Suppose \(K\) is a field. Let \(K(x)\) denote the field of rational functions in one variable \(x\) over \(K\), and let \(G\) be the subgroup of the automorphism group of \(K(x)\) over \(K\) generated by \(\sigma\), where \(\sigma(x) = 1 - (1/x)\). Find a specific \(u \in K(x)\) such that the fixed field of \(G\) is \(K(u)\), and find the minimal polynomial of \(x\) over \(K(u)\).
Solution
One may take \[ \boxed{u=\frac{x^3-3x+1}{x^2-x}}. \] Then \(K(x)^G=K(u)\), and the minimal polynomial is \[ \boxed{T^3-uT^2+(u-3)T+1\in K(u)[T]}. \] All integers in these formulas are interpreted in \(K\).
1. The automorphism \(\sigma\) has order \(3\).
Direct substitution gives \[ \sigma(x)=\frac{x-1}{x},\qquad \sigma^2(x)=1-\frac{x}{x-1}=\frac1{1-x},\qquad \sigma^3(x)=1-(1-x)=x. \] The rational functions used here have nonzero denominators because \(x\) is transcendental. Also \(\sigma(x)\ne x\): equality would imply \(x^2-x+1=0\), a nonzero polynomial relation over \(K\). Since the automorphism fixes \(K\) and \(K(x)\) is generated by \(x\), it follows that \(\sigma^3=1\) and \(\sigma\ne1\). Its order is therefore exactly \(3\). In particular \(x,\sigma(x),\sigma^2(x)\) are distinct.
2. The element \(u\) is fixed by \(G\), and \(x\) satisfiesthe displayed cubic over \(K(u)\).
The sum of the three iterates is fixed by \(\sigma\), which cyclically permutes its summands. Combining the fractions gives \[ x+\frac{x-1}{x}+\frac1{1-x} =\frac{x^2(x-1)+(x-1)^2-x}{x(x-1)} =\frac{x^3-3x+1}{x^2-x}=u. \] Thus \(K(u)\subseteq K(x)^G\). Multiplying the formula for \(u\) by \(x^2-x\) and rearranging gives \[ x^3-ux^2+(u-3)x+1=0. \] This is a monic cubic relation. Hence \([K(x):K(u)]\leq3\).
3. The fixed field is exactly \(K(u)\), and the cubic isthe minimal polynomial of \(x\).
Put \(E=K(x)^G\). Since \(K(u)\subseteq E\subseteq K(x)\), step 2 makes \(K(x)/E\) finite of degree at most \(3\). Let \(m(T)\) be the minimal polynomial of \(x\) over \(E\). Every element of \(G\) fixes its coefficients, so applying \(1,\sigma,\sigma^2\) to \(m(x)=0\) shows that \(x,\sigma(x),\sigma^2(x)\) are roots of \(m\). Step 1 shows that they are distinct; consequently \(\deg m\geq3\). As \(E\) contains \(K\), one has \(E(x)=K(x)\), and therefore \([K(x):E]=3\).
The tower law now gives \[ 3\geq[K(x):K(u)]=[K(x):E][E:K(u)]=3[E:K(u)]. \] Thus \([E:K(u)]=1\), so \(E=K(u)\) and \([K(x):K(u)]=3\). The monic cubic from step 2 must be the minimal polynomial. Its roots are the three distinct iterates above, so it is separable as well. The argument never divides by \(2\) or \(3\) and applies in every characteristic.