Brian Bi
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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 $$-\sqrt{2}$$.