# 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.

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