[Math] Why is, for a group scheme of finite type, “smooth” (resp. irreducible) equivalent to “geometrically reduced” (resp. geometrically irreducible)

abelian-varietiesag.algebraic-geometryschemes

I have some questions about two statements from Bosch's "Algebraic Geometry and Commutative Algebra" about algebraic varieties (page 479). Since I still don't have the permission to add images I quote the relevant excerpt:

…The notion of properness has been introduced in 9.5/4. It means that the
structural morphism $p: A \to Spec(K)$ is of finite type, separated, and universally
closed. For the property of smoothness see 8.5/1. It follows from 8.5/15
in conjunction with 2.4/19 that all stalks $\mathcal{O}_{A,x}$ of a smooth $K$-group scheme $A$
are integral domains. Since abelian varieties are required to be irreducible, they
give rise to integral schemes. Also let us mention that for $K$-group schemes of
finite type smooth is equivalent to geometrically reduced, which means that all
stalks of the structure sheaf of $A×_K \bar{K}$ are reduced. In addition, let us point out
that for $K$-group schemes of finite type the property irreducible can be checked
after base change with $\bar{K}/K$ so that we may replace irreducible by geometrically
irreducible…

We fix an abelian variety $A$ over field $K$. By definition an abelian variety over $K$ is a proper smooth $K$-group scheme that is irreducible.

Following two questions:

  1. Why is for a $K$-group scheme of finite type smooth equivalent to geometrically reduced?

  2. Why under same conditions as in 1. (so $K$-group scheme of finite type) the property irreducible is equivlaent to geometrically irreducible?

Remark: Here I previously asked this question in MSE: https://math.stackexchange.com/questions/3136827/abelian-varieties

Best Answer

Let $G/K$ be a group scheme of finite type.

  1. $G/K$ is smooth if and only if $\bar G / \bar K$ is smooth. Suppose $\bar G$ is reduced, then it has a smooth $\bar K$-point $x$ (because we are over an algebraically closed field). But $\bar G(\bar K)$ acts transitively on itself, so now every closed point of $\bar G$ is smooth, so $\bar G$ is smooth. (And of course if $\bar G$ is smooth then it is reduced.)

  2. The point is that $G$ comes with a section, the neutral element $e\in G(K)$. Suppose that $\bar G$ is reducible, then since $\bar G^{\rm red}$ is reduced and hence smooth, we see that $\bar G$ is disconnected. If $\bar G^\circ$ is the connected component of the neutral element $e$, then since the Galois group ${\rm Gal}(\bar K/K)$ acts on $\bar G$ preserving $e$, it has to preserve $\bar G^\circ$, and so $\bar G^\circ$ descends to give a component of $G$, so $G$ is disconnected.

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