0
$\begingroup$

Consider the definition of group scheme in Stack Project [022R]. In the paragraph following definition 39.4.1, it is said that

We have morphisms of schemes over $S$: (identity) $e:S\rightarrow G$ and (inverse) $i:G\rightarrow G$ such that for every $T$ the quadruple $(G(T)=Hom(T,G),m,e,i)$ satisfies the axioms of a group listed above.

As for my intuition, the identity is the identity of $G(S)$, and the inverse is the inverse of $id : G\rightarrow G$ in $G(G)$. But I don't know how to check that $(G(T),m,e,i)$ is a group.

Improve this question
$\endgroup$
8
  • $\begingroup$ I think we must add one more condition : $G(\cdot)$ is a functor from category of schemes to category of groups. $\endgroup$
    – XT Chen
    Commented Feb 8, 2021 at 16:15
  • 1
    $\begingroup$ Functoriality is automatic from the other conditions. What does it mean how to ask whether a definition is satisfied when you don't have an example you want to check? $\endgroup$
    – LSpice
    Commented Feb 8, 2021 at 16:24
  • $\begingroup$ @LSpice Sorry to ask. How to show the functoriality? And forgive my pool English reading ability, I don't know what "What does it mean how to ask whether a definition is satisfied when you don't have an example you want to check?" means. $\endgroup$
    – XT Chen
    Commented Feb 8, 2021 at 16:29
  • $\begingroup$ @LSpice I can show that now. Case closed. $\endgroup$
    – XT Chen
    Commented Feb 8, 2021 at 16:41
  • $\begingroup$ You said "I don't know how to check that $(G(T), m, e, i)$ is a group", but the way you check that depends on what $G$ is. So you can't check it until you've got an example in mind. $\endgroup$
    – LSpice
    Commented Feb 8, 2021 at 17:30

1 Answer 1

Reset to default
0
$\begingroup$

You need to use the condition that $m$ is a morphism of scheme. By that we can deduce $G(\cdot)$ is a functor.

Improve this answer
$\endgroup$
1
  • $\begingroup$ Are you now satisfied with this answer, or do you want something more? $\endgroup$
    – Will Sawin
    Commented Feb 10, 2021 at 16:37

Not the answer you're looking for? Browse other questions tagged .