How can you prove that any automorphism of $\displaystyle \mathbb{Q} (\sqrt[3]{2})$ leaves $\displaystyle \mathbb{Q}$ fixed?
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Originally Posted by syme.gabriel How can you prove that any automorphism of $\displaystyle \mathbb{Q} (\sqrt[3]{2})$ leaves $\displaystyle \mathbb{Q}$ fixed? The automorphism of any field leaves its prime subfield fixed. Since Q is the prime subfield, it says fixed. If you do not like that proof, I have another one.
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