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Let $\phi: C \rightarrow \mathbb P^1$ a morphism (over a field of characteristic 0) from a rational curve $C$ to $\mathbb P^1$ of degree 3. By the Riemann-Hurwitz formula the degree of the ramification divisor is 4, hence $\phi$ can ramify in three possible ways: (a) two points of ramification index 3; (b) one point of ramification index 3 and two point of ramification index 2; (c) four points of ramification index 2.

The morphism $\phi$ corresponds to an extension of fields $k(t) \subset k(C)$ of degree 3. This extension can be a Galois extension or not.

(1) In which cases (a), (b) or (c) can this extension be Galois?

Suppose it is not, and suppose that its Galois closure $k(t) \subset k(C) \subset k(B)$ corresponds to the function field of another curve $B$, so that we have a morphism $\psi: B \rightarrow C$.

(1) What is the degree of the extension $k(C) \subset k(B)$ or, in other words, what is the degree of $\psi$?

(2) What is the genus of $B$?

(3) What are the possible ramification indexes of the composite morphism $\phi \circ \psi$?

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(1) If $k(C)/k(t)$ were Galois then the ramification index of a place $P$ of $k(C)$ in this extension would equal the size of the inertia group of $P$; since this inertia group is a subgroup of the Galois group, its size divides the size of the Galois group, namely $3$ in this case. So if the extension were Galois then there cannot be a point of ramification index $2$.

Next, if there are two points of ramification index $3$ then the extension can be Galois or not. For instance, if $x^3=t$ then $\mathbf{Q}(x)/\mathbf{Q}(t)$ is a non-Galois degree-$3$ extension which has two ramification points ($x=0$ and $x=\infty$), both of which have ramification index $3$. On the other hand, $\mathbf{C}(x)/\mathbf{C}(t)$ is Galois and has the same ramification as $\mathbf{Q}(x)/\mathbf{Q}(t)$.

We can say more if there are two ramification points and they are both $k$-rational. Since $C$ is a rational curve, its function field $k(C)$ is isomorphic to $k(x)$. Now if the ramification points are $k$-rational then they correspond to two places of the form $x=\alpha$ with $\alpha\in k\cup\{\infty\}$. By replacing $x$ by $\mu(x)$ for some degree-one $\mu(X)\in k(X)$, we may assume that these two places are $x=0$ and $x=\infty$. By replacing $t$ by $\nu(t)$ for some degree-one $\nu(X)\in k(X)$, we may assume that the places under $x=0$ and $x=\infty$ are $t=0$ and $t=\infty$, and also that the place $t=1$ lies under $x=1$. It follows that $t=f(x)$ for some degree-$3$ rational function $f(X)\in k(X)$ such that $f^{-1}(\infty)=\{\infty\}$, $f^{-1}(0)=\{0\}$, and $f(1)=1$. Thus $f(X)=X^3$. So in this case the function field extension is $k(x)/k(x^3)$, which is Galois if and only if $k$ contains a nontrivial cube root of unity.

(1) Next you assume that $k(C)/k(t)$ is not Galois. Since it is a degree-$3$ extension of fields of characteristic zero, its Galois closure $L$ is a degree-$2$ extension of $k(C)$. There are two possibilities: either $k$ is algebraically closed in $L$, or it isn't. If $k$ is not algebraically closed in $L$, then let $\ell$ be the algebraic closure of $k$ in $L$, and we must have $L=\ell(C)$. For the rest of your questions, you assume that $k$ is algebraically closed in $L$, so that $L=k(B)$ for some curve $B$ over $k$, and the extension $L/k(C)$ corresponds to a degree-$2$ morphism $\psi\colon B\to C$ defined over $k$.

(2) The extension $k(B)/k(t)$ is Galois with Galois group $G\cong S_3$. I'm going to replace $k$ by its algebraic closure $\bar{k}$, so that places correspond to points in the terminology of your question. This replacement doesn't change the ramification or the genus of the involved fields. The first thing to observe is that $\bar k(B)/\bar (t)$ only ramifies over places of $\bar k(t)$ which ramify in $k(C)/k(t)$. How to prove this depends on your background: you could consider completions, or consider inertia groups, or use Abhyankar's lemma, or... Anyway, taking this for granted, then use the fact that since $\bar k(B)/\bar k(t)$ is Galois, all places of $\bar k(B)$ which lie over a prescribed place of $\bar k(t)$ must have the same ramification index as one another, and moreover this ramification index equals the size of the inertia group, which is a cyclic subgroup of $G$ (and hence has order $1$, $2$ or $3$). It follows that if a place $Q$ of $\bar k(t)$ is totally ramified in $\bar k(C)/\bar k(t)$, then (since ramification indices are multiplicative in towers) the ramification index in $\bar k(B)/\bar k(t)$ of any place $P$ lying over $Q$ must have order divisible by $3$, and hence has order $3$, so this place $P$ is unramified in $\bar k(B)/\bar k(t)$. Likewise if a place $Q$ of $\bar k(t)$ lies under two places of $\bar k(C)$, one (say $R$) with ramification index $2$ and one (say $S$) with ramification index $1$, then the ramification index in $\bar k(B)/\bar k(t)$ of any place $P$ lying over $Q$ must be $2$, so that $R$ is unramified in $\bar k(B)/\bar k(C)$ but $S$ has ramification index $2$ in $\bar k(B)/\bar k(C)$. Therefore the number of places of $\bar k(B)$ which ramify in $\bar k(C)/\bar k(B)$ equals the number of points with ramification index $2$ under $\phi$, namely, $0$, $2$ or $4$ in your cases (a), (b), (c). Actually case (a) can't happen, by what I said at the end of my answer to the first question (1). Now use Riemann-Hurwitz: $\bar k(B)/\bar k(C)$ is a degree-$2$ extension of a genus-$0$ field which has either $2$ or $4$ ramification points, so its genus is $0$ in case (b) and $1$ in case (c).

(3) The possible ramification indices of the composite morphism $\phi\circ\psi$ were computed in my answer to (2). Again, case (a) can't happen under the hypothesis that $k$ is algebraically closed in $L$, which your phrasing suggests you are assuming. In case (b) there is one point of $\mathbb{P}^1$ which has two preimages in $B$ of ramification index $3$, and two points of $\mathbb{P}^1$ which each have three preimages in $B$ of ramification index $2$, and there is no further ramification in $\phi\circ\psi$. In case (c) there are four points of $\mathbb{P}^1$ which have three preimages in $B$ of ramification index $2$, and there is no further ramification in $\phi\circ\psi$.

Finally, in a comment you asked for references. For the function field perspective, Stichtenoth's book "Algebraic Function Fields and Codes" is a very accessible introduction. For a combination of geometric and algebraic, you can see Miranda's book "Algebraic Curves and Riemann Surfaces". There are many other excellent references as well.

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  • $\begingroup$ Thank you Michael, I'll give it a try with your hints and post the answer when I find it! $\endgroup$ – Davide Cesare Veniani May 5 '14 at 19:17
  • $\begingroup$ I guess: (1) The extension cannot be Galois in cases (b) and (c) since there cannot be an inertia subgroup of order 2 in a group of order 3. (is it always Galois in case (a)?) (2) In case the extension is not Galois, taking any primitive element in $k(C)$, its minimal polyomial over $k(t)$ has degree 3, so we need to add only one more root to get to the Galois closure, hence the degree of $\psi$ is 2. (3) I guess one can calculate it by Riemann Hurwitz. (4) Some intuitive answers, but still no proof. @Michael Zieve, could you please give me some good references? $\endgroup$ – Davide Cesare Veniani May 29 '14 at 9:41

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