Let L be the splitting field of $ x^3- 2$ over $ \mathbb{Q}$. Then $ G=\operatorname{Gal}(L/K) \cong S_3$. Let $\sigma\in G$ such that the fixed field of $ \sigma$ is $\mathbb{Q}(2^{1/3})$. Let $x,y\in L $ be $Q $ independent and $x/y \in Q(2^{1/3}\omega)$ then can we always find a $\lambda\in \mathbb{Q}$ such that the expression $$ (y\sigma(y))\lambda^2 + (y\sigma(x) + x \sigma(y))\lambda + x\sigma(x)\in \mathbb{Q}.$$
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2$\begingroup$ $\sigma$ 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? $\endgroup$– Emil JeřábekCommented Feb 29 at 7:42
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2$\begingroup$ 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). $\endgroup$– Emil JeřábekCommented Feb 29 at 8:12
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$\begingroup$ I have edited something. Pls See the new question. $\endgroup$– SkyCommented Feb 29 at 16:11
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