# How does an affine algebraic group become a group scheme?

Both in the sense of schemes, and in the sense of classical algebraic varieties, we have a notion of an algebraic group.

A group scheme $$G/S$$ is an $$S$$-scheme together with $$S$$-morphisms $$\mu: G\times_S G$$, $$\beta: G\to G$$, $$e: S\to G$$ which satisfy the "usual identities". We call it an algebraic group, if $$S=spec(k)$$ and $$G$$ is of finite type (and perhaps smooth) over $$k$$ [see Mumford-Fogarty-Kirwan, GIT].

An algebraic group in the sense of varieties is a group $$G$$, that has the structure of a variety (i.e., an irreducible zero locus of polynomials over $$\mathbb{C}$$), such that multiplication and inversion are regular maps.

This answer explains that you can obtain a group from a group scheme by passing on to $$S$$-rational points.

I would like to understand how/whether the converse is true in some sense. Let $$G$$ be an algebraic group variety over $$\mathbb{C}$$, with multiplication $$G\times G\to G$$, how do we get an algebraic group scheme from this? Of course, we get a scheme $$G'$$ of finite type over $$k$$ that corresponds to $$G$$. (In the affine case, this would simply be $$spec$$ instead of $$specmax$$, generally, this a functor as described in Hartshorne's first chapter). But the morphism $$\mu: G'\times_{spec\mathbb C} G' \to G'$$ is bugging me, as fibre products are not necessarily products of topological spaces.

The fibre product is not over $$\mathbb{C}$$, but over $$\mathrm{Spec}(\mathbb{C})$$, which is just a point.
• Yes, a fibre product over $\mathrm{Spec}(\mathbb{C})$ is just a Cartesian product. May 23, 2019 at 15:15