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Let $K$ be a number field, and let $X_K$ be a $K$-variety.

The etale fundamental group of $X_K$, as defined in SGA1, classifies the automorphisms of finite etale covers of $X_K$. Some of these etale covers are not geometric. For example, if $L$ is a finite field extension of $K$, then $X_L\rightarrow X_K$ is a finite etale cover. A cover $Y$ of $X_K$ is called regular (think: geometric) if it doesn't have an extension of scalars (to be precise, if $K$ is algebraically closed in the function field of every connected component of $Y$).

I know, however, that there is no "regular etale fundamental group" for $X_K$, that classifies the automorphisms of just the regular finite etale covers of $X_K$. I vaguely recall hearing in a conference that this is because the regular etale covers don't form a Galois category. Is that correct? What fails for them to form a Galois category? This question has been in the back of my mind ever since I started working with the etale fundamental group.

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  • $\begingroup$ Surely the answer is going to be that fiber products takes you out of your category of "regular" covers. $\endgroup$
    – John Pardon
    Commented Oct 4, 2012 at 16:17
  • $\begingroup$ unknown, it sounds like you have an example in mind. $\endgroup$
    – James D. Taylor
    Commented Oct 4, 2012 at 16:19
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    $\begingroup$ Is it fanciful to think of this as being a bit like the non-existence of a maximal totally ramified extension of a local field? $\endgroup$
    – David Loeffler
    Commented Oct 4, 2012 at 17:16
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    $\begingroup$ Just take the fiber product of a regular cover with itself. The correct "relative" substitute for the fundamental group of X is Deligne's relative fundamental groupoid, as defined in “Le Groupe Fondamental de la Droite Projective Moins Trois Points” math.ias.edu/files/deligne/GaloisGroups.pdf. $\endgroup$
    – Angelo
    Commented Oct 4, 2012 at 18:32
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    $\begingroup$ The simplest reason is that there many regular covers whose Galois closure is not regular. For example $\Q(t^{1/4})/\Q(t)$. $\endgroup$
    – Manish Kumar
    Commented Oct 4, 2012 at 18:45

1 Answer 1

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$K = \mathbb{Q}$, $X_K = \mathrm{Spec} \mathbb{Q}[t, t^{-1}]$. Let $Y$ and $Z$ be the covers of $X_K$ corresponding to adjoining $\sqrt{t}$ and $\sqrt{-t}$ respectively. They are both regular covers, but their composite contains $\sqrt{-1}$ and thus is not.

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