Variety of Connected Components

Question

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.

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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.

Answer 1

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.

Answer 2

Your construction might not be correct (though there may be a not obvious equivalence between your construction and the true construction, or maynot and with an obvious counter-example). First problem is that it may not well-defined that what is the large field in $A$. For example $A=k'\times k''$ for $k'\ncong k''$ non-trivial field extensions. And also there is no obvious indication that your $K$ is finite type over $k$.

I don't know whether Milne thought the construction is absolutely trivial. I must complain that it causes lots of difficulty to readers by not writing down the construction or even one reference.

The construction is based on Stack 0316.

Let $\bar{k}$ be the separable closure of $k$ and $\pi_0(X_\bar{k})$ be the set of irreducible components of $X_\bar{k}$. Then $\pi_0(X_\bar{k})$ has an $\mathrm{Gal}(\bar{k}/k)$ on it. Since we are talking about scheme of finite type, $\pi_0(X_\bar{k})$ is finite. For $X_1, X_2 \subset X$ disjoint, the orbit of $\pi_0((X_1)_\bar{k})$ and $\pi_0((X_2)_\bar{k})$ are disjoint. In fact by Lemma 038B the union of an orbit of $\mathrm{Gal}(\bar{k}/k)$ can be descended to a subscheme of $X$. Hence orbits correspond exactly to connected components of $X$.

We now look at one orbit and assume $X$ is connected. In view of Lemma 038D, we can take the kernel $N$ of $\mathrm{Gal}(\bar{k}/k) \rightarrow \mathrm{Perm}(\pi_0(X_\bar{k}))$. Then $\bar{k}^N=:k'$ satisfies that $X_{k'}$ has a trivial Galois action on it. By Lemma 038D(1), trivial Galois action is equivalent to geometrically connectedness in each components. So $k'$ is the minimal field extension that makes $X$ geometrically connected in each components and is the $\pi_0(X)$ we want to construct. In fact the action of $\mathrm{Gal}(\bar{k}/k)$ on $\pi_0(X_\bar{k})$ can be identified with the action on $\bar{k}\otimes_kk'$.

In view of Lemma 038D(2) and the fact that the action of $\mathrm{Gal}(\bar{k}/k)$ on $\pi_0(X_\bar{k})$ can be identified with the action on $\bar{k}\otimes_kk'$, the condition (2) of varieties of connected components is easy to shown. In fact let $k_1, k_2$ correspond to $\pi_0$ of $X$ and $Y$, then the Galois action on $X\times Y$ can be identified with that on $\bar{k}\otimes k_1 \otimes k_2$. So we get (3).

I find a gap in the construction. We must have a morphism from $X$ to $\pi_0(X)$. $X_\bar{k} = \coprod_{\sigma \in \pi_0(X_\bar{k})} (X_\bar{k})_\sigma$ is a decomposition into connected components. Every $(X_\bar{k})_\sigma$ is naturally a $\bar{k}$-scheme. Then we have a morphism $X_\bar{k} \rightarrow \coprod_{\sigma \in \pi_0(X_\bar{k})} (\bar{k})_\sigma$. Note that $\coprod_{\sigma \in \pi_0(X_\bar{k})} (\bar{k})_\sigma$ has a Galois action on it, consisting of the action on $\pi_0(X_\bar{k})$ and on $\bar{k}$ it self. Note that the morphism is compatible with the Galois action since we use the 'natural' $\bar{k}$ structure on $X_\bar{k}$. I explain the naturality in the following.

The natural structure can be seen locally. Assume $X = \mathrm{Spec}(A)$. Then $X_\bar{k} = \mathrm{Spec}(A\otimes\bar{k})$. Let $A\otimes\bar{k} = \prod A_i$. Then multiplication of $\bar{k}$ on $A\otimes \bar{k}$ gives the multiplication of $\bar{k}$ on each $A_i$. This gives the natural $\bar{k}$-algebra structure on each $A_i$ and each $\bar{k} \rightarrow A_i$ summing up to be $\prod (\bar{k})_i \rightarrow \prod A_i$. Let $\sigma \in \mathrm{Gal}(\bar{k}/k)$. Since $\sigma(la)= \sigma(l) \sigma(a)$ for $l \in \bar{k}$ and $\sigma(a) \in A$ and $(l)_i(a)_i = la$ for $a \in A_i= A_i \times 0 \times \cdots \times 0 \subset A$ where we write $(l)_i$ (resp. $(a)_i$) to indicate that it is considered as an element in the $i$-th component of $\prod (\bar{k})_i$ (resp. $\prod A_i$), we have that the morphism is compatible with the Galois action.

As the construction, $\prod_{\sigma \in \pi_0(X_\bar{k})} (\bar{k})_\sigma \simeq \pi_0(X)\otimes_k\bar{k}$ with the Galois action acting on the right tensor product component. Hence when you do Galois descent, you get $\pi_0(X)$, i.e. when you do Galois descent, $X_\bar{k} \rightarrow \coprod_{\sigma \in \pi_0(X_\bar{k})} (\bar{k})_\sigma$ becomes $X \rightarrow \pi_0(X)$.

You can come up with the idea by yourself once your read through that stack thread.