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Coming up with examples of $D_8$-covers of $\mathbb{C}(x)$ is easy. For example:

$Quot(\mathbb{C}(x)[y,z]/(y^2=x(x-7), z^4=(y+\sqrt{-6})^2(y-\sqrt{-6})^2(y+\sqrt{-10})(-y+\sqrt{-10})^3))$

defines a $D_8$ extension of $\mathbb{C}(x)$. If you view this field extension as an extension of the corresponding smooth projective models, it is branched at $x=0,1,2,7$.

I wish to control the inertia groups (=decomposition groups, in this case) associated to the ramification.

I'll use the presentation $D_8= \langle a,b | a^4, b^2, ab=ba^{-1} \rangle$.

Take $\mathbb{P}^1_{\mathbb{C}} \setminus 0,7,1,2$. Its fundamental group is $\langle \gamma_1, \gamma_2, \gamma_3, \gamma_4| \gamma_1\gamma_2\gamma_3\gamma_4=1 \rangle$ (where $(\gamma_1, \gamma_2, \gamma_3, \gamma_4)$ are respectively the loops from a fixed base point around the points $(0,7,1,2)$). The surjection: $(\gamma_1, \gamma_2, \gamma_3, \gamma_4)$ goes to $(b, ab, a^2, a^3)$, gives a $D_8$-Galois cover of $\mathbb{P}^1_{\mathbb{C}}$.

I wish to find the equations that define THAT $D_8$ cover. The equations I wrote in the beginning won't work because there the inertia groups of the ramification points above $x=0$ and above $x=7$ are the same groups; whereas if done through the description, since no conjugate of $b$ is equal to any conjugate of $ab$, no inertia group of any ramification point above $x=0$ will equal the inertia group of any ramification point above $x=7$.

For some reason I easily confuse myself in such computations.

If you mod out the description by $\langle a \rangle$ you see that the description of the $D_8 / \langle a \rangle$ sub-cover of the original cover, has description $(b\langle a \rangle, b\langle a \rangle, \langle a \rangle, \langle a \rangle)$, and therefore must be given by $y^2=x(x-7)$ (if it's, say, $y'^2=\frac{x}{x-7}$, then $y'=\frac{y}{x-7}$). But I can't find the right other equation that would make this work.

Note that this should be easy, because this $D_8/\langle a \rangle$-cover is a $\mathbb{P}^1_{\mathbb{C}}$ with parameter $y$, so there's no funny business. And yet I'm having some trouble finishing this example up.

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2 Answers 2

up vote 4 down vote accepted

As you say, the fixed field by $\langle a\rangle$ is $\mathbb{C}(u)$ with $u^2=-5x/2(x-7)$, and the fixed field by $\langle a^2,b \rangle$ is similarly $\mathbb{C}(v)$ with $v^2=7(x-2)/2(x-7)$ (constants chosen so that $u^2+v^2=1$). So their compositum, the fixed field by $\langle a^2\rangle$, has genus zero and is $\mathbb{C}(t)$, where $u=2t/(t^2+1)$, $v=(t^2-1)/(t^2+1)$. This means that the last quadratic step of your tower is determined entirely by its ramification points, which are the 2 points above $x=2$ (which are the points $t=\pm 1$) and the 4 points above $x=1$, call them $t=c_1,\dots,c_4$. So your desired field is $\mathbb{C}(t,w)$ where $w^2=(t-c_1)(t-c_2)(t-c_3)/(t-c_4)(t^2-1)$.

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I can summarize now what I neglected to notice. Every time I reached a genus 0 curve I should have found the parameter and worked with it instead (like you implicitly did). Great. The next examples will be much easier to do on my own. Thanks! –  H. Hasson Aug 10 '10 at 13:10

Whenever I need to do something explicit involving a Galois extension of a field whose Galois group is a small 2-group (and it does happen sometimes!) I always find what I need in the paper "Groups of order 16 as Galois groups," by Grundman, Smith, and Swallow.

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