# Viewing a finite group as a group scheme

I've read books which have this statement, without explanation : 'every finite group is an algebraic group'. I'm trying to understand what exactly they mean. The definition I have in my mind of a group scheme $$G$$ over a base field $$k$$ is a representable functor $$G : (\text{Sch}/k)^{op} \rightarrow (\text{Grp})$$ I discussed this with my professor and here's what he explained to me (which I find very beautiful):

Suppose $$G$$ is a finite group. Consider the affine scheme $$X_G = \coprod_{g \in G} \text{spec} (k) = \text{spec} ( \prod_{g \in G} k)$$ which is the disjoint union of $$|G|$$ copies of $$\text{spec} (k)$$. This gives the corresponding Yoneda functor and for any $$k$$-scheme $$S$$, we have $$X_G(S) = \text{Hom} (S, X_G) = \text{Hom}_{k} \left( \prod_{g \in G} k, \Gamma(S, \mathcal{O}_S) \right)$$ Now, a $$k$$-algebra morphism from $$\prod_{g \in G} k$$ to $$\Gamma(S, \mathcal{O}_S)$$ is determined by knowledge of the images of the idempotents $$e_g \in \prod_{g \in G} k$$. So $$X_G(S) = \{ (s_g)_{g \in G} | s_g \in \Gamma(S, \mathcal{O}_S) , \sum s_g = 1, s_g s_{g'} = 0, (s_g)^2 = s_g \}$$ The group structure is given by $$(s_g) \cdot (t_g) = (u_g)$$ where $$u_g = \sum_{hk=g} s_h t_k$$. The fact that $$(u_g)$$ also satisfies the property of being in $$X_G(S)$$ is clear from computation. $$X_G(S)$$ indeed becomes a group with identity given by $$(a_g) : a_e = 1$$ and $$a_g = 0$$ when $$g \neq e$$ and the inverse of $$(s_g)$$ is $$(t_g)$$ with $$t_g = s_{g^{-1}}$$.

So my (poorly phrased) question: Is this the only way to view a finite group as a group scheme or are there other ways too?

For context, this really came up in a discussion about quotients in algebraic groups. If $$G$$ is $$GL_n$$ and $$T$$ is a maximal torus with normalizer $$N_G(T)$$, then $$N_G(T)/T = S_n$$, the symmetric group on $$n$$ letters.

• This is not the only way, but it's the way that's meant unless something else is stated. For example, $\mu_n$ is finite, but it doesn't arise this way. The significance of your construction is that the scheme is constant: i.e., its group of rational points is $G$, regardless of the ground field (whereas, for example, $\mu_3(\mathbb R)$ is trivial while $\mu_3(\mathbb C)$ has order $3$). (I think—this is my naïve understanding, as a mere user of and not expert in group schemes, and hopefully someone will correct me if I've used the words incorrectly.) – LSpice Feb 15 '20 at 3:35

Let $$S$$ be any base scheme. If $$G$$ is any set, define the functor $$G(-) : (\mathsf{Sch}/S)^{\mathrm{op}} \to \mathsf{Set}$$ by $$G(X) := \{f : X \to G \text{ locally constant}\}.$$ The action on morphisms is clear. The functor $$G(-)$$ is representable by the $$S$$-scheme $$\coprod_{s \in G} S$$, since a locally constant map $$X \to G$$ corresponds to a partition $$X = \coprod_{g \in G} X_g$$ into disjoint open subschemes, which corresponds to an $$S$$-morphism $$X \to \coprod_{g \in G} S$$ (since $$\mathsf{Sch}$$ is extensive).
If $$G$$ is a group, then the functor $$G(-)$$ factors over $$\mathsf{Grp}$$. It follows that $$\coprod_{g \in G} S$$ carries the structure of a group scheme over $$S$$. (There is no need to write down the multiplication map etc. We simply get it from the functorial characterization of group schemes.) If $$G$$ is finite and $$S$$ is affine, then $$\coprod_{g \in G} S$$ is affine and of finite type over $$S$$.