The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

38.9 Abelian varieties

An excellent reference for this material is Mumford's book on abelian varieties, see [AVar]. We encourage the reader to look there. There are many equivalent definitions; here is one.

Definition 38.9.1. Let $k$ be a field. An abelian variety is a group scheme over $k$ which is also a proper, geometrically integral variety over $k$.

We prove a few lemmas about this notion and then we collect all the results together in Proposition 38.9.11.

Lemma 38.9.2. Let $k$ be a field. Let $A$ be an abelian variety over $k$. Then $A$ is projective.

Proof. This follows from Lemma 38.8.7 and More on Morphisms, Lemma 36.43.1. $\square$

Lemma 38.9.3. Let $k$ be a field. Let $A$ be an abelian variety over $k$. For any field extension $K/k$ the base change $A_ K$ is an abelian variety over $K$.

Proof. Omitted. Note that this is why we insisted on $A$ being geometrically integral; without that condition this lemma (and many others below) would be wrong. $\square$

Lemma 38.9.4. Let $k$ be a field. Let $A$ be an abelian variety over $k$. Then $A$ is smooth over $k$.

Proof. If $k$ is perfect then this follows from Lemma 38.8.2 (characteristic zero) and Lemma 38.8.4 (positive characteristic). We can reduce the general case to this case by descent for smoothness (Descent, Lemma 34.20.27) and going to the perfect closure using Lemma 38.9.3. $\square$

Lemma 38.9.5. An abelian variety is an abelian group scheme, i.e., the group law is commutative.

Proof. Let $k$ be a field. Let $A$ be an abelian variety over $k$. By Lemma 38.9.3 we may replace $k$ by its algebraic closure. Consider the morphism

\[ h : A \times _ k A \longrightarrow A \times _ k A,\quad (x, y) \longmapsto (x, xyx^{-1}y^{-1}) \]

This is a morphism over $A$ via the first projection on either side. Let $e \in A(k)$ be the unit. Then we see that $h|_{e \times A}$ is constant with value $(e, e)$. By More on Morphisms, Lemma 36.38.6 there exists an open neighbourhood $U \subset A$ of $e$ such that $h|_{U \times A}$ factors through some $Z \subset U \times A$ finite over $U$. This means that for $x \in U(k)$ the morphism $A \to A$, $y \mapsto xyx^{-1}y^{-1}$ takes finitely many values. Of course this means it is constant with value $e$. Thus $(x, y) \mapsto xyx^{-1}y^{-1}$ is constant with value $e$ on $U \times A$ which implies that the group law on $A$ is abelian. $\square$

Lemma 38.9.6. Let $k$ be a field. Let $A$ be an abelian variety over $k$. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ A$-module. Then there is an isomorphism

\[ m_{1, 2, 3}^*\mathcal{L} \otimes m_1^*\mathcal{L} \otimes m_2^*\mathcal{L} \otimes m_3^*\mathcal{L} \cong m_{1, 2}^*\mathcal{L} \otimes m_{1, 3}^*\mathcal{L} \otimes m_{2, 3}^*\mathcal{L} \]

of invertible modules on $A \times _ k A \times _ k A$ where $m_{i_1, \ldots , i_ t} : A \times _ k A \times _ k A \to A$ is the morphism $(x_1, x_2, x_3) \mapsto \sum x_{i_ j}$.

Proof. Apply the theorem of the cube (More on Morphisms, Theorem 36.29.6) to the difference

\[ \mathcal{M} = m_{1, 2, 3}^*\mathcal{L} \otimes m_1^*\mathcal{L} \otimes m_2^*\mathcal{L} \otimes m_3^*\mathcal{L} \otimes m_{1, 2}^*\mathcal{L}^{\otimes -1} \otimes m_{1, 3}^*\mathcal{L}^{\otimes -1} \otimes m_{2, 3}^*\mathcal{L}^{\otimes -1} \]

This works because the restriction of $\mathcal{M}$ to $A \times A \times e = A \times A$ is equal to

\[ n_{1, 2}^*\mathcal{L} \otimes n_1^*\mathcal{L} \otimes n_2^*\mathcal{L} \otimes n_{1, 2}^*\mathcal{L}^{\otimes -1} \otimes n_1^*\mathcal{L}^{\otimes -1} \otimes n_2^*\mathcal{L}^{\otimes -1} \cong \mathcal{O}_{A \times _ k A} \]

where $n_{i_1, \ldots , i_ t} : A \times _ k A \to A$ is the morphism $(x_1, x_2) \mapsto \sum x_{i_ j}$. Similarly for $A \times e \times A$ and $e \times A \times A$. $\square$

Lemma 38.9.7. Let $k$ be a field. Let $A$ be an abelian variety over $k$. Let $\mathcal{L}$ be an invertible $\mathcal{O}_ A$-module. Then

\[ [n]^*\mathcal{L} \cong \mathcal{L}^{\otimes n(n + 1)/2} \otimes ([-1]^*\mathcal{L})^{\otimes n(n - 1)/2} \]

where $[n] : A \to A$ sends $x$ to $x + x + \ldots + x$ with $n$ summands and where $[-1] : A \to A$ is the inverse of $A$.

Proof. Consider the morphism $A \to A \times _ k A \times _ k A$, $x \mapsto (x, x, -x)$ where $-x = [-1](x)$. Pulling back the relation of Lemma 38.9.6 we obtain

\[ \mathcal{L} \otimes \mathcal{L} \otimes \mathcal{L} \otimes [-1]^*\mathcal{L} \cong [2]^*\mathcal{L} \]

which proves the result for $n = 2$. By induction assume the result holds for $1, 2, \ldots , n$. Then consider the morphism $A \to A \times _ k A \times _ k A$, $x \mapsto (x, x, [n - 1]x)$. Pulling back the relation of Lemma 38.9.6 we obtain

\[ [n + 1]^*\mathcal{L} \otimes \mathcal{L} \otimes \mathcal{L} \otimes [n - 1]^*\mathcal{L} \cong [2]^*\mathcal{L} \otimes [n]^*\mathcal{L} \otimes [n]^*\mathcal{L} \]

and the result follows by elementary arithmetic. $\square$

Lemma 38.9.8. Let $k$ be a field. Let $A$ be an abelian variety over $k$. Let $[d] : A \to A$ be the multiplication by $d$. Then $[d]$ is finite locally free of degree $d^{2\dim (A)}$.

Proof. By Lemma 38.9.2 (and More on Morphisms, Lemma 36.43.1) we see that $A$ has an ample invertible module $\mathcal{L}$. Since $[-1] : A \to A$ is an automorphism, we see that $[-1]^*\mathcal{L}$ is an ample invertible $\mathcal{O}_ X$-module as well. Thus $\mathcal{N} = \mathcal{L} \otimes [-1]^*\mathcal{L}$ is ample, see Properties, Lemma 27.26.5. Since $\mathcal{N} \cong [-1]^*\mathcal{N}$ we see that $[d]^*\mathcal{N} \cong \mathcal{N}^{\otimes n^2}$ by Lemma 38.9.7.

To get a contradiction $C \subset X$ be a proper curve contained in a fibre of $[d]$. Then $\mathcal{N}^{\otimes d^2}|_ C \cong \mathcal{O}_ C$ is an ample invertible $\mathcal{O}_ C$-module of degree $0$ which contradicts Varieties, Lemma 32.43.14 for example. (You can also use Varieties, Lemma 32.44.9.) Thus every fibre of $[d]$ has dimension $0$ and hence $[d]$ is finite for example by Cohomology of Schemes, Lemma 29.21.1. Moreover, since $A$ is smooth over $k$ by Lemma 38.9.4 we see that $[d] : A \to A$ is flat by Algebra, Lemma 10.127.1 (we also use that schemes smooth over fields are regular and that regular rings are Cohen-Macaulay, see Varieties, Lemma 32.25.3 and Algebra, Lemma 10.105.3). Thus $[d]$ is finite flat hence finite locally free by Morphisms, Lemma 28.45.2.

Finally, we come to the formula for the degree. By Varieties, Lemma 32.44.11 we see that

\[ \deg _{\mathcal{N}^{\otimes d^2}}(A) = \deg ([d]) \deg _\mathcal {N}(A) \]

Since the degree of $A$ with respect to $\mathcal{N}^{\otimes d^2}$, respectively $\mathcal{N}$ is the coefficient of $n^{\dim (A)}$ in the polynomial

\[ n \longmapsto \chi (A, \mathcal{N}^{\otimes nd^2}),\quad \text{respectively}\quad n \longmapsto \chi (A, \mathcal{N}^{\otimes n}) \]

we see that $\deg ([d]) = d^{2 \dim (A)}$. $\square$


Lemma 38.9.9. Let $k$ be a field. Let $A$ be an abelian variety over $k$. Then $[d] : A \to A$ is étale if and only if $d$ is invertible in $k$.

Proof. Observe that $[d](x + y) = [d](x) + [d](y)$. Since translation by a point is an automorphism of $A$, we see that the set of points where $[d] : A \to A$ is étale is either empty or equal to $A$ (some details omitted). Thus it suffices to check whether $[d]$ is étale at the unit $e \in A(k)$. Since we know that $[d]$ is finite locally free (Lemma 38.9.8) to see that it is étale at $e$ is equivalent to proving that $\text{d}[d] : T_{A/k, e} \to T_{A/k, e}$ is injective. See Varieties, Lemma 32.16.8 and Morphisms, Lemma 28.34.16. By Lemma 38.6.4 we see that $\text{d}[d]$ is given by multiplication by $d$ on $T_{A/k, e}$. $\square$

Lemma 38.9.10. Let $k$ be a field of characteristic $p > 0$. Let $A$ be an abelian variety over $k$. The fibre of $[p] : A \to A$ over $0$ has at most $p^ g$ distinct points.

Proof. To prove this, we may and do replace $k$ by the algebraic closure. By Lemma 38.6.4 the derivative of $[p]$ is multiplication by $p$ as a map $T_{A/k, e} \to T_{A/k, e}$ and hence is zero (compare with proof of Lemma 38.9.9). Since $[p]$ commutes with translation we conclude that the derivative of $[p]$ is everywhere zero, i.e., that the induced map $[p]^*\Omega _{A/k} \to \Omega _{A/k}$ is zero. Looking at generic points, we find that the corresponding map $[p]^* : k(A) \to k(A)$ of function fields induces the zero map on $\Omega _{k(A)/k}$. Let $t_1, \ldots , t_ g$ be a p-basis of $k(A)$ over $k$ (More on Algebra, Definition 15.45.1 and Lemma 15.45.2). Then $[p]^*(t_ i)$ has a $p$th root by Algebra, Lemma 10.152.2. We conclude that $k(A)[x_1, \ldots , x_ g]/(x_1^ p - t_1, \ldots , x_ g^ p - t_ g)$ is a subextension of $[p]^* : k(A) \to k(A)$. Thus we can find an affine open $U \subset A$ such that $t_ i \in \mathcal{O}_ A(U)$ and $x_ i \in \mathcal{O}_ A([p]^{-1}(U))$. We obtain a factorization

\[ [p]^{-1}(U) \xrightarrow {\pi _1} \mathop{\mathrm{Spec}}(\mathcal{O}(U)[x_1, \ldots , x_ g]/(x_1^ p - t_1, \ldots , x_ g^ p - t_ g)) \xrightarrow {\pi _2} U \]

of $[p]$ over $U$. After shrinking $U$ we may assume that $\pi _1$ is finite locally free (for example by generic flatness – actually it is already finite locally free in our case). By Lemma 38.9.8 we see that $[p]$ has degree $p^{2g}$. Since $\pi _2$ has degree $p^ g$ we see that $\pi _1$ has degree $p^ g$ as well. The morphism $\pi _2$ is a universal homeomorphism hence the fibres are singletons. We conclude that the (set theoretic) fibres of $[p]^{-1}(U) \to U$ are the fibres of $\pi _1$. Hence they have at most $p^ g$ elements. Since $[p]$ is a homomorphism of group schemes over $k$, the fibre of $[p] : A(k) \to A(k)$ has the same cardinality for every $a \in A(k)$ and the proof is complete. $\square$


Proposition 38.9.11. Let $A$ be an abelian variety over a field $k$. Then

  1. $A$ is projective over $k$,

  2. $A$ is a commutative group scheme,

  3. the morphism $[n] : A \to A$ is surjective for all $n \geq 1$,

  4. if $k$ is algebraically closed, then $A(k)$ is a divisible abelian group,

  5. $A[n] = \mathop{\mathrm{Ker}}([n] : A \to A)$ is a finite group scheme of degree $n^{2\dim A}$ over $k$,

  6. $A[n]$ is étale over $k$ if and only if $n \in k^*$,

  7. if $n \in k^*$ and $k$ is algebraically closed, then $A(k)[n] \cong (\mathbf{Z}/n\mathbf{Z})^{\oplus 2\dim (A)}$,

  8. if $k$ is algebraically closed of characteristic $p > 0$, then there exists an integer $0 \leq f \leq \dim (A)$ such that $A(k)[p^ m] \cong (\mathbf{Z}/p^ m\mathbf{Z})^{\oplus f}$ for all $m \geq 1$.

Proof. Part (1) follows from Lemma 38.9.2. Part (2) follows from Lemma 38.9.5. Part (3) follows from Lemma 38.9.8. If $k$ is algebraically closed then surjective morphisms of varieties over $k$ induce surjective maps on $k$-rational points, hence (4) follows from (3). Part (5) follows from Lemma 38.9.8 and the fact that a base change of a finite locally free morphism of degree $N$ is a finite locally free morphism of degree $N$. Part (6) follows from Lemma 38.9.9. Namely, if $n$ is invertible in $k$, then $[n]$ is étale and hence $A[n]$ is étale over $k$. On the other hand, if $n$ is not invertible in $k$, then $[n]$ is not étale at $e$ and it follows that $A[n]$ is not étale over $k$ at $e$ (use Morphisms, Lemmas 28.34.16 and 28.33.15).

Assume $k$ is algebraically closed. Set $g = \dim (A)$. Proof of (7). Let $\ell $ be a prime number which is invertible in $k$. Then we see that

\[ A[\ell ](k) = A(k)[\ell ] \]

is a finite abelian group, annihilated by $\ell $, of order $\ell ^{2g}$. It follows that it is isomorphic to $(\mathbf{Z}/\ell \mathbf{Z})^{2g}$ by the structure theory for finite abelian groups. Next, we consider the short exact sequence

\[ 0 \to A(k)[\ell ] \to A(k)[\ell ^2] \xrightarrow {\ell } A(k)[\ell ] \to 0 \]

Arguing similarly as above we conclude that $A(k)[\ell ^2] \cong (\mathbf{Z}/\ell ^2\mathbf{Z})^{2g}$. By induction on the exponent we find that $A(k)[\ell ^ m] \cong (\mathbf{Z}/\ell ^ m\mathbf{Z})^{2g}$. For composite integers $n$ prime to the characteristic of $k$ we take primary parts and we find the correct shape of the $n$-torsion in $A(k)$. The proof of (8) proceeds in exactly the same way, using that Lemma 38.9.10 gives $A(k)[p] \cong (\mathbf{Z}/p\mathbf{Z})^{\oplus f}$ for some $0 \leq f \leq g$. $\square$

Comments (0)

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0BF9. Beware of the difference between the letter 'O' and the digit '0'.