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.

  • $\begingroup$ Surely the answer is going to be that fiber products takes you out of your category of "regular" covers. $\endgroup$ Oct 4, 2012 at 16:17
  • $\begingroup$ unknown, it sounds like you have an example in mind. $\endgroup$ Oct 4, 2012 at 16:19
  • 1
    $\begingroup$ 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? $\endgroup$ Oct 4, 2012 at 17:16
  • 1
    $\begingroup$ 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
    Oct 4, 2012 at 18:32
  • 2
    $\begingroup$ The simplest reason is that there many regular covers whose Galois closure is not regular. For example $\Q(t^{1/4})/\Q(t)$. $\endgroup$ Oct 4, 2012 at 18:45

1 Answer 1


$K = \mathbb{Q}$, $X_K = \mathrm{Spec} \mathbb{Q}[t, t^{-1}]$. 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.