Question

Find [Q(\sqrt[3]{2},\omega)\vdots Q] where \omega^3 = 1, \omega \neq 1

03 Feb 2024
Answer :
Word Count : 276

To find whether the field extension \([ \mathbb{Q}(\sqrt[3]{2}, \omega) : \mathbb{Q} ]\) is divisible, where \(\omega\) is a complex cube root of unity, we need to examine the structure of the extension and determine if it can be expressed as a tower of extensions, each one being a finite extension over the previous one.

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