Excuse me for the specificity of this question, but this is a silly computation that's been giving me trouble for some time.

I want to explicitly realize the order 21 Frobenius group over ℂ(x), as ℂ(x,y,z) where y3=g(x) and z7=h(x,y). The order 21 Frobenius group is C7⋊C3, where the generator of C3 acts by taking the generator of C7 to its square. Or in other words < a,b|a3=1, b7=1, ba=a(b2) >. Furthermore, I want it to branch at exactly three points, two of which will have 7 preimages, each with ramification 3, and the third will have 3 points over it with ramification 7 each.

This can easily be shown to exist: Take ℙ1 minus three points (say x=6,5,2); look at its algebraic fundamental group (=the profinite completion of < c, d, t| cdt = 1 >), and map it to the Frobenius group surjectively by c goes to a, d goes to b, and t goes to b-1a-1. This gives you a Galois cover of ℙ1, with said ramification (because order(a)=3, order(b)=7, and order(b-1a-1)=3), and group the 21 order Frobenius group.

Of course, this construction is extremely difficult to track because of the topology involved. It would be much easier to deduce the field extension from the ramification behavior.

So: this can be broken down to two cyclic Galois extensions. The first, a ℂ(x,y), of the form y3=(x-2)2(x-6) is pretty easy to deduce (I need it to ramify at x=2 and x=6 and nowhere else -- this must be the equation up to change of variables). The second, a z7=h(x,y) is tricky. I want it to ramify only above x=5. There's some Abhyankar's lemma things going on here, and that makes the guesswork difficult, and my life much harder.

I should note that the distinction of the Frobenius group of order 21, and the reason that I'm at all interested in this example, is that it is the only order 21 group which isn't cyclic. Geometrically, it means that h is a function of both x and y, and not just x.

Thanks in advance.

  • 6
    $\begingroup$ I can't help with the answer, but I think your question is well-written, especially in how it provides the detailed background, and very suitable for Math Overflow. $\endgroup$ Jan 3 '10 at 23:01

I'd like to change your numbers slightly. (EDIT: Slight adjustment to make the formula nicer and address the correction in the comments, nothing to see here)

One solution is to set $y^3 = x$, triply ramified only over $0$ and $\infty$, and if we want the 7-fold ramification over $x=1$ (which has solutions $y=1,\omega,\omega^2$) we set $z^7 = (1-y)(1 - \omega y)^2(1-\omega^2 y)^4$, which only ramifies over the three preimages. To show this is Galois, it suffices to show that the automorphism $y \mapsto \omega^2 y$ lifts to an automorphism of the whole field. This automorphism maps $(1-y)(1-\omega y)^2(1-\omega^2 y)^4$ to $(1-\omega^2 y)(1- y)^2(1- \omega y)^4$, which has seventh root $z^2/(1-\omega^2 y)$.

  • $\begingroup$ This is a little nitpicky, but you meant that (y-1)(y-w)^2(y-w^2)^4 goes to w(y-w^2)(y-1)^2(y-w)^4. It doesn't detract from the argument, of course. $\endgroup$
    – H. Hasson
    Jan 5 '10 at 0:32
  • 6
    $\begingroup$ By the way, this covering is the same as the tower of modular curves X(7) --> X_1(7) --> X_0(7). In particular, it can be extended to an even larger tower X(7) --> X(1) with Galois group PSL_2(F_7), in which the upper triangular matrices form the order-21 group. $\endgroup$ Jan 5 '10 at 2:26
  • $\begingroup$ Ah, shoot. Yes, 7 is congruent to 1 mod 3. Thanks for the correction. $\endgroup$ Jan 19 '10 at 5:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.