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>



Why is, for a group scheme of finite type, "smooth" (resp. irreducible) equivalent to "geometrically reduced" (resp. geometrically irreducible)?
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.


