If $K/\mathbb{Q}$ is an extension of degree $2$, then necessarily it is of the form $\mathbb{Q}(\sqrt{d})$. So, I am wondering if a similar philosophy holds for extension over $\mathbb{Q}$ of degree 3.
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@RhythmInk, I am asking that if $K/\mathbb{Q}$ is an extension s.t $[K:\mathbb{Q}]=3$ then there exists $d$ cubfree s.t $K=\mathbb{Q}(\sqrt[3]{d})$? – AZMEH Oct 26 '18 at 22:53
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My mistake. I don't know what I was thinking. It's been deleted. – RhythmInk Oct 26 '18 at 22:58
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Would you consider $\mathbb{Q}\left(e^{\frac{2i\pi}{3}}\right)$ to be of the given form $\left(e^\frac{2i\pi}{3}\right)^3 = 1$, but we generally wouldn't say it's $\sqrt[3]{1}$. – memerson Oct 26 '18 at 23:04
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Your answer I think should be negative because if, for example, $\theta$ is the real root of $x^3+x+1=0$ then $K=\mathbb Q(\theta)$ is of degree $3$ but $\theta\approx -0.6823$ being such that $\theta^3+\theta+1=0$ is not of the form you asked.
Piquito
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How do you know $K=\mathbb{Q}(\theta)\neq \mathbb{Q}(\sqrt[3]{d})$ for some $d\in\mathbb{Q}$? – cut May 31 '23 at 10:11
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$\mathbb{Q}(\sqrt[3]{d})$ is not a Galois extension if $d$ is not a power of $3$. In fact, it is not normal, there exists Galois extensions of degree $3$,therefore they cannot be isomorphic to $\mathbb({Q}\sqrt[3]{d})$.
Tsemo Aristide
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