Let \[ K=\mathbb Q(\sqrt[10]{2}). \] Prove that \([K:\mathbb Q]=10\) and that the automorphism group of \(K\) has order \(2\).
Solution
Let \[ \alpha=\sqrt[10]{2}, \qquad K=\QQ(\alpha). \]
1. The polynomial \(x^{10}-2\) is irreducible over \(\QQ\).
Eisenstein’s criterion at the prime \(2\) applies: every nonleading coefficient is divisible by \(2\), while the constant term \(-2\) is not divisible by \(4\). Thus \(x^{10}-2\) is irreducible in \(\QQ[x]\).
2.\[ [K:\QQ]=10. \]
By step 1, the minimal polynomial of \(\alpha\) over \(\QQ\) is \(x^{10}-2\), which has degree \(10\). Therefore \[ [\QQ(\alpha):\QQ]=10. \]
3. Every \(\QQ\)-automorphism \(\sigma\) of \(K\) sends \(\alpha\) to either\(\alpha\) or \(-\alpha\).
Since \(\sigma\) fixes \(\QQ\), \[ \sigma(\alpha)^{10} = \sigma(\alpha^{10}) = 2. \] Thus \(\sigma(\alpha)\) is a root of \(x^{10}-2\). Also \(K\subseteq\RR\), because \(K\) is generated over \(\QQ\) by the real number \(\alpha\). The real roots of \(x^{10}-2\) are exactly \(\alpha\) and \(-\alpha\). Hence \[ \sigma(\alpha)\in\{\alpha,-\alpha\}. \]
4. Both assignments\[ \alpha\longmapsto\alpha \qquad\text{and}\qquad \alpha\longmapsto-\alpha \]define \(\QQ\)-automorphisms of \(K\).
The first assignment is the identity. For the second, define \[ \tau\bigl(p(\alpha)\bigr)=p(-\alpha) \] for \(p\in\QQ[x]\). Since \(\alpha\) and \(-\alpha\) have the same minimal polynomial \(x^{10}-2\), evaluation at \(-\alpha\) induces a well-defined field isomorphism \[ \QQ[x]/(x^{10}-2)\longrightarrow\QQ(-\alpha)=K. \] Under the standard identification of the source with \(K\), this is a \(\QQ\)-automorphism satisfying \(\tau(\alpha)=-\alpha\).
5.\[ \abs{\operatorname{Aut}_{\QQ}(K)}=\boxed{2}. \]
An automorphism of \(K=\QQ(\alpha)\) is determined by the image of \(\alpha\). Step 3 gives at most two possibilities, and step 4 realizes both.
6. Q.E.D.
Steps 2 and 5 prove the two requested assertions.