# 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 Mordell--Weil 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

$$[n]^*L=L^{(n^2+n)/2} \otimes [-1]^* L^{(n^2-n)/2}.$$

In particular, if $L$ is symmetric ($[-1]^*L=L$) then $[n]^*L=L^{n^2}$, while if $L$ is anti-symmetric ($[-1]^*L=L^{-1}$) then $[n]^*L=L^n$.

Proof. Applying the previous corollary to the maps $[n]$, $[1]$, and $[-1]$, we find

$$[n]^*L \otimes [n+1]^*L^{-1} \otimes [n-1]^*L^{-1} \otimes [n]^*L \otimes L \otimes [-1]^* L$$

is trivial. In other words,

$$[n+1]^*L = [n]^*L^2 \otimes [n-1]^*L^{-1} \otimes L \otimes [-1]^*L.$$

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 point-set $\Pic^0(A)$. 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 2-cocycle 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 Mordell--Weil theorem

Theorem (Mordell--Weil). 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 so-called weak Mordell--Weil 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 inflation--restriction 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 semi-simple 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 semi-simple 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 $k$.)