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$(R)$.
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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$(k)$)
Best Answer
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.]