It is known, that the étale fundamental group of a normal connected scheme equals the galois group of the maximal unramified extension of its function field.

This is not true for integral schemes in general. The point is somehow, that one cannot construct flat finite covers from a function field extension as one does with normalization. So my questions are as follows:

  1. What are examples for integral schemes, where the fundamental group is not a galois group of the function field? Or equivalently: A scheme with non-isomorphic étale covers but isomorphic function fields?

  2. Is there already an example which is one-dimensional, e.g. an algebraic curve or an order in a number field? Somehow the étale covers should look different over the singularities. What can happen there?

  3. In the case of curves it is known how to construct singular curves from normal ones. Is there a way of constructing étale covers of a singular curve by taking its normalization in a function field extension and then singularizing it again?

I thank you very much for your answers.

  • $\begingroup$ As you can see from ACL's example, by losing normality, you lose the correspondence between connected covers and field extensions. The cover gives you an étale algebra over the function field, whose spectrum is disconnected, but the gluing makes the cover itself a connected scheme. $\endgroup$
    – S. Carnahan
    May 5, 2013 at 13:35
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    $\begingroup$ It would be instructive to address the related issue of whether a connected finite etale cover of a non-normal irreducible (noetherian) $X$ is determined up to isomorphism by its generic fiber. In ACL's answer, the connected finite etale covers are determined up to isomorphism by their generic fiber. $\endgroup$
    – user29283
    May 5, 2013 at 15:18
  • $\begingroup$ No. Just take the self-product of two copies of ACL's example. There will be many etale covers of each degree. $\endgroup$
    – Will Sawin
    May 5, 2013 at 17:22
  • $\begingroup$ How does this answer xuhan's question? Taking the self-product of two degree n covers from ACL should give a disjoint union of n copies of it. More generally the product of two such covers of degree n and m receives a morphism from the cover of degree lcm(n,m). It is then isomorphic to gcd(n,m) many copies of that. So still every connected component is determined by its degree/generic fibre. $\endgroup$ May 5, 2013 at 18:39

1 Answer 1


Take a nodal cubic $C$, that is a projective line of which you identify two points $0$ and $\infty$. This curve has connected étale covers of any degree: take $n$ copies of the projective line, numbered circularly, and identify the $\infty$ of each of them with the $0$ of the next one. This is a $\mathbf Z/n\mathbf Z$-Galois cover of $C$.

(Taken from Hartsorne's Algebraic Geometry, page 276.)

  • $\begingroup$ Thanks. Is there a similar construction for cusps? $\endgroup$ May 5, 2013 at 18:10
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    $\begingroup$ @AndreasMihatsch, I believe the following is true: the normalization morphism for a cuspidal curve is a universal homeomorphism (being surjective, integral, and radicial), and thus induces an equivalence of etale sites. So the coverings of the cuspidal curve should be in bijection with the coverings of the affine line. $\endgroup$ Feb 10, 2016 at 18:08

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