat the case that k is algebraically closed and to verify that $f^*:H^1(P,\mathbb {Z}_\ell )\rightarrow H^1(Y,\mathbb {Z}_\ell )$ becomes bijective after tensoring with $\mathbb {Q}_\ell $ , and we may replace the coefficient sheaf $\mathbb {Z}_\ell $ by the Tate twist $\mathbb {Z}_\ell (1)$ . Let A be the maximal abelian subvariety in $\mathrm{Pic}^\tau _{Y/k}$ . One may identify A with the maximal abelian subvariety in $\mathrm{Pic}^\tau _{P/k}$ , compare [Reference Laurent and Schröer40, Definition 8.1 and Proposition 8.3]. Arguing as in the proof of Lemma 2.5, we obtain a commutative diagram

where the maps from the left to the right are injective, with finite cokernel. Thus, $f^*\otimes \mathbb {Q}_\ell $ is bijective.

Proof. Fix an ample invertible sheaf $\mathscr {L}_0$ on A. Then $\mathscr {L}=\bigotimes _{\sigma \in G}\sigma ^*(\mathscr {L}_0)$ is ample, and its class in $\mathrm{Pic}(A)$ is G-fixed. Write $A^\vee =\mathrm{Pic}^0_{A/k}$ for the dual abelian variety, and consider the homomorphism

$$ \begin{align*}f:A\longrightarrow A^\vee,\quad a\longmapsto \tau_a^*(\mathscr{L})\otimes\mathscr{L}^\vee, \end{align*} $$

where $\tau _a(x)=x+a$ denotes translation. By contravariance, the G-action on A induces an action of the opposite group $G^{\mathrm{op}}$ on the dual $A^\vee $ , which is converted back to a G-action via $\sigma \cdot \mathscr {N}=(\sigma ^{-1})^*(\mathscr {N})$ . The above map is equivariant with respect to these G-actions: First observe

$$ \begin{align*}(\sigma^{-1})^*\circ\tau_a^* = (\tau_a\circ\sigma^{-1})^*=(\sigma^{-1}\circ\tau_{\sigma(a)})^*=\tau_{\sigma(a)}^*\circ(\sigma^{-1})^* \end{align*} $$

for $\sigma \in G$ and $a\in A(k)$ . Together with $(\sigma ^{-1})^*(\mathscr {L})\simeq \mathscr {L,}$ this gives

$$ \begin{align*}\sigma\cdot f(a)=(\sigma^{-1})^*(\tau_a^*\mathscr{L})\otimes(\sigma^{-1})^*(\mathscr{L}^\vee) = \tau_{\sigma(a)}^* (\mathscr{L}) \otimes\mathscr{L}^\vee=f(\sigma\cdot a). \end{align*} $$

Furthermore, the homomorphism $f:A\rightarrow A^\vee $ has finite kernel, because the invertible sheaf $\mathscr {L}$ is ample.

Now fix a prime $\ell>0$ not dividing $2p\cdot \deg (f)\cdot |G|$ . This has three consequences: First, the induced G-equivariant map $f:A[\ell ]\rightarrow A^\vee [\ell ]$ is an isomorphism. With this identification, the Weil pairing $A[\ell ]\times A^\vee [\ell ] \rightarrow \mu _\ell $ becomes a G-fixed symplectic form on $A[\ell ]$ . Second, Serre’s result ensures that the homomorphism $G\rightarrow \mathrm{Aut}({A[\ell ]})$ is injective ([Reference Grothendieck23, Appendix] or [Reference Mumford46, Section 21, Theorem 5]). Choosing a symplectic basis of $k^{\mathrm{sep}}$ -valued points in the finite étale group scheme $A[\ell ]$ , we obtain an inclusion $G\subset \mathrm {Sp}_{2n}(\mathbb {F}_\ell )$ . Now recall that the finite symplectic group has order

$$ \begin{align*}|\mathrm{Sp}_{2n}(\mathbb{F}_\ell)| = \ell^{n^2}\cdot\prod_{i=1}^n(\ell^{2i}-1) \end{align*} $$

(see, e.g., [Reference Grove31, Theorem 3.1]). As a third consequence, we see that the order $|G|$ must be a divisor of $\prod _{i=1}^n(\ell ^{2i}-1)$ . This holds for almost all primes $\ell>0$ , and the assertion follows from the very definition of ${\mathrm{MN}} (n)$ .

Proof. Without loss of generality, we may assume that $f:A\rightarrow X$ is Galois. Write P for the underlying para-abelian variety of A. The finite group $H=\mathrm{Aut}(P/X)$ is a subgroup of the semi-direct product $\mathrm{Aut}(P)=A(k)\rtimes \mathrm{Aut}(A)$ . Set $N=H\cap A(k)$ . The quotient $B=A/N$ is another abelian variety, and the induced map $g:B\rightarrow X$ is a finite étale Galois covering, with relative automorphism group $G=H/N$ . By construction, we have $G\subset \mathrm{Aut}(B)$ , and Lemma 8.2 ensures that $\deg (B/X)=|G|$ divides the integer ${\mathrm{MN}} (n)$ .

Proof of Theorem 8.1

By assumption, we have a discrete valuation ring R whose residue field ${k=R/\mathfrak {m}_R}$ is separably closed of characteristic $p>0$ , and whose field of fraction $F=\mathrm {Frac}(R)$ contains the rational numbers, together with an algebraic space $\mathfrak {X}$ and a proper flat morphism to $S=\mathrm{Spec}(R)$ whose closed fiber $X=\mathfrak {X}\otimes k$ is smooth of dimension $n\geq 0$ , with $h^0(\mathscr {O}_X)=1$ , and Chern classes $c_1=0$ and $c_2=0$ . Furthermore, $p\geq 2n+2$ . Our task is to find a finite étale covering $A\rightarrow X$ by some para-abelian variety A. Indeed, by Bertini’s theorem, there is a closed point $e\in A$ where the residue field extension $k\subset \kappa (e)$ is separable. The inclusion is an equality, by assumption (iv), so A is indeed an abelian variety.

We also may assume that R is Henselian, and contained in the field of complex numbers. Furthermore, by [Reference Grothendieck30, Expos\’e IX, Theorem 4.10], it suffices to construct the covering after passing to a finite extension $k'$ of our separably closed field k, and we are thus free to replace R by the normalization in some finite extension $F\subset F'$ .

The $\ell $ -adic local systems $R^ih_*(\mathbb {Q}_\ell )$ are constant, because R is strictly Henselian, and it follows that the relative Chern classes $c_i(\Omega ^1_{\mathfrak {X}/S}) $ vanish for $i\leq 2$ . So, the generic fiber $U=\mathfrak {X}\otimes _RF$ and the complex fiber $V=\mathfrak {X}\otimes _R\mathbb {C}$ both satisfy $c_1=0$ and $c_2=0$ . As discussed at the beginning of the section, there is a finite étale covering $V'\rightarrow V$ by some abelian variety $V'$ . In light of the preceding paragraph and [Reference Grothendieck30, Expos\’e X, Corollary 1.8], we may assume that it arises from U by base-change. With Proposition 8.3, we obtain a finite étale Galois covering $U'\rightarrow U$ by some abelian variety $U'$ with the additional property that $d=\deg (U'/U)$ divides the integer ${\mathrm{MN}} (n)$ . From Formula (10) and assumption (iii), we see that $p\nmid d$ , with a possible exception for $p=2n+1$ .

However, the case $p\mid d$ and $p=2n+1$ is also impossible, as the referee pointed out: First note that we may assume that $G=\mathrm{Aut}(U'/U)$ , which belongs to $U'(k)\rtimes \mathrm{Aut}(U')$ , projects injectively to $\mathrm{Aut}(U')$ . Then G contains an element $g=(a,h)$ with $\mathrm{ord}(g)=\mathrm{ord}(h)=p$ . The induced representation of $h^*$ on $H^1(U',\mathbb {Q}_\ell )$ has a primitive pth root of unity as eigenvalue, hence the characteristic polynomial $P\in \mathbb {Q}[T]$ is divisible by $T^{p-1}+T^{p-2} + \cdots +1$ . So $p-1\leq \deg (P) = 2n = p-1$ , and thus $P(T)= T^{p-1}+T^{p-2} + \cdots +1$ . This precludes $\lambda =1$ as eigenvalue, consequently ${\mathrm{id}}-h:U'\rightarrow U'$ must be surjective. By Lemma 4.2, the automorphism $g:U'\rightarrow U'$ has a fixed point, contradiction.

Summing up, $p\nmid d$ . Hence, by the theory of the specialization of the algebraic fundamental group [Reference Grothendieck30, Expos\’e X, Corollary 3.9], we see that $U'\rightarrow U$ extends to some finite étale covering $\mathfrak {A}\rightarrow \mathfrak {X}$ . The schematic closure of the origin $e\in U'(F)$ defines a section $e:S\rightarrow \mathfrak {A}$ . This turns $\mathscr {A}\rightarrow S$ into a family of abelian varieties, by [Reference Mumford, Fogarty and Kirwan47, Theorem 6.14], and the closed fiber yields the desired $A\rightarrow X$ .