Timeline for Degree 6 Galois extension over $\mathbb{Q} $
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
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| Feb 29 at 16:11 | comment | added | Sky | I have edited something. Pls See the new question. | |
| Feb 29 at 16:08 | history | edited | Sky | CC BY-SA 4.0 |
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| Feb 29 at 9:11 | history | edited | YCor | CC BY-SA 4.0 |
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| Feb 29 at 8:12 | comment | added | Emil Jeřábek | So, the answer is negative for $x=2^{1/3}\omega$, $y=2^{1/3}\overline\omega$. Then any linear combination $z$ is $2^{1/3}$ times an element of $\mathbb Q(\sqrt3i)$, hence $z\sigma(z)$ is a rational multiple of $2^{2/3}$, which is not rational (if nonzero). | |
| Feb 29 at 7:42 | comment | added | Emil Jeřábek | $\sigma$ is just complex conjugation. Anyway, the question boils down to: does every $\mathbb Q$-linear subspace of $L$ of dimension $2$ contain a nonzero $z$ such that $z\sigma(z)$ is rational? | |
| Feb 29 at 6:49 | history | asked | Sky | CC BY-SA 4.0 |