Let \(\mathbb{Q}(x)\) be the field of rational functions of one variable over \(\mathbb{Q}\). Let \(i \colon \mathbb{Q}(x) \to \mathbb{Q}(x)\) be the unique field automorphism such that \(i(x) = x^{-1}\). Prove that the subfield of elements fixed by \(i\) is equal to \(\mathbb{Q}(x + x^{-1})\).
Solution
Set \[ K\coloneqq\QQ(x), \qquad y\coloneqq x+x^{-1}, \] and let \[ F\coloneqq\{r\in K:i(r)=r\} \] be the fixed field.
1. One has \(\QQ(y)\subseteq F\).
Since \(i\) fixes \(\QQ\) pointwise and \[ i(y) = i(x)+i(x)^{-1} = x^{-1}+x = y, \] the automorphism \(i\) fixes every rational function in \(y\) with coefficients in \(\QQ\).
2. One has\[ [K:\QQ(y)]=2. \]
The element \(x\) satisfies \[ x^2-yx+1=0, \] so \([K:\QQ(y)]\leq2\). If the degree were \(1\), then \(x\in\QQ(y)\). Step 1 would then imply \(i(x)=x\), but \[ i(x)=x^{-1}\neq x \] in the rational function field \(\QQ(x)\). Hence the degree is exactly \(2\).
3. The fixed field \(F\) is a proper subfield of \(K\).
The element \(x\) is not fixed by \(i\), because \(i(x)=x^{-1}\neq x\). Thus \(x\notin F\), so \(F\neq K\).
4. The fixed field is\[ \boxed{F=\QQ(x+x^{-1})}. \]
By step 1, \[ \QQ(y)\subseteq F\subseteq K. \] Step 2 and the tower formula give \[ 2 = [K:\QQ(y)] = [K:F][F:\QQ(y)]. \] Step 3 gives \([K:F]>1\). Therefore \[ [K:F]=2, \qquad [F:\QQ(y)]=1, \] and hence \(F=\QQ(y)=\QQ(x+x^{-1})\).
5. Q.E.D.
Step 4 is the required description of the fixed subfield.