# Finite extensions of rational functions

I know that finite extensions of $\mathbb{C}(x)$ correspond to finite branched covers of $\mathbb{P}^1$, and this leads to an abstract characterization of the absolute Galois group of $\mathbb{C}(x)$ as the profinite completion of a free group of rank $\# \mathbb{C}$ via Riemann existence (or some algebraic analogue).

My question is much more concrete: say I am given a polynomial $f \in \mathbb{C}[x,y]$, irreducible over $\mathbb{C}(x)$ (for example, $f(x,y) = y^4 + xy^2 + x$). Does this geometric perspective say anything helpful about $\mathbb{C}(x)[y]/(f)$ or the splitting field of $f$? In particular, I would be happy to see

1. a geometric criterion for the extension $\mathbb{C}(x)[y]/(f)$ to be Galois, and
2. a geometric description of the Galois group of $f$.
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maybe you can get elements of the Galois group by letting $x \in \mathbb{C}$ vary along the complex plane in closed paths, and see how the roots get permuted? –  john mangual Aug 30 '12 at 7:25

## 1 Answer

If you do a loop around a branch point, the branchs around the point get permuted. The Galois group is the group generated by those permutations.

The extension is Galois when the Galois group acts transitively and without fixpoints, which means that the covering is connected (which happens if and only if your polynomial is irreducible), and that if there is some loop that return a point to its original position on its original branch, then the loop doesn't do anything neither on all the other branches : all the branches have to behave the same, and looking at only one of them is enough to deduce what happens on all the others.
If one root $y$ is fixed by an element of the Galois group, while some other root $z$ isn't, it means that $z$ is not in the subfield generated by $y$, and that $k \subset k(y)$ is not Galois.

Taking your example, the branch points are at $x=0$ where $f(0,y) = y^4$, where the four branches are cyclically permuted, and at $x=4$ where $f(4,y) = (y^2+2)^2$, where each pair of branch is switched. Thus the Galois group is generated by a $4$-cycle $\sigma$ and a double transposition $\tau$.

Then, there are two possibilities for our group. Either $\tau = \sigma^2$, in which case the covering and the extension are Galois, either it's not, in which case the Galois group is the dihedral group of order $8$, and the extension is not Galois.

With $y^4+xy^2+x$, we are in the second case : Algebraically, we have $\mathbb C(x) = k \subset k(y^2) \subset k(y)$, where $\operatorname{Gal}_k(k(y^2))$ is generated by $\rho : y^2 \mapsto -x-y^2$ ; $Gal_{k(y^2)}(k(y))$ is generated by $\theta : y \mapsto -y$. Since $-x-y^2$is not a square in $k(y)$, the Galois closure is obtained by adjoining $z$ such that $z^2 = -x-y^2$, so that $\rho$ can extend to $k(y,z)$ as $\sigma : (y,z) \mapsto (z,- y)$ and $\tau : (y,z) \mapsto (z,y)$.

Now we can observe that indeed, the group generated by $\sigma$ and $\tau$ has order $8$ while $y$ may only take the $4$ values $y,z,-y,-z$, and that to witness the group completely, it is necessary to observe its action on two roots (two branches) simultaneously instead of just one.

When you look at the covering of $L \otimes_K L$, this is exactly what you look at : the behaviour of pairs of roots/pairs of branches. If $L$ was Galois, every branch would move in the same way, so there would be $d$ components, each isomorphic to the original covering. Here instead, we get only two components, that are both Galois coverings showing the Galois group of $f$ (because luckily the group is not $S_4$ and we don't need to look at $3$ branches simultaneously)

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Beautiful! This is what I was looking for. I'm still not very good at visualizing things like "the four branches are cyclically permuted," but I thought such an explanation would exist. –  Justin Campbell Oct 5 '12 at 17:13
@ Justin : Close to 0, it looks pretty much like the function "y = 4th root of x" (and it is what you get when you localize at x=0). When x does one loop around 0, y (and the other 4th roots) does a quarter of a loop (it gets multiplied by i) : it permutes each root in a circular manner, and if x does four complete loops around zero, then y (and all the other 4th roots) do a complete loop around 0 and return to their original values. –  mercio Oct 6 '12 at 1:08