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.

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    $\begingroup$ 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.) $\endgroup$
    – LSpice
    Feb 15, 2020 at 3:35

1 Answer 1


I don't think that there is a different natural construction of a group scheme associated to a finite group. You can assume that authors mean this construction unless otherwise stated.

Notice, however, that this construction can be also done in the functorial setup, and a little bit more general and elegant as well:

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 $S$.


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