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.


1 Answer 1


The fibre product is not over $\mathbb{C}$, but over $\mathrm{Spec}(\mathbb{C})$, which is just a point.

  • $\begingroup$ So is the scheme corresponding to an algebraic set that is constructed as a product is always the fibre product? $\endgroup$
    – Lukas
    May 23, 2019 at 15:00
  • $\begingroup$ Yes, a fibre product over $\mathrm{Spec}(\mathbb{C})$ is just a Cartesian product. $\endgroup$
    – Spenser
    May 23, 2019 at 15:15

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