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

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

• I took the liberty of changing the title of your question. – Piotr Achinger Mar 7 at 19:43

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.
• Thank you for your answer. One penible question: When you talk about the ${\rm Gal}(\bar K/K)$-action on $\bar G = G \otimes \bar{K}$ do you implicitely mean the action on "points" $\bar{G}(\bar{K})= Hom(\bar{K}, \bar{G})$ via composing $\phi \mapsto \phi \circ g$ for a $g \in {\rm Gal}(\bar K/K)$ or the action via "base change" namely that $g$ induces automorphism on $\bar{G}$ via $id \otimes g$? – KarlPeter Mar 7 at 21:06
• @Karl_Peter I think the latter action induces the former on $\bar K$-points. Here it is enough to say that the profinite group ${\rm Gal}(\bar K/K)$ acts continuously on the underlying topological space $|\bar G|$ of $\bar G$ (or just the subspace of closed points $\bar G(\bar K)$) and the quotient space is identified with $|G|$. – Piotr Achinger Mar 7 at 22:56
• But what is concretely the "canonical" action of ${\rm Gal}(\bar K/K)$ on the underlying topological space $|\bar G|$? – KarlPeter Mar 7 at 23:36
• The argument in the answer that the identity component of $G$ over the algebraic closure of $K$ is defined over $K$ applies only in characteristic zero. Otherwise the proof is more difficult. – anon Mar 9 at 7:01