Lecture 4: Abelian varieties (algebraic theory)
This lecture covers the basic theory of abelian varieties over arbitrary fields. I begin with the basic results such as commutativity and the structure of torsion. Then I discuss the dual abelian variety. Next I prove the weak MordellWeil theorem, as the same ideas will be important for us later on. Last is Poincaré irreducibility, and its interpretation in terms of the isogeny category.
A good reference for today is Mumford's "Abelian varieties" (MR0282985) or Milne's notes
General facts about abelian varieties
Fix a field $k$. Many of the results about abelian varieties over $\bC$ continue to hold over $k$. However, the proofs are quite different and more complicated. We give some indications as to how the theory is developed, but omit most of the arguments.
Commutativity
We begin by explaining the most basic fact: commutativity. One can establish this using an argument similar to the one we used in the complex case. We present a different argument here, which provides a more general result. It is based on the following general fact:
Lemma (Rigidity lemma). Let $X$ be a complete variety, let $Y$ and $Z$ be arbitrary varieties, and let $f \colon X \times Y \to Z$ be a map of varieties. Suppose there exists $x_0 \in X$ and $y_0 \in Y$ such that the restriction of $f$ to each of $X \times \{y_0\}$ and $\{x_0\} \times Y$ is constant. Then $f$ is constant.
Corollary. Let $X$ and $Y$ be abelian varieties and let $f \colon X \to Y$ be any map of varieties such that $f(0)=0$. Then $f$ is a morphism of abelian varieties, i.e., it respects the group structure.
Proof. Consider the map $h \colon X \times X \to Y$ given by $(x,y) \mapsto f(x+y)f(x)f(y)$. Then $h(x, 0)=h(0,x)=0$ for all $x \in X$, and so by the Rigidity Lemma $h=0$, i.e., $f$ is a homomorphism. ◾
Corollary. An abelian variety is commutative.
Proof. The map $x \mapsto x$ takes 0 to 0 and is therefore a homomorphism, which implies commutativity. ◾
Theorem of the cube
Theorem (Theorem of the cube). Let $X$, $Y$, and $Z$ be varieties, with $X$ and $Y$ complete, and let $x_0 \in X$, $y_0 \in Y$, and $z_0 \in Z$ be points. Let $L$ be a line bundle on $X \times Y \times Z$, and suppose the restrictions of $L$ to each of $X \times Y \times \{z_0\}$, $X \times \{y_0\} \times Z$, and $\{x_0\} \times Y \times Z$ are trivial. Then $L$ is trivial.
Remark. This can be thought of as a version of the rigidity lemma for maps to the stack $\mathrm{B}\bG_m$.
Corollary. Let $A$ be an abelian variety. Let $p_i \colon A \times A \times A \to A$ denote the projection map, and let $p_{ij}=p_i+p_j$ and $p_{123}=p_1+p_2+p_3$. Let $L$ be a line bundle on $A$. Then the line bundle
$$ p_{123}^*L \otimes p_{12}^*L^{1} \otimes p_{13}^*L^{1} \otimes p_{23}^*L^{1} \otimes p_1^*L \otimes p_2^*L \otimes p_3^*L $$on $A \times A \times A$ is trivial.
Proof. This follows immediately from the theorem of the cube. For example, if we restrict to $A \times A \times \{0\}$ then $p_{123}^*L=p_{12}^*L$, $p_{13}^*L=p_1^*L$, and $p_3^*L=1$, so all factors cancel. ◾
Corollary. Let $A$ be an abelian variety, let $X$ be any variety, let $f,g,h \colon X \to A$ be maps, and let $L$ be a line bundle on $A$. Then the line bundle
$$ (f+g+h)^*L \otimes (f+g)^*L^{1} \otimes (f+h)^*L^{1} \otimes (g+h)^*L^{1} \otimes f^*L \otimes g^*L \otimes h^*L $$on $X$ is trivial.
Proof. This follows from the previous corollary by considering the map $X \to A \times A \times A$ given by $(f,g,h)$. ◾
Structure of torsion
Proposition. Let $L$ be a line bundle on an abelian variety $A$. Then
$$ [n]^*L=L^{(n^2+n)/2} \otimes [1]^* L^{(n^2n)/2}. $$In particular, if $L$ is symmetric ($[1]^*L=L$) then $[n]^*L=L^{n^2}$, while if $L$ is antisymmetric ($[1]^*L=L^{1}$) then $[n]^*L=L^n$.
Proof. Applying the previous corollary to the maps $[n]$, $[1]$, and $[1]$, we find
$$ [n]^*L \otimes [n+1]^*L^{1} \otimes [n1]^*L^{1} \otimes [n]^*L \otimes L \otimes [1]^* L $$is trivial. In other words,
$$ [n+1]^*L = [n]^*L^2 \otimes [n1]^*L^{1} \otimes L \otimes [1]^*L. $$The result now follows by induction. ◾
Proposition. The map $[n]$ is an isogeny, i.e., it is surjective with finite kernel.
Proof. One can show that abelian varieties are projective. Let $L$ be an ample line bundle on $A$. Replacing $L$ by $L \otimes [1]^*L$, we can assume $L$ is symmetric. Since $[n]^*L=L^{n^2}$, it is ample. However, the restriction of this to the $n$torsion is obviously trivial. Since the $n$torsion is a complete variety on which the trivial bundle is ample, it must be finite. This implies that $[n]$ is surjective, by reasoning with dimension. ◾
Proposition. The degree of $[n]$ is $n^{2g}$.
Proof. Let $f \colon X \to Y$ be a finite map of complete varieties of degree $d$. If $D_1, \ldots, D_n$ are divisors on $Y$, where $n=\dim(X)=\dim(Y)$, then there is an equality of intersection numbers:
$$ (f^* D_1 \cdots f^* D_n) = d (D_1 \cdots D_n). $$Now, let $D$ be an ample divisor such that $[1]^*D$ is linearly equivalent to $D$ (e.g., the divisor associated to the line bundle used above). Then $[n]^*D$ is linearly equivalent to $n^2 D$. We thus find
$$ \deg([n]) (D \cdots D) = ( (n^2 D) \cdots (n^2 D) ) = n^{2g} (D \cdots D). $$Since $D$ is ample, $(D \cdots D) \ne 0$, and thus $\deg([n])=n^{2g}$. ◾
One can show that $[n] \colon A \to A$ induces multiplication by $n$ on the tangent space. This shows that $[n]$ is separable if and only if $n$ is prime to the characteristic. Combined with the above (and the usual induction argument), we see that:
Corollary. If $n$ is prime to the characteristic, then $A[n](\ol{k})$ is isomorphic to $(\bZ/n\bZ)^{2g}$.
Corollary. If $\ell$ is a prime different from the characteristic then $T_{\ell} A$ is isomorphic to $\bZ_{\ell}^{2g}$.
Since $[p]$ is not separable, $A[p](\ol{k})$ must have fewer than $p^{2g}$ points. In fact, when we study group schemes we will see that it can have at most $p^g$ points.
Theorem of the square
Theorem (Theorem of the square). Let $L$ be a line bundle on an abelian variety $A$, and let $x$ and $y$ be two points of $A$. Then
$$ t_{x+y}^* L \otimes L = t_x^* L \otimes t_y^* L. $$Here $t_x$ denotes translation by $x$.
Proof. Apply the $f$, $g$, $h$ proposition with $f=x$ (constant map), $g=y$, and $h=\id$. ◾
Define $\Pic(A)$ to be the set of isomorphism classes of line bundles on $A$. For a line bundle $L$, let $\phi_L \colon A(k) \to \Pic(A)$ be the map $\phi_L(x)=t_x^*L \otimes L^{1}$. The theorem of the square exactly states that $\phi_L$ is a group homomorphism.
The dual variety
Over the complex numbers, we can write an abelian variety $A$ as $V/M$, where $V$ is a complex vector space and $M$ is lattice. We defined the dual abelian variety $A^{\vee}$ as $\ol{V}^*/M^{\vee}$. We would like to be able to define the dual variety over any field, but this definition obviously does not carry over. The key idea is to reinterpret $A^{\vee}$ in terms of line bundles.
Recall that over $\bC$ we showed that the set $A^{\vee}$ was canonically in bijection with the set $\Pic^0(A)$. Furthermore, although our definition of $\Pic^0(A)$ was originally topological (and does not generalize to other fields), we characterized $\Pic^0(A)$ as the translation invariant line bundles (which does generalize to other fields). We therefore have a possible method of defining the dual.
Definition of the dual
Let $k$ be an arbitrary field, and let $A$ be an abelian variety over $k$. We defined $\Pic(A)$ above to be the set of isomorphism classes of line bundles on $A$. We now define $\Pic^0(A)$ to be the subgroup consisting of those line bundles $L$ which are translation invariant, i.e., which satisfy $t_x^*(L) \cong L$ for all $x \in A$. Motivated by the complex case, we want to define $A^{\vee}$ to be an abelian variety with pointset $\Pic^0(A)$. However, it is not good enough to just define the points of a variety over a field: we must define its functor of points.
For a variety $T$, let $F(T)$ be the of isomorphism classes of line bundles $L$ on $A \times T$ satisfying the following two conditions: (a) for all $t \in T$, the restriction of $L$ to $A \times \{t\}$ belongs to $\Pic^0(A)$; and (b) the restriction of $L$ to $\{0\} \times T$ is trivial. Thus $F(k)=\Pic^0(A)$. We define the dual abelian variety $A^{\vee}$ to be the variety that represents $F$, if it exists. If it does exist, then it automatically comes with a universal bundle $\cP$ on $A \times A^{\vee}$, which is called the Poincaré bundle.
Construction of the dual
Let $L$ be an ample bundle on $A$. We then have the map $\phi_L \colon A \to \Pic^0(A)$. (The theorem of the square implies the image is in $\Pic^0$.) Over the complex numbers, we saw that this map was an isogeny of tori. In general, one can prove the it is surjective, and has finite kernel $K(L)$. In fact, one can give $K(L)$ a natural scheme structure. This suggests that $A^{\vee}$ should be the quotient $A/K(L)$, and one can show that this is indeed the case.
Another approach to the dual
Let $L$ be in $\Pic^0(A)$. Then, by definition, $t_x^*(L)$ and $L$ are isomorphic for all $x \in A$. Choose an isomorphism $\phi_x$. Then $\phi_y t_y^*(\phi_x)$ and $\phi_{x+y}$ are two isomorphisms $t_{x+y}^*(L) \to L$, and thus differ by an element $\alpha_{x,y}$ of $\Aut(L)=\bG_m$. It is obvious that $\alpha$ is a 2cocycle of $A$ with coefficients in $\bG_m$, and thus defines a central extension $\cG(L)$ of $A$ by $\bG_m$. In fact, $\cG(L)$ is a commutative group.
Here is a different construction of $\cG(L)$. One can show that $L$ being translation invariant is equivalent to $p_1^* L \otimes p_2^* L$ being isomorphic to $m^* L$, where $m$ is the multiplication map $A \times A \to A$ and $p_i$ are the projection maps. The fiber at $(x,y)$ of this isomorphism is an identification $L_x \otimes L_y \to L_{x+y}$. In other words, there is a natural map $L \times L \to L$ (identifying $L$ with its total space) over the multiplication map on $A$. The group $\cG(L)$ is then just $L$ minus its zero section, with this multiplication.
We have thus constructed a map $\cG \colon \Pic^0(A) \to \Ext^1(A, \bG_m)$, where $\Ext^1$ is taken in the category of commutative group varieties. Serre showed (MR0103191, Chapter VII, Section 3) that this map is an isomorphism. Forming $\Ext^1$ in the category of fppf sheaves allows one to recover the functor of points of $A^{\vee}$.
The MordellWeil theorem
Theorem (MordellWeil). Let $A$ be an abelian variety over the number field $K$. Then $A(K)$ is a finitely generated abelian group.
The proof of this theorem usually proceeds in two steps: first, one shows that $A(K)/nA(K)$ is a finite group (the socalled weak MordellWeil theorem), and then one uses height functions to deduce the theorem. We will only discuss the proof of the first step. A complete proof, in the case of elliptic curves, is given in Chapter VII of Silverman’s “The arithmetic of elliptic curves” (MR0817210).
Consider the exact sequence
$$ 0 \to A[n](\ol{K}) \to A(\ol{K}) \stackrel{n}{\to} A(\ol{K}) \to 0. $$Taking Galois cohomology, one obtains an exact sequence
$$ 0 \to A(K)/nA(K) \to \rH^1(G_K, A[n](\ol{K})) \to \rH^1(G_K, A(\ol{K}))[n] \to 0. $$This is called the Kummer sequence, and is very important. To show that $A(K)/nA(K)$ is finite, it suffices to show that the middle cohomology group is finite. This is not quite true; however, one can show that the image of the first map only hits classes which are unramified outside a fixed finite set of places $S$, and so it's enough to establish finiteness for such classes, which is true. (The set $S$ can be taken to be the set of places of bad reduction for $A$, together with those places above $n$.)
Let $L/K$ be a finite Galois extension containing all the $n$torsion of $A$, and enlarge $S$ so that $L/K$ is unramified outside $S$. Then one has the inflationrestriction sequence:
$$ 0 \to \rH^1(\Gal(L/K), A[n](\ol{K})) \to \rH^1(G_{K,S}, A[n](\ol{K})) \to \rH^1(G_{L,S}, A[n](\ol{K})), $$and so to prove finiteness of the middle group it suffices to prove finiteness of the outer groups. Finiteness of the group on the left comes for free, since $\Gal(L/K)$ and $A[n](\ol{K})$ are both finite. Since $G_{L,S}$ acts trivially on $A[n](\ol{K})$, the right group is just $\Hom(G_{L,S}, \bZ/n\bZ)^{2g}$. Giving a map $G_{L,S} \to \bZ/n\bZ$ is (almost) the same as giving a $\bZ/n\bZ$ extension of $L$ unramified outside of $S$. Since there are only finitely many such extensions unramified, the finiteness result follows.
Structure of the isogeny category
Poincaré reducibility
Theorem (Poincaré reducibility). Let $A$ be an abelian variety, and let $B$ be an abelian subvariety. Then there exists an abelian subvariety $C$ such that $B \cap C$ is finite and $B \times C \to A$ is an isogeny.
Proof. Choosing polarizations on $A$ and $A/B$ to identify them with their duals, the dual to the quotient map $A \to A/B$ is a map $A/B \to A$. We let $C$ be its image. The properties are easy to verify. ◾
We say that an abelian variety $A$ is simple if the only abelian subvarieties are 0 and $A$.
Corollary. Every abelian variety is isogenous to a product of simple varieties.
The isogeny category
Define a category $\Isog$ as follows. The objects are abelian varieties. For two abelian varieties $A$ and $B$, we put $\Hom_{\Isog}(A, B) = \Hom(A, B) \otimes \bQ$. One can show that if $f \colon A \to B$ is an isogeny then there exists an isogeny $g \colon B \to A$ such that $gf=[n]$, for some $n$; it follows that $\tfrac{1}{n} g$ is the inverse to $f$ in $\Isog$. Thus isogenies become isomorphisms in $\Isog$.
It is not difficult to see that $\Isog$ is in fact an abelian category. The simple objects of this category are exactly the simple abelian varieties. Poincaré's theorem shows that $\Isog$ is semisimple as an abelian category.
From this formalism, and general facts about abelian varieties, we deduce two results:

The decomposition (up to isogeny) into a product of simple abelian varieties is unique (up to isogeny). (Reason: in any semisimple abelian category, the decomposition into simples is unique up to isomorphism.)

If $A$ is a simple abelian variety then $\End(A) \otimes \bQ$ is a division algebra over $\bQ$. (Reason: if $A$ is a simple object in an abelian category and $\End(A)$ contains a field $k$, then it is a division algebra over $k$.)