A connected group variety is geometrically connected I'm reading J.S. Milne's notes on abelian varieties. A current link is https://www.jmilne.org/math/CourseNotes/AV.pdf .
I have a question to following paragraph on page 14


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*My first question is why can we find a field $k'$ that is contained in $R$ and a proper superset of $k$, whenever $k$ is not algebraically closed in Quot$(R)$.


*In the comments to this question k is algebraically closed in field of fractions of k-domain that admits $k$ morphism to $k$. reuns gives an example, with $R=\mathbb{R}[t, i t]$ and $k=\mathbb{R}$. Is this a counter example?
( A connected group variety is a group scheme $V$ over Spec$(k)$ for $k$ a field such that $V$ is separated, of finite type and geometrically reduced over Spec$(k)$)
 A: There is a general result that if $X$ is a connected algebraic scheme over a field $k$ such that $X(k)$ is nonempty, then $X$ is geometrically connected. One way of seeing this is to note that the set $\pi_{0}(X_{k^{\mathrm{sep}}})$ of connected components of $X_{k^{\mathrm{sep} }}$ has an action of $\mathrm{Gal}(k^{\mathrm{sep}}/k)$, and so corresponds to a finite etale $k$-algebra $\pi(X)$ (Grothendieck-Galois theory), which can be identified with the largest etale $k$-subalgebra of $\Gamma (X,\mathcal{O}{}_{X})$. The factors of $\pi(X)$ correspond to the connected components of $X$. If $X$ is connected, then $\pi(X)$ is a field $k^{\prime}$ containing $k$, and if $X(k)$ is nonempty, then there is a $k$-algebra homomorphism $\Gamma(X,\mathcal{O}_{X})\rightarrow k$, and so $k^{\prime} =k$. Hence $\pi_{0}(X_{k^{\mathrm{sep}}})$ consists of a single element. [Unfortunately, I don't know a good reference for this.]
A: *

*If $\operatorname{Spec} R\to\operatorname{Spec} k$ is of finite type and connected but not geometrically connected, we can find a subfield $k\subset k'\subset R$ where the inclusion $k\subset k'$ is proper: if one could not do this, we would have a $k$-point, and for such $k$-schemes we know that they are geometrically connected.


*The spectrum of $A=\Bbb R[t,it]\cong\Bbb R[x,y]/(x^2+y^2)$ is geometrically connected - the base change up to $\Bbb C$ is $\Bbb C[x,y]/(x^2+y^2)\cong \Bbb C[t,u]/(tu)$ which has connected spectrum (two lines meeting at a point). On the other hand, $\Bbb R$ is not closed in $\operatorname{Frac}(A)\cong \Bbb C(t)$. So this is a counterexample to the claim cited by Milne as (AG 11.7). (In fact, this result has been removed from the latest edition of Milne's AG text.) Do note that it's not a counterexample to the point made in part 1, since $\Bbb R$ is algebraically closed in $A$ for degree reasons.
The fact that the claim in part 2 is incorrect is maybe not completely surprising - talking about $k(V)$ means we're dealing with some birational property, but it's possible for a connected scheme to only be geometrically connected because of some specific proper closed subset and indeed that's what this example is demonstrating. With additional assumptions, like the fact that $V$ is a group scheme, one may be able to get around this by using the fact that group schemes "look the same everywhere", but I leave this proof for someone else.
