\[ \DeclareMathOperator{\ker}{ker} \DeclareMathOperator{\im}{im} \DeclareMathOperator{\diag}{diag} \DeclareMathOperator{\char}{char} \DeclareMathOperator{\lcm}{lcm} \newcommand\divides{\mathbin |} \newcommand\ndivides{\mathbin \nmid} \newcommand\d{\mathrm{d}} \newcommand\p{\partial} \newcommand\C{\mathbb{C}} \newcommand\N{\mathbb{N}} \newcommand\Q{\mathbb{Q}} \newcommand\R{\mathbb{R}} \newcommand\Z{\mathbb{Z}} \newcommand\pref[1]{(\ref{#1})} \DeclareMathOperator{\Aut}{Aut} \DeclareMathOperator{\Gal}{Gal} \]

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Section 16.4. Isomorphisms of Field Extensions

Exercise 16.4.1(a) Every isomorphism of a field extension of \(\Q\) fixes \(\Q\). By Lemma 16.4.2(a), an automorphism of \(\Q(\sqrt[3]{2})\) must take \(\sqrt[3]{2}\) to another root of \(x^3 - 2\). But \(\sqrt[3]{2}\) is the only real root, so it must be mapped to itself. By Lemma 16.4.2(c), since \(x^3 - 2\) is irreducible, the \(\Q\)-automorphism of \(\Q(\sqrt[3]{2})\) is unique, therefore it can only be the identity.

Lemma 16.4.2(a) implies that an automorphism of \(\Q(\sqrt[3]{2},\omega)\) must map \(\sqrt[3]{2}\) to one of the three roots of the polynomial \(x^3 - 2\). Likewise \(\omega\), a root of the polynomial \(x^2 + x + 1\), must be mapped to either itself or the other root, namely \(\omega^2\). Lemma 16.4.2(b) implies that the automorphism will be completely determined by the choice of these two mappings; so there are at most six distinct automorphisms of \(\Q(\sqrt[3]{2},\omega)\).

Now, any \(\Q(\sqrt[3]{2})\)-automorphism will also be a \(\Q\)-automorphism, and since the polynomial \(x^2 + x + 1\) remains irreducible over \(\Q(\sqrt[3]{2})\), Lemma 16.4.2(c) tells us that there are two automorphisms of \(\Q(\sqrt[3]{2},\omega)\) that fix \(\Q(\sqrt[3]{2})\), one of which sends \(\omega\) to \(\omega^2\); these form a subgroup \(\{1, \sigma\}\) of the full automorphism group of \(\Q(\sqrt[3]{2},\omega)\). By similar reasoning, and the fact that \(x^3 - 2\) remains irreducible over \(\Q(\omega)\), we find that there are three automorphisms of \(\Q(\sqrt[3]{2},\omega)\) that fix \(\Q(\omega)\), which therefore form a subgroup \(\{1, \tau, \tau^2\}\) of order 3. The full automorphism group of \(\Q(\sqrt[3]{2},\omega)\) must therefore have order divisible by 6, therefore it is exactly 6, and the six automorphisms are \(1, \sigma, \tau, \tau\sigma, \tau^2, \tau^2\sigma\).

The problem doesn't ask us to identify the isomorphism class of the automorphism group. However, it's easy to do so. Say \(\tau(\sqrt[3]{2}) = \omega\sqrt[3]{2}\). Then \(\sigma(\tau(\sqrt[3]{2})) = \sigma(\omega\sqrt[3]{2}) = \sigma(\omega)\sigma(\sqrt[3]{2}) = \omega^2\sqrt[3]{2}\), and \(\tau(\sigma(\sqrt[3]{2})) = \tau(\sqrt[3]{2}) = \omega\sqrt[3]{2}\). Therefore \(\sigma\tau \ne \tau\sigma\), and the automorphism group is nonabelian; therefore it is isomorphic to \(S_3\).

Exercise 16.4.1(b) The given polynomial splits completely in \(\Q(\sqrt{2})\) since its four roots in \(\C\) are \(1 \pm \sqrt{2}\) and \(1 \pm 2\sqrt{2}\). It is also clear that \(\Q(1 + \sqrt{2}) = \Q(\sqrt{2})\), so \(\Q(\sqrt{2})\) is the splitting field. By Lemma 16.4.2(a) and (c), there are exactly two automorphisms of this field: the identity, and one other that sends \(\sqrt{2}\) to \(-\sqrt{2}\).