Abelian varieties: 3. Abelian schemes are commutative group schemes

This posting follows Matt Stevenson's notes for Math 731 (Fall 2017) at the University of Michigan. We also follow Mumford's book.

Convention. When we discuss an abelian scheme over a base scheme $S,$ we will assume that $S$ is nonempty.

We have discussed how an abelian variety $A$ over $\mathbb{C}$ is a commutative group scheme using analytic techniques. More specifically, we have gone through the following two steps:

Step 1. We have studied the conjugation action of each element of $A(\mathbb{C})$ on the tangent space $T_{e}A(\mathbb{C})$ at the identity $e$.

Step 2. We have made use of the exponential map $T_{e}A(\mathbb{C}) \rightarrow A(\mathbb{C}).$

We are going to show that an abelian variety over any field is a commutative group scheme. Step 1 works in algebraic setting, but Step 2 does not.

Remark. In characteristic $p > 0,$ it is known that there are examples of automorphisms that act trivially on the tangent space but not on the variety. (We might add some examples later.)

Let $A$ be an abelian variety over a field $k.$ The strategy we take is that we are going to consider the conjugations $c_{t} : A(k) \rightarrow A(k)$ as a flat family parametrized by $t \in A(k).$ We will show that $c_{t}$ is constant in $t.$

Our specific goal. Given an abelian scheme $A$ over $S,$ for $a \in A(S)$ (i.e., an $S$-scheme map $a : S \rightarrow A$), we define the left-translation by $a$ as the map $$l_{a} := m_{A} \circ (a \circ \pi_{A}, \mathrm{id}_{A}) : A \rightarrow A \times_{S} A \rightarrow A,$$ where $m_{A}$ is the multiplication map of the group $S$-scheme $A,$ and $\pi_{A} : A \rightarrow S$ is the structure map. Even though the above definition for $l_{a}$ is simple, it is not the best definition to work with. Another way to describe it is that given any $S$-scheme $T,$ the map $l_{a} : A \rightarrow A$ gives $l_{a, T} : A(T) \rightarrow A(T)$ defined by $x \mapsto \pi_{T}^{*}(a) \cdot x = (a \circ \pi_{T}) \cdot x,$ where $\pi_{T} : T \rightarrow S$ is the structure map. This allows us to show that $l_{e} = \mathrm{id}_{G}$ and $l_{g \cdot g'} = l_{g} \circ l_{g'}$ for all $g, g' \in G(S).$ In particular, we have $l_{g \cdot g^{-1}} = \mathrm{id}_{G} = l_{g^{-1} \cdot g}.$ This discussion works for any group scheme $G$ over $S.$ Details can be found in this posting.

Theorem. Let $A, B$ be abelian $S$-schemes with identities $e_{A}, e_{B}.$ If $S$ is a Noetherian scheme, given any $S$-scheme map $f : A \rightarrow B,$ there is a factorization $f = l_{f_{S}(e_{A})} \circ h$ for some map $h : A \rightarrow B$ of $S$-groups, where $f_{S}(e_{A}) = f \circ e_{A} : S \rightarrow A \rightarrow B$ as usual.

Corollary. If $S$ is a Noetherian scheme, any abelian $S$-scheme $A$ is a commutative group scheme.

Proof. Denote by $i_{A} : A \rightarrow A$ the inversion map. Since $i_{A}(e_{A}) = e_{A} : S \rightarrow A,$ we have $l_{i_{A}(e_{A})} = l_{e_{A}} = m_{A} \circ (e_{A} \circ \pi, \mathrm{id}_{A}) = \mathrm{id}_{A},$ where the last equality follows from a group scheme axiom for $A.$ Thus, applying Theorem for $f = i_{A},$ it shows that $i_{A}$ is a group scheme map. From here, it follows that $A$ is a commutative group scheme. $\Box$

Hence, it remains to show Theorem. First, we note that it is enough to show that $l_{f_{S}(e_{A})^{-1}} \circ f : A \rightarrow B$ is a map of $S$-groups. Since $$\begin{align*}(l_{f_{S}(e_{A})^{-1}} \circ f)_{S}(e_{A}) &= l_{f_{S}(e_{A})^{-1}} \circ f \circ e_{A} \\ &= l_{f_{S}(e_{A})^{-1}} \circ f_{S}(e_{A}) \\ &= l_{f_{S}(e_{A})^{-1},S}(f_{S}(e_{A})) \\ &= f_{S}(e_{A})^{-1} \cdot f_{S}(e_{A}) \\ &= e_{B}\end{align*}$$ in $B(S)$ because the $S$-scheme structure map for $S$ is $\mathrm{id}_{S}.$ Thus, this reduces the problem to the case where $f_{S}(e_{A}) = e_{B},$ and we desire to prove that $f : A \rightarrow B$ is an $S$-group map. This means that given any $S$-scheme $T$ and $a, a' \in A(T),$ we have $$f_{T}(m_{A,T}(a, a')) = f_{T}(a \cdot a') = f_{T}(a) \cdot f_{T}(a') = m_{B,T}(f_{T}(a), f_{T}(a')).$$ In terms of $S$-scheme map language, this precisely states that $$f \circ m_{A} = m_{B} \circ (f \times_{S} f) : A \times_{S} A\rightarrow B,$$ and thus this is what we need to prove. For each $S$-scheme $T,$ we may instead try to prove $$m_{B,T}(f_{T}(m_{A,T}(a, a')), i_{B}(m_{B,T}(f_{T}(a), f_{T}(a')))) = e_{B,T} \in B(T)$$ for all $a, a' \in A(T).$ In the $S$-scheme map language, this means that we want to prove $$m_{B} \circ (f \circ m_{A}, i_{B} \circ m_{B} \circ (f \times_{S} f)) = e_{B} \circ \pi_{A \times_{S} A} : A \times_{S} A \rightarrow B,$$ where again $e_{B} = e_{B,S} : S \rightarrow B$ is the identity element of the group $B(S)$ and $\pi_{A \times_{S} A} : A \times_{S} A \rightarrow S$ is the $S$-scheme structure map. The map on the left-hand side is the composition $$A \times_{S} A \rightarrow B \times_{S} B \rightarrow B$$ of the relevant maps.

Reduction of the problem. We write $$g := m_{B} \circ (f \circ m_{A}, i_{B} \circ m_{B} \circ (f \times_{S} f)) : A \times_{S} A \rightarrow B.$$ We want to show that $g = e_{B} \circ \pi_{A \times_{S} A}.$ Note that given any $S$-scheme $T,$ the map $g_{T} : A(T) \times A(T) \rightarrow B(T)$ is given by $(a, a') \mapsto f_{T}(a \cdot a') \cdot (f_{T}(a) \cdot f_{T}(a'))^{-1}.$ Hence, we have $$\begin{align*}g \circ (\mathrm{id}_{A}, e_{A} \circ \pi_{A}) \circ a &= g_{T}(a, e_{A,T}) \\ &= f_{T}(a \cdot e_{A,T}) \cdot (f_{T}(a) \cdot f_{T}(e_{A,T}))^{-1} \\ &= f_{T}(a \cdot e_{A,T}) \cdot (f_{T}(a) \cdot e_{B,T})^{-1} \\ &= e_{B,T} \\ &= e_{B} \circ \pi_{T}\end{align*}$$ because, using $f_{S}(e_{A}) = e_{B},$ we have $$\begin{align*}f_{T}(e_{A,T}) &= f \circ e_{A} \circ \pi_{T}\\ &= f_{S}(e_{A}) \circ \pi_{T}\\ &= e_{B} \circ \pi_{T} \\&= e_{B,T}.\end{align*}$$ Hence, if we prove that $g : A \times_{S} A \rightarrow B$ is a constant map, then it will follow that for every $a, a' \in A(T),$ we have $g_{T}(a, a') = e_{B, T}.$ The only $S$-scheme map $A \times_{S} A \rightarrow B$ such that $A(T) \times A(T) \rightarrow B(T)$ is giving the constant value of $e_{B,T}$ is $e_{B} \circ \pi_{A \times_{S} A},$ so this will finish the proof. Thus, it remains to show that $g : A \times_{S} A \rightarrow B$ is constant.

To finish the proof of Theorem, we use the following "rigidity" lemma:

Lemma (Rigidity). Fix any Noetherian scheme $S.$ Let  $\pi_{X} : X \rightarrow S$ be a proper flat $S$-scheme such that $\dim_{\kappa(s)}(H^{0}(X_{s}, \mathscr{O}_{X_{s}})) = 1$ for all $s \in S,$ where $$X_{s} = X \times_{S} \mathrm{Spec}(\kappa(s)),$$ the fiber at $s.$ If $\phi : X \rightarrow Y$ is any $S$-scheme map such that there exists any $s \in S$ such that the restriction $X_{s} \rightarrow Y_{s}$ of $\phi$ is constant, then $\phi$ is constant on the connected component of $s$ in $S.$

Remark. An $S$-scheme map $X \rightarrow Y$ is said to be constant if it factors through the structure map of $X.$ In particular, in such situation, for any $S$-scheme $T,$ the induced map $X(T) \rightarrow Y(T)$ is constant in set-theoretic sense. Given a field $k,$ a constant map of $k$-schemes $X \rightarrow Y$ necessarily give a constant map on the underlying topological spaces, as it factors as $X \rightarrow \mathrm{Spec}(k) \rightarrow Y.$ I don't think the converse is true: there are probably many examples where a $k$-scheme map is topologically constant but not a constant map according to the definition we are using. However, I have not thought deeply about such examples. (I may add them later if there are such though.)

Proof of Theorem. We use the notations given before Lemma. Write $$p_{1}^{-1}(e_{A}) := S \times_{A \times_{S} A} A,$$ with respect to $e_{A} : S \rightarrow A$ and $p_{1} : A \times_{S} A \rightarrow A.$ Given any $S$-scheme $T,$ the $S$-scheme map $$p_{1}^{-1}(e_{A}) \rightarrow A \times_{S} A$$ induces the set map $$(p_{1}^{-1}(e_{A}))(T) \rightarrow A(T) \times A(T),$$ which necessarily maps to elements of the form $(a, e_{A,T})$ where $a \in A(T)$ may vary. We have checked before that $g_{T}(a, e_{A,T}) = e_{B,T}$ constantly regardless of the choice of $a,$ so this implies that the composition $p_{1}^{-1}(e_{A}) \rightarrow A \times_{S} A \xrightarrow{g} B$ is constant. This implies that the following composition is also constant: $$p_{1}^{-1}(e_{A}) \rightarrow A \times_{S} A \xrightarrow{(p_{1}, g)} A \times_{S} B.$$ Then now note that $(p_{1}, g)$ is an $A$-scheme map that is constant on the fiber at any point in $A$ in the image of $e_{A} : S \rightarrow A.$ Now note that

• $A$ is connected,
• both $A \times_{S} A$ and $A \times_{S} B$ are proper and smooth over $A$ (base change) and hence also flat over $A$ (e.g., 25.2.2. (iii) in Vakil), and
• for $s \in A,$ we have $(A \times_{S} A)_{s} = p_{1}^{-1}(s) \simeq \mathrm{Spec}(\kappa(s)) \times_{A} A \times_{S} A \simeq A_{s},$ which is geometrically connected, so $\dim_{\kappa(s)} (A \times_{S} A)_{s} = 1.$

Thus, we may apply Rigidity Lemma to conclude that that $(p_{1}, g)$ is constant. Since $g$ is the composition $$A \times_{S} A \xrightarrow{(p_{1}, g)} A \times_{S} B \rightarrow B,$$ where the latter one is the projection map onto $B,$ the fact that $(p_{1}, g)$ is constant implies that $g$ is constant. This finishes the proof $\Box$