Is a group object in the category of algebraic schemes over $k$ an honest group?

Question

Let $G$ be an algebraic scheme over a field $k$ (in the sense of these notes http://www.jmilne.org/math/CourseNotes/iAG200.pdf, I can try to explain more if anyone needs), and $m: G \times G \rightarrow G$ a morphism in this category. Here (I think) you can put a sheaf on the literal cartesian product $G \times G$ to make $G \times G$ into an algebraic scheme which is a categorical product with respect to the restriction maps (which will also be morphisms).

The pair $(G,m)$ is called a group object if there exist morphisms $e: \ast \rightarrow G$ and $\iota:G \rightarrow G$ which satisfy a bunch of diagram commutativity properties. Here $\ast$ is the one point scheme $\textrm{Max } k$. Since morphisms of schemes are also continuous functions on the underlying topological spaces, if I consider $m$ as a continuous function on the cartesian product $G \times G$, it seems like a group object is literally a group, whose identity is the image of the function $e$.

Is this the case? In the notes I'm reading, it seems like Milne is not saying this.

Answer 1

There's some confusion arising from the conventions in Milne's notes. As Alex G. remarks, the underlying set of the product of schemes is not equal to the cartesian product of the underlying sets in general. However, Milne works only with closed points (the underlying set of an affine scheme is the maximal spectrum instead of the prime spectrum) and only considers schemes of finite type over $k$.

If $k$ is algebraically closed, and $A$ and $B$ are schemes of finite type over $k$, then the set of closed points of the product of $A$ and $B$ is equal to the cartesian product of the sets of closed points of $A$ and $B$. This is also true for fibered products.

So using these conventions, when $k$ is algebraically closed, the "underlying set" (set of closed points) of an "algebraic group scheme" (group scheme of finite type over $k$) is an honest group.

Answer 2

It can't be in general, since the product of schemes $G \times G$ is not the Cartesian product as a set.

Answer 3

If $G$ is a group object in a category $C$ with finite products, $D$ is another category with finite products, and $F : C \to D$ is a functor that preserves finite products, then $F(G)$ is a group object in $D$. However, the underlying set functor from schemes (thought of as locally ringed spaces) to sets does not preserve products.

Things are better if you think of schemes in terms of their functors of points. Then you can say that a group object in a category $C$ is the same thing as an object $G$ such that $\text{Hom}(-, G)$ has a natural group structure (in the sense that it lifts to a functor taking values in groups). (The weaker statement that a group object $G$ gives such a group-valued functor can be thought of as the special case of the above observation for $F$ the Yoneda embedding.)

In the case of schemes, this means that the set $G(R)$ of $R$-points naturally forms a group, for any commutative $k$-algebra. If $k$ is a field then it's natural to take $R$ to be, say, the separable closure $k_s$ of $k$, in which case we can even hope that the functor $G \mapsto G(k_s)$ is faithful, and so attempt to think of algebraic groups in terms of their $k_s$-points. This is, as I understand it, the classical approach.