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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.

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3 Answers 3

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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.

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  • $\begingroup$ It should be $\sigma(t^k) = (-1)^kt^k$ , isn't it? $\endgroup$
    – Daniel
    May 24, 2015 at 20:45
  • $\begingroup$ I just made the correction for @Spooky $\endgroup$
    – Lubin
    May 24, 2015 at 22:11
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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.

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  • $\begingroup$ Is it clear that $\frac{t+t^3}{t-t^3}$ is contained in $\mathbb R(t^2)$? $\endgroup$
    – Spooky
    May 24, 2015 at 20:21
  • $\begingroup$ @Spooky, that rational function is not the quotient of two odds, nor of two evens, so is not in $\Bbb R(t^2)$. $\endgroup$
    – Lubin
    May 24, 2015 at 22:13
  • $\begingroup$ @Lubin $t+t^3$ and $t-t^3$ are both odd, and $\frac{t+t^3}{t-t^3}=\frac{1-t^2}{1+t^2}\in\Bbb R(t^2)$. $\endgroup$
    – anon
    May 25, 2015 at 0:26
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    $\begingroup$ Oh sorry, blame an old man’s poor vision. I thought those were $1$’s up top, not $t$’s to the power $1$. Then to @Spooky I would precisely echo anon’s comment. (Again, sorry!) $\endgroup$
    – Lubin
    May 25, 2015 at 16:29
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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 $\ne 2$.

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