A computation in the field of rational functions.

Mathematics Asked by Jake Mirra on December 20, 2020

In Dummit and Foote 3 ed., Chapter 14, Section 2, Exercise 30, I am asked the following:

Let $ k $ be a field, $ k(t) $ the field of rational functions in the variable $ t $. Define the maps $ sigma $ and $ tau in Aut(k(t)/k) $ by
sigma f(t) = f left( frac{1}{1-t} right) quad tau f(t) = f left( frac{1}{t} right)

for $ f(t) in k(t) $. Prove that the fixed field of $ langle tau rangle $ is $ k left( t + frac{1}{t} right) $, the fixed field of $ langle tau sigma^2 rangle $ is $ k(t(1-t)) $; determine the fixed field of $ langle tau sigma rangle $ and $ langle sigma rangle $.

The only part of this I am struggling with is the fixed field of $ langle sigma rangle $. Call this fixed field $ E = k(s) $, where $ s = P(t) / Q(t) in k(t) $ is some rational function. Note, I am making an assumption here that $ E $ is of the form $ k(s) $, and so far cannot justify this a priori. I have shown in a previous exercise from the last chapter that $ [k(t) : k(s)] = max left{ deg P(t), deg Q(t) right} $, so, since $ k(t)/k(s) $ is a Galois extension ($k(s)$ being the fixed field of a subgroup of automorphisms), I expect
max left{ deg P(t), deg Q(t) right} = [k(t) : k(s)] = |langle sigma rangle| = 3

All I have been able to accomplish at this point was brute-force equation-solving by computer, setting
s = frac{a_3 t^3 + a_2 t^2 + a_1 t + a_0}{b_3 t^3 + b_2 t^2 + b_1 t + b_0}

and solving the equations resulting from $ sigma s = s $. I thereby found the element $ s = frac{t^3 – 3t + 1}{t(t-1)} $. Hence I am inclined to conclude that $ k left( frac{t^3 – 3t + 1}{t(t-1)} right) $ is the fixed field of $ langle sigma rangle $. This approach feels inelegant, and I would like to know what tools I might have used to avoid an unsatisfying and opaque computer-search.

One Answer

For $G$ a finite subgroup of $Aut(k(t)/k)$ then the fixed subfield is $k(t)^G=k(a_0(t),ldots,a_{|G|-1}(t))$ where $prod_{gin |G|} (X-g(t))=sum_{m=0}^{|G|} a_m(t) X^m$.

Then take any non-constant coefficient $a_m(t)$, because each $g(t) = frac{e_g t+b_g}{c_g t+d_g}$ is a Möbius transformation we get that $a_m(t)$ has at most $|G|$ poles counted with multiplicity (including the pole at $infty$), thus $[k(t):k(a_m(t))]le |G|=[k(t):k(t)^G]$ which implies that $$k(t)^G=k(a_m(t))$$

Edit by OP: for this problem, the technique produces the element $ a_2(t) = frac{t^3 - 3t + 1}{t(t-1)} $, reifying the computer calculations.

Correct answer by reuns on December 20, 2020

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