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I know that a group scheme is an $S$-scheme $G$ equipped with one of the equivalent sets of data

  • a triple of morphisms $μ$: $G$ ×S $G$ → $G$, $e$: $S$ → $G$, and $ι$: $G$ → $G$, satisfying the usual compatibilities of groups (namely associativity of $μ$, identity, and inverse axioms)
  • a functor from schemes over $S$ to the category of groups, such that composition with the forgetful functor to Set is equivalent to the presheaf corresponding to $G$ under the Yoneda embedding.

But I can't understand the equivalence. I understand that if we give the triple morphisms i have the other, but given a Functor how can i can the triple morphisms.

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    $\begingroup$ I think most of these details can be found in Jantzen's Representation of Algebraic groups (I don't have it right here so I will have to check later). Basically, the functor going to the category of groups provides a structure of a Hopf algebra on the algebra of regular functions, and this structure then yields the required morphisms. $\endgroup$ Mar 16, 2014 at 14:52

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Given the representable functor, one way you can find the morphisms and prove their various compatibilities is by looking at judicious ``test objects."

More precisely, suppose that $G$ is an $S$-scheme whose functor of points $T\mapsto G(T)$ factors through the category of groups, i.e., $G(T)$ is a group for each $S$-scheme $T$, and if $T_1\to T_2$ is an $S$-morphism, $G(T_2)\to G(T_1)$ is a group homomorphism.

The multiplication is supposed to be a morphism $m:G\times_S G\to G$, i.e., an element of $G(G\times_SG)$. Do we know any arrows in this group? The only completely obvious ones are the projections $p_1,p_2:G\times_SG\to G$. So a reasonable thing to try is $m:=p_1*p_2$, where $*$ denotes the multiplication on the group $G(G\times_SG)$. For the identity section, one takes the identity element of the group $G(S)$, and for the inverse, look at the inverse of $\mathrm{id}_G$ in $G(G)$.

Another, perhaps more direct way to do this, is to observe that the functoriality in the functor of points definition says precisely that there is a natural transformation of functors $m:h_G\times h_G\to h_G$ which gives, for each $S$-scheme $T$, the group operator $G(T)\times G(T)\to G(T)$. Similarly there is a natural transformation $\epsilon:h_S\to h_G$ such that the image of the unique point of $S(T)$ in $G(T)$ is the identity element for the group operator given by $m$, for all $S$-schemes $T$. Likewise for the inversion. All these natural transformations, by Yoneda's lemma, are induced by uniquely determined morphisms of $S$-schemes, and the various compatibilities (associativity and so forth) hold because they hold on $T$-points for every $S$-scheme $T$.

EDIT: As Matt E indicates in the comments, the real content of this answer is the "perhaps more direct way." What I'm doing in the first few paragraphs is explicitly finding the morphisms inducing given natural transformations of the functor of points $h_G$, i.e., implementing Yoneda's lemma explicitly (for morphisms from $G\times G$ to $G$, from $S$ to $G$, and from $G$ to $G$, respectively). So it's not really a different way of doing it. Also, as Kevin Carlson remarks, and this is definitely worth pointing out, this has nothing to do with schemes. One can work with group objects in any category with products and do the same thing.

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    $\begingroup$ Dear Keenan, I think the "perhaps more direct way" is really the correct answer. Plugging in the test objects (as in your first couple of paragraphs) is then just going through the proof of Yoneda in this particular case. Cheers, $\endgroup$
    – Matt E
    Mar 16, 2014 at 15:15
  • $\begingroup$ It may be worth pointing out that none of this has anything to do with schemes! $\endgroup$ Mar 16, 2014 at 15:15
  • $\begingroup$ Dear @Matt E, You're absolutely right. Thank you for pointing this out. $\endgroup$ Mar 16, 2014 at 15:21
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    $\begingroup$ Dear @Kevin Carlson, That's a good point. Thank you! $\endgroup$ Mar 16, 2014 at 15:22

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