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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.

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  • $\begingroup$ 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? $\endgroup$ – Zhen Lin Oct 16 '14 at 18:59
  • $\begingroup$ @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. $\endgroup$ – Slade Oct 16 '14 at 19:04
  • $\begingroup$ $\pi_0(X)$ is finite and étale over $k$. $\endgroup$ – Cantlog Oct 16 '14 at 19:31
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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.

This might also resolve your issues with 3). The basic point is a fibered product of geometrically reduced things is geometrically reduced, and a base change of something geometrically reduced is geometrically reduced.

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  • $\begingroup$ Thanks! This certainly helps. Do you know why the product of two geometrically reduced things is geometrically reduced (and why the product of a geometrically reduced scheme with a reduced scheme is reduced)? Also, now that we know that taking fiber products and base extensions gives us examples of finite etale schemes over the base field (to borrow the fancier terms from the comments above), do you know why they are the "largest" (or universal) examples? $\endgroup$ – DCT Oct 17 '14 at 12:57
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    $\begingroup$ @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. $\endgroup$ – Slade Oct 17 '14 at 13:34
  • $\begingroup$ @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. $\endgroup$ – Slade Oct 17 '14 at 22:28
  • $\begingroup$ 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). $\endgroup$ – DCT Oct 17 '14 at 22:41
  • $\begingroup$ @Dtseng Oh, I see. Yes, somehow I didn't realize that the arrows go the wrong way. $\endgroup$ – Slade Oct 17 '14 at 22:51

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