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Let $K$ be the field $\mathbb{Q}(\sqrt[3]{2})$.

I want to construct an explicit morphism from $K$ to the fraction field of

$\mathbb{Q}[X,Y,Z]/(X^3 + 2Y^3 + 4Z^3 - 6XYZ)$

but this doesn't seem to be that easy. Can someone help me?

Of course I just have to send $\sqrt[3]{2}$ to some explicit polynomial/rational function in $X$, $Y$ and $Z$...?

EDIT. Should I start thinking that it is impossible?

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    It is some kind of norm for the extension $K/\mathbb{Q}$. However I don't see how this makes things easier!2011-10-01

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Let $L = \mathbb{Q}(Y, Z)$, and $f(X) = X^3 - 6XYZ + 2Y^3 + 4Z^3$. You're asking if $L[X] / f(X) \cong L(\sqrt[3]{2})$, or equivalently if $f(X)$ has a root in $L(\sqrt[3]{2})$. At first glance it seems a little unlikely, but maybe it can work out. The rational root theorem (which I think still applies here) says it would have to be a factor of $2Y^3 + 4Z^3$, which we can factor as being the sum of two cubes. Can you take it from here?

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    Yes, I realized that I made a mistake in the calculations, that's why I deleted the comment later. But ok, I got it now, thanks a lot for your help!2011-10-01