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 $\Box$

Group schemes: basic definitions

I realized that I am still not quite comfortable with basic properties of group schemes, so I have decided to give myself a self-exercise to organize the material in my own words. I am not directly following but taking a look at Vistoli's notes for this.

Let $S$ be any scheme. An $S$-scheme $G$ is called a group scheme over $S$ (or an $S$-group) if the functor $h_{G} = \mathrm{Hom}_{\textbf{Sch}_{S}}(-, G) : \textbf{Sch}_{S}^{\mathrm{op}} \rightarrow \textbf{Set}$ factors as $$\textbf{Sch}_{S}^{\mathrm{op}} \rightarrow \textbf{Grp} \rightarrow \textbf{Set},$$ where the second functor is the forgetful functor.

This means that for each $S$-scheme $T,$ the set $G(T) = \mathrm{Hom}_{\textbf{Sch}_{S}}(T, G)$ has a group structure. In particular, we have the multiplication (set) map $$m_{T} : G(T) \times G(T) \rightarrow G(T)$$ and the inverse map $$i_{T} : G(T) \rightarrow G(T).$$ Moreover, both $m_{T}$ and $i_{T}$ are functorial in $T.$ To see this, fix any $S$-scheme map $\phi : T' \rightarrow T.$ The induced map $\phi^{*} : G(T) \rightarrow G(T')$ is a group map, so we must have $$\phi^{*} \circ m_{T} = m_{T'} \circ \phi^{*},$$ which shows the functoriality of $m_{-}.$ The functoriality of $i_{-}$ holds for the same reason. Since $i_{-} : h_{G} \rightarrow h_{G}$ is the map of functors, using the Yoneda embedding $\textbf{Sch}_{S} \hookrightarrow \textbf{PSh}_{\textbf{Set}}(\textbf{Sch}_{S})$ given by $T \mapsto h_{T} = T(-),$ there is a unique $S$-scheme map $i : G \rightarrow G$ that maps to $i_{-} : G(-) \rightarrow G(-).$ We remark that this uniqueness is up to an isomorphism of $S$-schemes. Given any $S$-scheme $T,$ we have $$i_{T} : G(T) \rightarrow G(T)$$ in $\textbf{Set},$ given by $f \mapsto i \circ f.$ We need slightly more for the multiplication map. First, given any $S$-scheme $T,$ we have $$G(T) \times G(T) \simeq (G \times_{S} G)(T)$$ given by $(x, y) \mapsto (x, y),$ where the left-hand side means a pair of two $S$-scheme maps $x, y : T \rightarrow G,$ while the right-hand side means the map $$(x, y) : T \rightarrow G \times_{S} G$$ of $S$-schemes induced by the fiber product. That is, if we denote by $\pi_{1}, \pi_{2} : G \times_{S} G \rightarrow G$ two projections, then

• $x = \pi_{1} \circ (x, y)$ and 
• $y = \pi_{2} \circ (x, y).$ 

Given any $S$-scheme map $\phi : T' \rightarrow T,$ we have

• $x \circ \phi = \pi_{1} \circ (x, y) \circ \phi$ and 
• $y \circ \phi = \pi_{2} \circ (x, y) \circ \phi,$

so $(x, y) \circ \phi = (x \circ \phi, y \circ \phi),$ so we see that $G(T) \times G(T) \simeq (G \times_{S} G)(T)$ is functorial in $T.$ Hence, we may realize $m_{-}$ as a map $(G \times_{S} G)(-) \rightarrow G(-),$ and using the Yoneda embedding, we may find a unique $S$-scheme map $$m : G \times_{S} G \rightarrow G$$ mapping to $m_{-}.$ We note that $m_{T} : (G \times_{S} G)(T) \rightarrow G(T)$ is given by $f \mapsto m \circ f,$ and if we look at $m_{T}$ as a map $G(T) \times G(T) \rightarrow G(T),$ the map is given by $(x, y) \mapsto m \circ (x, y),$ where $(x, y)$ on the right-hand side means the map $G \rightarrow G \times_{S} G.$

Since each $G(T) = \mathrm{Hom}_{\textbf{Sch}_{S}}(T, G)$ is a group, we have the identity element $e_{T} \in G(T)$ for any $S$-scheme $T.$ Any $S$-scheme $T$ comes with its structure map $\pi_{T} : T \rightarrow S,$ and $\pi_{T}^{*} : G(S) \rightarrow G(T)$ is a group map, so we must have $e_{S} \mapsto e_{T}.$ This implies that $e_{T} = e_{S} \circ \pi_{T} : T \rightarrow S \rightarrow G.$ We thus denote by $e := e_{S} : S \rightarrow G,$ as we can recover any $e_{T}$ from it for any $S$-scheme $\pi_{T} : T \rightarrow S.$

The fact that $e_{T}$ is the identity of the group $G(T)$ means that for any $x \in G(T),$ we have $$m \circ (e_{T}, x) = x = m \circ (x, e_{T}).$$ In other words, we have $m_{T}(e_{T}, x) =x = m_{T}(x, e_{T}),$ when we look at $m_{T}$ as a map $G(T) \times G(T) \rightarrow G(T).$ We can make this condition present in $\textbf{Sch}_{S}.$ First, note that $S(T) = \{\pi_{T}\}.$ Then we have the set map $S(T) \rightarrow G(T)$ given by $\pi_{T} \mapsto e_{T},$ which gives $$e_{T} \times \mathrm{id}_{G(T)} : S(T) \times G(T) \rightarrow G(T) \times G(T)$$ given by $(\pi_{T}, x) \mapsto (e_{T}, x).$ Given any $S$-scheme map $\phi : T' \rightarrow T$, we have $\pi_{T} \circ \phi = \pi_{T'}$ and $$\phi_{T} \circ \phi = \phi^{*}(e_{T}) = e_{T'},$$ where the last equality uses that $\phi^{*} : G(T) \rightarrow G(T')$ is a group map. We see that $S(T') \times G(T') \rightarrow G(T') \times G(T')$ gives $$(\pi_{T} \circ \phi, x \circ \phi) = (\pi_{T'}, x \circ \phi) \mapsto (e_{T'}, x \circ \phi) = (e_{T} \circ \phi, x \circ \phi).$$ Hence, the map $e_{T} \times \mathrm{id}_{G(T)}$ is functorial in $T.$ Now, the property that $e_{T}$ is the identity of $G(T)$ is equivalent to saying that $$m_{T} \circ (e_{T} \times \mathrm{id}_{G(T)}) \circ \pi_{2,T} = \mathrm{id}_{G(T)} = m_{T} \circ (\mathrm{id}_{G(T)} \times e_{T}) \circ \pi_{2,T}$$ where the left-hand side is given by the following chain of the compositions $$G(T) \simeq S(T) \times G(T) \rightarrow G(T) \times G(T) \rightarrow G(T)$$ with $\pi_{2,T}$ being the first functorial bijection in $T$ given by the projection onto $G(T).$ Note that the map $e_{-} \times \mathrm{id}_{G(-)} : S(-) \times G(-) \rightarrow G(-) \times G(-)$ of functors corresponds to $e \times_{S} \mathrm{id}_{G} : S \times_{S} G \rightarrow G \times_{S} G,$ and likewise $\mathrm{id}_{G(-)} \times e_{-}$ corresponds to $\mathrm{id}_{G} \times_{S} e.$ Therefore, saying that $e_{T}$ is the identity element of $G(T)$ for every $T$ is equivalent to saying $$m \circ (e \times_{S} \mathrm{id}_{G}) \circ \pi_{2} = \mathrm{id}_{G} = m \circ (\mathrm{id}_{G} \times_{S} e) \circ \pi_{2},$$ which is often called the identity axiom for the group scheme $G$ over $S.$

Now, let's think about associativity of $G(T)$ given an $S$-scheme $T.$ Note that the following are equivalent:

• $(x \cdot y) \cdot z = x \cdot (y \cdot z)$ for all $x, y, z \in G(T)$;
• $m_{T} \circ (m_{T} \times \mathrm{id}_{G(T)}) = m_{T} \circ (\mathrm{id}_{G(T)} \times m_{T})$;
• $m \circ (m \times_{S} \mathrm{id}_{G}) = m \circ (\mathrm{id}_{G} \times_{S} m).$

The last condition is called the associative axiom for $G.$

Given $g \in G(T),$ the inverse $g^{-1} = i_{T}(g)$ satisfies $$m_{T}(g, i_{T}(g)) = e_{T} = m_{T}(i_{T}(g), g).$$ Giving such condition for all $g \in G(T)$ is equivalent to saying that $$m_{T}(g, i_{T}(g)) = e_{T} = m_{T}(i_{T}(g), g),$$ which is equivalent to saying that $$m \circ (\mathrm{id}_{G}, i) = e \circ \pi_{T} = m \circ (i, \mathrm{id}_{G}).$$ This is called the inverse axiom for $G.$

Left multiplication map. Let $G$ be an $S$-group. Given any $g \in G(S),$ for any $S$-scheme $\pi_{T} : T \rightarrow S,$ we have $\pi_{T}^{*}(g) = g \circ \pi_{T} \ \in G(T).$ We then have the map $l_{g, T} :G(T) \rightarrow G(T)$ given by $x \mapsto (g \circ \pi_{T}) \cdot x = m_{T}(g \circ \pi_{T}, x),$ which is functorial in $T$ because given any $S$-scheme map $\phi : T' \rightarrow T,$ we have $$\phi^{*}(\pi_{T}^{*}(g) \cdot x) = \phi^{*}(\pi_{T}^{*}(g)) \cdot \phi^{*}(x) = (\pi_{T} \circ \phi)^{*}(g) \cdot \phi^{*}(x) = \pi_{T'}^{*}(g) \cdot \phi^{*}(x)$$ for all $x \in G(T),$ which implies that $\phi^{*} \circ l_{g,T} = l_{g,T'} \circ \phi^{*}.$ Thus, there must be a unique $S$-scheme map $l_{g} : G \rightarrow G$ such that for any $x \in G(T),$ we have $$l_{g} \circ x = l_{g, T}(x) \in G(T).$$ Since $\pi_{T} = \pi_{G} \circ x = \pi_{T}$ and $x = \mathrm{id}_{G} \circ x,$ we have $$(g \circ \pi_{G}, \mathrm{id}_{G}) \circ x : T \rightarrow G \times_{S} G,$$ so $$\begin{align*}l_{g, T}(x) &= m_{T}(g \circ \pi_{T}, x) \\ &= m \circ (g \circ \pi_{T}, x) \\ &= m \circ (g \circ \pi_{G}, \mathrm{id}_{G}) \circ x\end{align*}$$ for every $x \in G(T).$ This implies that $$l_{g} = m \circ (g \circ \pi_{G}, \mathrm{id}_{G}).$$ Of course, it is quite easy to define $l_{g} : G \rightarrow G$ as the $S$-scheme map given by this formula, but without understanding all these functorial descriptions, it seems difficult to prove something like following:

Theorem. Let $G$ be an $S$-group. Given $g, g' \in G(S),$ we have $$l_{g \cdot g'} = l_{g} \circ l_{g'} : G \rightarrow G.$$

Proof. Fix any $S$-scheme $T.$ For each $x \in G(T),$ we have $$\begin{align*}l_{g \cdot g'} \circ x &= l_{g \cdot g', T}(x) \\ &= \pi_{T}^{*}(g \cdot g') \cdot x \\ &= \pi_{T}^{*}(g) \cdot \pi_{T}^{*}(g') \cdot x \\ &= l_{g,T}(l_{g',T}(x)) \\ &= l_{g} \circ (l_{g',T}(x)) \\ &= l_{g} \circ (l_{g'} \circ x) \\ &= (l_{g} \circ l_{g'}) \circ x.\end{align*}$$ Hence, take $T = G$ and $x = \mathrm{id}_{G}$ in the above computation to finish the proof. $\Box$