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. $\Box$