By "Galois morphism" I mean a morphism $f: Y \to X$ such that $Y \times_X Y$ is a disjoint union of schemes isomorphic to $Y$.

Let $X$ be reduced curve over afield of char. 0. I wonder what is a simple example of an etale morphism $f: Y \to X$ which is not Galois in the above sense.

update: What about a "geometric" example? (i.e. the base field is algebraically closed).


You may take (over the characteristic zero field $\mathbb Q$) the morphism $$f:Y=\mathbb A_{\mathbb Q(\sqrt [3]2)}^1 \to X=\mathbb A_{\mathbb Q}^1$$ corresponding to the inclusion $\mathbb Q[T] \hookrightarrow \mathbb Q(\sqrt [3]2)[T]$


At Dima's request, here is a completely explicit geometric example.

Let $A=\frac {\mathbb C[T,Z]}{(Z^3-3Z+2T)} $ and $g:Spec (A)\to \mathbb A^1_\mathbb C$ the étale morphism corresponding to the inclusion $\mathbb C[T] \hookrightarrow A$

Then $g$ is not Galois because the field extension $\mathbb C(T)\hookrightarrow K=Frac (A)=\mathbb C(T)(z) \;$ is not Galois since the discriminant of $ \; Z^3-3Z+2T\;$ is $\;108(1-T^2)\;$ and it is not a square in $\mathbb C(T)$ .
However $g$ is not étale either since it is ramified over $T=\pm1$.
The remedy is simple: delete these two points i.e. corestrict $g$ to $$X=A^1_\mathbb C\setminus \lbrace 1,-1\rbrace=Spec(\mathbb C[T,\frac {1}{T^2-1}])$$ and restrict $g$ to $$Y=g^{-1}(X)=Spec(B)$$ where $B=A[\frac {1}{T^2-1}]=\mathbb C[T,\frac {1}{T^2-1}](z)$.

Finally the required étale non Galois morphism is the restriction $$f=res(g):Y\to X $$

A final remark
Actually there exist no non-trivial étale covers of the affine line $\mathbb A^1_k$ over an algebraically closed field $k$ of characteristic zero . (The affine line is then said, unsurprisingly, to be simply connected) .

However in characteristic $p\gt 0$, surprisingly, there exist many étale covers of the affine line .
The simplest is the Artin-Schreyer cover $\; Spec \frac {\mathbb k[T,Z]}{(Z^p-Z-T)}\to \mathbb A^1_k=Spec (k[T]) \;$but there are loads more: this is still a topic of current research.

  • $\begingroup$ what about a "geometric" example, i.e. when $Y$ and $X$ are defined over an algebrically closed field? $\endgroup$ Apr 6 '12 at 8:18
  • $\begingroup$ I have added such an example over $\mathbb C$ in an Edit. $\endgroup$ Apr 6 '12 at 12:02
  • $\begingroup$ Thank you, Georges! I didn't know about that argument using the discriminant. $\endgroup$ Apr 7 '12 at 10:40
  • 1
    $\begingroup$ Georges, sorry for pinging you on such an old post. I don't understand your final remark. I thought that if $k$ had characteristic zero then $\pi_1(\mathbb{A}^1_k)=\pi_1(k)$. So, in particular, if $k$ is not algebraically closed, then there are non-trivial finite etale covers of $\mathbb{A}^1_k$--they are exactly $\mathbb{A}_\ell^1$ for $\ell/k$ finite. $\endgroup$ Feb 22 '14 at 8:34
  • $\begingroup$ Dear @Alex: you are right . I have modified that statement.Thanks for your attention. $\endgroup$ Feb 22 '14 at 9:09

Let $C$ be a curve of genus $2$, say over $\mathbb C$. Think of $C$ as a Riemann surface for a moment. Then $\pi_1(C)$ is generated by four elements $a,b,c,d$ satisfying the relation $[a,b][c,d] = 1$. Let $p: \pi_1(C) \to S_3$ be the surjection that takes $a$ and $d$ to $(12)$, and $b$ and $c$ to $(123)$, and let $H \subset \pi_1(C)$ be the preimage under $p$ of a subgroup of order $2$ in $S_3$, so $H$ has index $3$ in $\pi_1(C)$ and is not normal.

Covering space theory shows that $H$ corresponds to a degree $3$ cover $C' \to C$ of Riemann surfaces which is not Galois. Now the Riemann existence theorem shows that $C'$ has a unique structure of algebraic curve over $\mathbb C$ so that $C' \to C$ is an etale morphism, which will not be Galois.

This is the simplest sample in some strict sense: $\pi_1$ of a genus zero Riemann surface is trivial, while $\pi_1$ of a genus one curve is abelian (so all subgroups are normal), and all index two subgroups of a group are normal; thus we have to go to genus $2$ and a degree $3$ cover in order to find an example, and this is what I have done. [Added: I should add that this is the simplest example if one wants an etale morphism of projective curves; Georges found a simpler example in his answer by considering non-projective curves.]

Of course, one could write down examples with explicit algebraic equations, but I would have to begin with a genus $2$ curve, which is of the form $y^2 = f(x)$ for some degree $5$ or $6$ equation, and then write down $C'$ explicitly. By Riemann--Hurwitz, $C'$ has genus $4$, so I would then have to write down an equation for a genus $4$ curve, and find an explict degree $3$ map to $C$. I haven't tried to do this; it's probably a good exercise, though.

  • $\begingroup$ Dear Matt, $\pi_1(C)$ is not free: I have just asked on MO. If you feel like commenting there, I'd be delighted: you'll understand why by looking at Mark Sapir's remark... $\endgroup$ Apr 6 '12 at 19:10
  • $\begingroup$ @Georges: Dear Georges, Oops, and thanks! I wrote this too early in the morning. Editing will ensue. Best wishes, $\endgroup$
    – Matt E
    Apr 6 '12 at 23:50
  • $\begingroup$ Why would the cover $C'\to C$ will be etale? Can't it be ramified? We can remove the ramification points, and an unramified finite morphism with smooth destination should be flat. We would thus end up with a non-projective example though. $\endgroup$ Apr 8 '12 at 9:27
  • $\begingroup$ @Dima: Dear Dima, I chose the cover $C' \to C$ to be a covering space (in the sense of topology) on complex points. This is precisely what etale means. Regards, $\endgroup$
    – Matt E
    Apr 8 '12 at 14:57

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