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I am reading some materials about group schemes.I am very confused about some definitions.I know what is a group scheme and also know what is an affine group scheme over $k$, $k$ is an commutative ring with identity.But I can not find definition about the finite group scheme over $k$.

I guess that a finite group scheme over $k$ is like $Spec A$, $A$ is $k$-algebra of finite type. Is it right? Could you give me a strict definition? And does finite group scheme have relation with group?

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A map of affine schemes $Spec(B) \to Spec(A)$ is finite if the corresponding homomorphism $A \to B$ makes $B$ into a finitely generated module over $A$. An (affine) finite group scheme over $k$ is a group scheme $Spec(A) \to Spec(k)$ such that $k \to A$ makes a $A$ a finitely generated $k$-module. In the case that $k$ is a field, then $A$ is a finite dimensional vector space, and so has only finitely many points.

If you don't want to restrict yourself to affine group schemes $G$, you can just imagine that the morphism $G \to Spec(k)$ need only locally look like the one above, i.e. $G$ has a cover by affines $Spec(A_{\alpha})$ such that each $A_{\alpha}$ is a finitely generated module over $k$.

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  • $\begingroup$ Thank you very much.One more question about commutative group schemes.Could you show me one example that an affine group scheme not commutative? $\endgroup$ – Mike Mar 30 '18 at 23:56
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    $\begingroup$ @YuShen Take your favorite non-abelian finite group and embed it into $GL_n$ via the standard representation $\endgroup$ – leibnewtz Mar 31 '18 at 0:12
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Group schemes are (usually?) considered over some fixed scheme $S$. A finite group scheme $G$ is a group scheme which is finite over $S$, which is not the same as being of finite type over $S$. It means that locally, e.g. for $G=\operatorname{Spec}(A)$ and $S=\operatorname{Spec}(k)$, the ring $A$ is finitely generated as a $k$-module. If $k$ is a field, it means that $A$ is a finite dimensional vector space. Hence if $S$ is the spectrum of an algebraically closed field, then the condition immediately implies that $G$ is a finite set of points when considered as a variety.

does finite group scheme have relation with group?

Yay for Algebraic Geometry. Of course it does! A group scheme is a generalization of a group. Take my example above, i.e. assume $S$ to be the spectrum of an algebraically closed field and view $G$ as a $k$-variety. Then we can identify $G$ with its closed points and the scheme morphisms that give it its group scheme structure also turn this set of closed points into a finite group.

An example of a non-abelian group scheme is therefore easily constructed from a non-abelian group. Embed $\mathfrak S_3 \subseteq \Bbb C^{3\times 3}$ as permutation matrices and let $I\subseteq\Bbb C[x_{ij}\mid1\le i,j\le3]$ be the ideal with $Z(I)=\mathfrak S_3$. Then, $I$ also defines a closed subscheme $G$ of $\Bbb A^{3\times 3}$ which becomes a nonabelian group scheme by restricting the matrix mulitplication morphism to it.

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  • $\begingroup$ Thank you very much.One more question about commutative group schemes.Could you show me one example that an affine group scheme not commutative? $\endgroup$ – Mike Mar 31 '18 at 0:09
  • $\begingroup$ @YuShen, basically what leibenwtz said. See my edit. $\endgroup$ – Jesko Hüttenhain Mar 31 '18 at 6:05
  • $\begingroup$ @YuShen Or for an example which is less like just a finite group: Take the group scheme $SL_n$ over a field of characteristic $p>0$. The map that raises each entry of a matrix to the power $p$ is a homomorphism. The kernel of this map is a finite group scheme. $\endgroup$ – Tobias Kildetoft Mar 31 '18 at 6:41

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