Finding the fixed Field of $\sigma \in Aut(\mathbb{R}(t)/\mathbb{R})$

Question

Finding the fixed Field of $\sigma \in Aut(\mathbb{R}(t)/\mathbb{R})$

Let $\sigma$ be such that $\sigma(t)=-t$.

• I assume there is only one automorphism like this, I am not sure exactly why...
• How do I find the fixed field of $<\sigma>$?

I am new at this field, I would appreciate your explanations, thanks.

Answer 1

If $\sigma(t)=-t$, then $\sigma(t^k) = (-1)^k t^k$, so a polynomial $\sum a_n t^n$ is fixed by $\sigma$ iff $a_i=-a_i$ when $i$ is odd. So a rational function is fixed iff both numerator are fixed or both are negated. So it should be the field generated by quotients of even polynomials and quotients of odd polynomials.

Answer 2

The answer given by Spooky is excellent. Note that the fixed field, as described, is simply $\mathbb{R}(t^2)$.

Regarding uniqueness of $\sigma$. Since $\mathbb{R}(t)$ is generated by the single element $t$, once we choose $\sigma(t)$ we determine $\sigma$ completely.

Answer 3

You can always arrange for a fraction $r(t) = \frac{P(t)}{Q(t)}$ to have an even denominator $r(t) = \frac{P(t) Q(-t)}{Q(t)Q(-t)}= \frac{P_1(t)}{Q_1(t)}$. An even polynomial like $Q_1(t)$ is invariant under $\sigma$. So if $r(t)$ is invariant, $P_1(t)$ must also be even.

One may need to do some reductions to get an invariant fraction as a quotient of two even polynomials.

Added: In fact, as @Spooky has showed, a fraction in lowest terms is invariant if and only if the numerator and denominator are polynomials in $t^2$ ( the odd case cannot be in lowest terms). Note this works for any field of characteristic $\ne 2$.