= Solution
Every <automorphism> fixes zero and one, and preserves sums, products, negatives and multiplicative inverses of nonzero elements. Consequently elements fixed by every member of $G$ are closed under these operations and form a <field>. <Artin fixed-field theorem> gives $\boxed{[L:L^G]=|G|}$ for a finite <automorphism> <group>.
Here $\mathbb Q(t)$ is the fraction <field> of $\mathbb Q[t]$, with $t$ transcendental; its elements are fractions of rational-coefficient <polynomials>, with nonzero denominator and the usual equality by cross-multiplication. Substitution of $1-t$ or $1/t$ defines an <automorphism>, each its own inverse. With $\sigma\tau$ meaning first apply $\tau$ to a <field> element and then $\sigma$, its image on $t$ is $1/(1-t)$. Iterating gives $(t-1)/t$ and then $t$, so its order is three. The relations reduce every word to at most six possibilities, and the six resulting images are
$$
t,\quad1-t,\quad1/t,\quad1/(1-t),\quad(t-1)/t,\quad t/(t-1).
$$
They are distinct <rational functions>, so the <group> has six elements. Its faithful <permutation> action on $\{0,1,\infty\}$ identifies it with $\boxed{S_3}$.
For the first <fixed field> put $u=t(1-t)$. It is fixed by $\sigma$, and $t$ satisfies $X^2-X+u=0$ over $\mathbb Q(u)$. The degree is exactly two because the nontrivial <automorphism> $\sigma$ fixes that <field>. The fixed-<field> degree is also two, so
$$
\boxed{\mathbb Q(t)^{\langle\sigma\rangle}=\mathbb Q(t(1-t)).}
$$
For the inversion put $v=t+1/t$. It is fixed and $t$ satisfies $X^2-vX+1=0$. The same degree argument gives
$$
\boxed{\mathbb Q(t)^{\langle\tau\rangle}=\mathbb Q(t+1/t).}
$$
The source requests <polynomial> generators in both cases. The second requirement is impossible. If a nonconstant <polynomial> $p(t)=a_dt^d+\cdots+a_0$ were fixed by inversion, $p(t)=p(1/t)$ would equate a <Laurent polynomial> with highest exponent $d>0$ to one with no positive exponent, a contradiction. Thus the only <polynomial> invariants of inversion are constants, which cannot generate its nonconstant <fixed field>. The <fixed field of inversion in a rational function field> therefore has the second displayed rational-function generator.
Back to article page
