Dtjytyk

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 $mathcalO_A,x$ 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 $A×_K barK$ 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 $barK/K$ 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

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 $ein G(K)$. 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.