Milne's definition of component group

Question

I am trying to understand the following definition from Milne's Algebraic Groups. Let $G$ be an algebraic group over $k$ and let $k^s$ be the separable closure of $k$. Let $\pi_0(G)$ be the etale $k$-scheme of connected components of $G$ and let $G\to \pi_0(G)$ be the canonical morphism of algebraic schemes. Taking the $k^s$ points of this morphism, one gets $G(k^s)\to \pi_0(G_{k^s})$. Milne then claims that because the identity component $G^\circ$ is normal in $G$, then there exists a unique group structure on $\pi_0(G_{k^s})$ such that the map $G(k^s)\to \pi_0(G_{k^s})$ is a group homomorphism respecting the Galois action, which would then make $\pi_0(G)$ an etale group scheme over $k$.

Does this mean that the map $G(k^s)\to \pi_0(G_{k^s})$ is a surjection? And the fibres of this map are the connected components of $G(k^s)$ (topology induced from $G_{k^s}$) and $G^\circ (k^s)$ is the connected component of the identity in $G(k^s)$? If so, why are these true?

I know that the map $G_{k^s}\to \pi_0(G_{k^s})$ is surjective, the fibres are the connected components of $G_{k^s}$. I am not sure of the situation when $G_{k^s}$ is replaced by $G(k^s)$. Could $G(k^s)$ be empty? For example, in the book of Qing Liu, existence of an element of $G(k^s)$ seems to require $G$ be geometrically reduced which is not generally satisfied by algebraic groups as defined by Milne.

Answer 1

I think this could be a legitimate mistake. Namely, Milne does seem to implicitly be saying that the map $G(k^\mathrm{sep})\to \pi_0(G_{k^\mathrm{sep}})$ is a surjection. But, this does not have to be the case.

Example: Let $k$ be a separably closed, but not perfect field of characterististic $2$ (e.g. $k=\mathbb{F}_2(\!(t)\!)^\mathrm{sep}$). As $k$ is not perfect, there exists some $\alpha \in k$ which is not a square. Consider the group functor

$$G\colon \mathbf{Alg}_k\to \mathbf{Grp},\qquad A\mapsto \{a\in A: a^4=\alpha a^2\},$$

which one can consider as a group under addition. This is obviously a finite type group $k$-scheme represented by $\mathrm{Spec}(k[x]/(x^4-\alpha x^2)$. Let us then observe that

$$G_{\overline{k}}\cong \mathrm{Spec}(\overline{k}[x]/(x^4-\alpha x^2)\cong \mathrm{Spec}(\overline{k}[x]/((x^2-\beta x)^2),$$

where $\beta$ is a square root of $\alpha$ in $\overline{k}$. In particular, as thickenings by nil-ideals are homeomorphisms (see Tag 0BR6) we see that

$$|G_{\overline{k}}|\cong |\mathrm{Spec}(\overline{k}[x]/(x^2-\beta x))|\cong |\mathrm{Spec}(\overline{k}[x]/(x-\beta))|\sqcup |\mathrm{Spec}(\overline{k}[x]/(x)|\cong |\mathrm{Spec}(\overline{k})|\sqcup |\mathrm{Spec}(\overline{k})|,$$

(where $|\cdot|$ denotes the underlying topological space). In particular, we see that $\pi_0(G_{\overline{k}})$ has two elements.

The upshot here is that as $\overline{k}/k$ is a totally inseparable extension, the map $G_{\overline{k}}\to G$ is a homeomorphism (see Tag 0BRD) and so the map $\pi_0(G_{\overline{k}})\to \pi_0(G)$ is a bijection. So, $\pi_0(G)$ has two elements. That said, $G(k)$ has one element as if $a^4=\alpha a^2$ where $a$ is non-zero, then by dividing by $a$ we'd deduce that $\alpha$ is a square, which is a contradiction. Thus, the map $G(k)\to \pi_0(G)$ cannot be a surjection. $\blacksquare$

This is not really an issue though. If $G$ is an algebraic group over any field $k$, one can still define a group $k$-scheme structure on $\Pi_0(G)$ (where I am using $\Pi_0$ opposed to $\pi_0$ just to emphasize the group scheme versus the group: so $\Pi_0(G)(k'):=\pi_0(G_{k'})$) such that $\Pi_0(G)$ is finite etale and the map $G\to \Pi_0(G)$ is a group homomorphism (and, in fact, the initial one to etale group schemes). This just follows essentially by applying the functor $\Pi_0$ to the maps

$$m\colon G\times_k G\to G,\quad i\colon G\to G, \quad e\colon \mathrm{Spec}(k)\to G$$

and using the fact that $\Pi_0$ commutes with fiber products over $k$. This is explained in Chapter II, §5, № 1 of Demazure--Gabriel's book Groupes Algebriques, Tome I.

Of course, if $k$ is perfect, then what Milne says is a shortcut to this.