Let \(K\subseteq\mathbb{R}\) be a field of characteristic \(0\), and let \(f(x) \in K[x]\) be an irreducible polynomial of degree \(n \ge 3\) such that \(f(x)\) has exactly one real root \(u \in \mathbb{R}\) and \(n - 1\) non-real complex roots. (1) Prove that the extension degree \([K(u) : K] = n\). (2) Prove that \(K(u)/K\) is not a Galois extension. (3) Prove that the automorphism group \(\operatorname{Aut}(K(u)/K)\) is trivial (\(\operatorname{Aut}(K(u)/K) = \{\operatorname{id}\}\)).
Solution
Because \(f\) is irreducible and \(f(u)=0\), it is the minimal polynomial of \(u\) over \(K\). Hence \[ [K(u):K]=\deg f=n. \]
Since \(K\subseteq\RR\) and \(u\in\RR\), we have \[ K(u)\subseteq\RR. \] All roots of \(f\) other than \(u\) are nonreal, so none lies in \(K(u)\). Thus \(f\) does not split over \(K(u)\). In characteristic \(0\) the extension is separable, but it is not normal; consequently \[ K(u)/K\quad\text{is not Galois}. \]
Finally let \[ \sigma\in\Aut_K(K(u)). \] Since \(\sigma\) fixes the coefficients of \(f\), \[ 0=\sigma(f(u))=f(\sigma(u)). \] Thus \(\sigma(u)\) is a root of \(f\) lying in \(K(u)\). The only such root is \(u\), so \(\sigma(u)=u\). Since \(K(u)\) is generated by \(K\) and \(u\), this forces \(\sigma=\mathrm{id}\).
Therefore \[ \boxed{[K(u):K]=n,\qquad K(u)/K\text{ is not Galois},\qquad \Aut_K(K(u))=1.} \]