# Variety of Connected Components

In Milne's text http://www.jmilne.org/math/CourseNotes/iAG.pdf (A71), he introduces the "variety of connected components" of a finite type scheme $X$ over $k$ as the universal example of a zero dimensional variety $\pi_0(X)$ with a map $X\rightarrow \pi_0(X)$. In particular, the fibers of the maps will be the connected components of $X$. (Milne's definition of a variety is a finite type $k$ scheme that is also geometrically reduced and separable.)

In particular, he claims that 1) $\pi_0(X)$ exists 2) the map $X\rightarrow \pi_0(X)$ commutes with extension of the base field 3) $\pi_0(X\times_k Y) = \pi_0(X)\times_k\pi_0(Y)$. This is used to show that a connected algebraic group is irreducible by allowing us to reduce to the case $k=\overline{k}$ by base change (after which our group would still be connected).

However, these facts are not obvious to me, and I wondered if there was a reference or an explanation.

Attempts: To approach 1) I thought above associating a field ${\rm Spec}K$ to each connected component of $X$. If $X$ is connected, here $K$ is the largest field, separated over $k$ with a map $K\rightarrow \Gamma(X,\mathscr{O}_X)$. (If you have two such fields, you can take the composite, so there's a unique maximal one.)

However, then 2) and 3) are not obvious. In particular, for 2), if I take an inseparable extension $L$ of $k$ and base change by that, $\pi_0(X)\times_k L$ might have points that aren't separable over $k$ (so not geometrically reduced), which means something went wrong.

If I assume that he meant for $\pi_0(X)$ to just be a scheme and not geometrically reduced, then I have to think of what the right nonreduced structure is, and I'm not sure how to do that.

Anyways, I think I spent more time on a small technical detail than I should have, and I should just ask for help.

• Isn't it clear that $\pi_0 (X)$ should be a disjoint union of $n$ copies of $\operatorname{Spec} k$, where $n$ is the number of connected components? – Zhen Lin Oct 16 '14 at 18:59
• @ZhenLin The trouble is that this is not universal. For example, the $\mathbb{R}$-scheme $\operatorname{Spec} \mathbb{C}$ certainly has a map to $\operatorname{Spec} \mathbb{R}$, but it factors through the map to $\operatorname{Spec} \mathbb{C}$. So it's important to take the biggest separable thing possible. – Slade Oct 16 '14 at 19:04
• $\pi_0(X)$ is finite and étale over $k$. – Cantlog Oct 16 '14 at 19:31

This inspired me to slog through quite a few definitions, but I think the point is this: Being geometrically reduced is relative to the base field. Once we've passed to $L$, we no longer need to worry about elements inseparable over $k$, because we'll be doing our base changes with respect to $L$.
To make it crystal clear: if I take $k = \mathbb{F}_2 (t)$, $L = \mathbb{F}_2 (\sqrt{t})$, then $L$ is certainly not geometrically reduced over $k$, because $L\otimes_k L \cong L[T]/(T^2 - t) \cong L[T]/(T-\sqrt{t})^2$ is not reduced. But this problem disappears when we ask whether $L$ is geometrically reduced over itself, because $L\otimes_L L \cong L$ is perfectly well reduced.
• @Dtseng Geometrically reduced means that the base change to any field is still reduced. But the base change of $X\times_k Y$ along $k\subset k'$ is just $X_{k'} \times_{k'} Y_{k'}$. But the product of a geometrically reduced $k'$-scheme with a reduced $k'$-scheme is reduced. I think you check this stalkwise, but in any case there's a proof at the Stacks Project. – Slade Oct 17 '14 at 13:34
• @Dtseng Also, I'm not sure I follow the question about universality, but your choice of $K$ gives universality for the map $X\to \pi_0 (X)$, and this universality is preserved by various other universal things. – Slade Oct 17 '14 at 22:28
• I just mean that, if $X\rightarrow \pi_0(X), Y\rightarrow \pi_0(Y)$, then $X\times Y\rightarrow \pi_0(X)\times \pi_0(Y)$ is induced. However, I don't know if $\pi_0(X)\times \pi_0(Y)$ is as big as possible. Why should $\pi_0(X)\times \pi_0(Y)\rightarrow k$ factor through $\pi_0(X\times Y)\rightarrow k$? I don't get it free from fiber products since the map I want is going the other way. I looked what was written in the Stacks project, and I think the key fact is, if $R$ is reduced and Noetherian, then $R$ embeds into a product of fields (its total quotient ring). – DCT Oct 17 '14 at 22:41