Realizing a finite group as a scheme

Question

Suppose $G$ is a finite group. I have seen in various sources, without explanation, that we can interpret $G$ as a scheme by letting $G:=\coprod_{g\in G}\operatorname{Spec}\mathbb{Z}$. Why and how can we make sense of $G$ in this way -- i.e. how is it compatible with the original group $G$? What is the group law given this interpretation?

If we for example let $G=\mathbb{Z}/2\mathbb{Z}$, then for a scheme $T$, we have that $G(T)=\coprod_{g\in G}\operatorname{Hom}(T,\operatorname{Spec\mathbb{Z}})$, which is a set with two elements since $\operatorname{Spec}\mathbb{Z}$ is the final object. But how can we impose a group structure that matches the original structure on $G$?

Answer 1

I will assume first we are working with schemes over a field $k$, so that we can think about this classically. That way, we can define the group structure as we would on a set, and look at how regular functions pull back.

Let's denote $X = \bigsqcup_{g \in G} \operatorname{Spec} k = \{P_g\:|\;g \in G\}$ and define a function $X \times X \to X$ by $(P_g, P_h) \mapsto P_{gh}$. Now, a regular function $\varphi$ on $X$ is just an element $(x_g)_{g \in G} \in k^{G}$. Similarly, regular functions on $X \times X$ are just given by an element $(x_{(g, h)})_{g,h \in G \times G} \in k^{G \times G}$. Hence, given such a function $(x_g)_{g \in G} \in k^{G}$, its pullback to $X \times X$ is the function which sends $(P_g, P_h) \mapsto P_{gh} \mapsto x_{gh}. $

We can then describe the associated homomorphism of rings $k^{G} \to k^{G \times G}$ to be given explicitly $(x_g)_{g \in G} \mapsto (x_{gh})_{(g, h) \in G \times G}$, and by construction, when we take the Spec of this map, we recover the usual group operation on $G$, when we identify $X$ and $G$ via $P_g \mapsto g$.

Now, notice that this homomorphism $k^G \to k^{G \times G}$ was natural and did not rely at all on $k$ being a field, so we can just as easily define such a map $\mathbb{Z}^G \to \mathbb{Z}^{G \times G} \cong \mathbb{Z}^G \otimes_{\mathbb{Z}} \mathbb{Z}^G$. When we base change to an arbitrary field $k$, we recover the same $X$ and the multiplication morphism defined before, so in this way, we can recover the group operation of $G$.

Answer 2

Another perspective is the following: There is a (product-preserving) functor $F\colon\mathrm{Set}^{op}\to\mathrm{Rings}$ which sends a set $A$ to the direct product $\prod_A\mathbb Z=\mathrm{Fun}(A,\mathbb Z)$, the space of functions $A\to\mathbb Z$ where products are given component-wise, and maps $A\to B$ are sent to pullbacks $\mathrm{Fun}(B,\mathbb Z)\to \mathrm{Fun}(A,\mathbb Z)$. There is also the standard (product-preserving) functor $\mathrm{Rings}^{op}\to\mathrm{Schemes}$. The composition gives a functor $\mathrm{Set}\to\mathrm{Schemes}$. Now a group $G$ is simply a group object in the category $\mathrm{Set}$, which is sent to a group object in schemes, i.e., a group scheme.