Let $X$ be a curve of genus three which is not hyperelliptic. Then $X$ is trigonal, i.e., there exists a finite morphism $X \to \mathbf P^1$ of degree $3$.

Let $Y\to X \to \mathbf P^1$ be a Galois closure of a trigonal curve. Then, unless $X\to \mathbf P^1$ is the Klein curve, the map $Y\to \mathbf P^1$ is of degree $6$.

Now, surely the latter map could have non-abelian Galois group, but I don't know any explicit examples. Can somebody give me an explicit example for which it is clear that the Galois group over $\mathbf P^1$ of the Galois closure is non-abelian?

Can we describe all such trigonal curves in the moduli space? How big is the dimension of the locus of curves which are hyperelliptic OR have a trigonal map $X \to \mathbf P^1$ whose Galois closure is cyclic (of degree 6) over $\mathbf P^1$?

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    $\begingroup$ If a field extension of degree 3 is not Galois, then its Galois closure has Galois group $S_3$, which is non-abelian. So either your morphism $X \to {\mathbb P}^1$ is already Galois (and then abelian), or else $Y \to {\mathbb P}^1$ has non-abelian Galois group. $\endgroup$ Mar 23, 2013 at 12:47
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    $\begingroup$ Every group of order 6 that acts faithfully on a set of 3 elements is the full symmetric group on that set. So, if the Galois closure does have degree 6, then it automatically has Galois group isomorphic to the symmetric group on 3 elements. $\endgroup$ Mar 23, 2013 at 12:53
  • $\begingroup$ @Michael: I didn't see your comment before I added my comment. $\endgroup$ Mar 23, 2013 at 12:54
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    $\begingroup$ What has the Klein curve to do with this? For any linear and quartic forms $P_1(x,y)$ and $P_4(x,y)$, the curve $X: z^3 P_1 = P_4$ admits a cyclic degree-3 map $y/x$ to the projective line, and $X$ is rarely (is it ever?) isomorphic with the Klein quartic. $\endgroup$ Sep 24, 2015 at 17:25

1 Answer 1


Turning my comment into an answer:

It is a general fact from Galois thory that a field extension of degree 3 is either Galois with cyclic Galois group, or else its Galois closure has non-abelian Galois group $S_3$. So unless your original morphism $X \to {\mathbb P}^1$ is a Galois covering, the Galois closure $Y \to {\mathbb P}^1$ will have non-abelian Galois group.

Assuming you work over $\mathbb C$ (or any other field containing cube roots of unity), Kummer theory tells us that the Galois coverings of ${\mathbb P}^1$ of degree 3 are exactly the superelliptic curves of the form $y^3 = f(x)$ (with map to ${\mathbb P}^1$ given by the $x$-coordinate), where $f$ is a cube-free polynomial and $X$ is taken to be the smooth projective model of the affine curve given by the equation.


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