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Prove $\mathbb{Q}(\sqrt[3]{12}) \cong \mathbb{Q}(\sqrt[3]{18})$

Hint: Look at $\frac{\sqrt[3]{12}^2}{2}$.

I know the hint is equal to $\sqrt[3]{18}$. I also know the extension $[\mathbb{Q}(\sqrt[3]{12}):\mathbb{Q}]=3.$

What is the best way to show the isomorphism? (without possibly defining the isomorphism itself and checking the details)

Walter
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  • By the hint, you have containment one way so just show they have the same degree over $\mathbb{Q}$. – Matt B Feb 23 '17 at 14:52

2 Answers2

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From the hint, $$ \sqrt[3]{18}=\frac{\sqrt[3]{12}^2}{2}\in\Bbb Q(\sqrt[3]{12})$$ and so $\Bbb Q(\sqrt[3]{18})\subseteq \Bbb Q(\sqrt[3]{12})$. Dually, $$ \sqrt[3]{12}=\frac{\sqrt[3]{18}^2}{3}\in\Bbb Q(\sqrt[3]{18})$$ shows $\Bbb Q(\sqrt[3]{12})\subseteq \Bbb Q(\sqrt[3]{18})$. We conclude that they are not only isomorphic, but in fact identical (provided possibly we agree on a $\bar{\Bbb Q}$)

Hagen von Eitzen
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  • Quick question: The actual question was part 4 of http://math.stackexchange.com/questions/2145608/show-that-mathbbq61-3-and-mathbbq121-3-have-the-same-degree . Does your answer still hold. It's a bit weird that question asks why they are isomorphic rather than equal. – Walter Feb 23 '17 at 15:15
  • Why is it weird? The cube root of $6$ gives a different number field, the cube root of $12$ gives the same as the cube root of $18$. So everything is alright. – Dietrich Burde Feb 23 '17 at 15:58
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Hint Define a $\mathbb Q$ homomorphism via $$f: \mathbb{Q}(\sqrt[3]{18}) \to \mathbb{Q}(\sqrt[3]{12}) \\ f(\sqrt[3]{18})=\frac{\sqrt[3]{12}^2}{2}$$

It is easy to argue that this is isomorphism.

As Hagen pointed, this is actually the identity mapping.

N. S.
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