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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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Do you recognize that polynomial? –  Qiaochu Yuan Sep 30 '11 at 23:59
    
It factors over $\mathbb{Q}(\sqrt[3]{2})$... –  Evariste Oct 1 '11 at 8:07
    
By "recognise" Qiaochu means: have you seen it somewhere before, maybe somehow in connection with $K$? –  Alex B. Oct 1 '11 at 10:42
    
It is some kind of norm for the extension $K/\mathbb{Q}$. However I don't see how this makes things easier! –  Evariste Oct 1 '11 at 12:20

1 Answer 1

up vote 1 down vote accepted

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 can see that taking $X = -\sqrt[3]{2}Y - \sqrt[3]{4}Z$ gives a root for $f(X)$. But it's still not clear to me how you'd construct your morphism. Writing $K = Q[T]/(T^3 - 2)$, which polynomial would be the image of (the class of) $T$? –  Evariste Oct 1 '11 at 5:41
    
Or which rational function? –  Evariste Oct 1 '11 at 5:42
    
Now that we know the fields are equal, I'm tempted to "brute force" out an answer -- rather than do anything clever, just set up a linear algebra problem and solve it. We know that $(1, X, X^2)$ and $(1, \sqrt[3]{2}, \sqrt[3]{2}^2)$ are bases for the same three-dimensional $L$-vector space, and you've written down two linear equations that relate them ($1=1$ is the other). You can find a third by algebra -- compute $X^2$. Then you'd have three independent equations in the three "unknowns" $(1, \sqrt[3]{2}, \sqrt[3]{2}^2)$. –  Hurkyl Oct 1 '11 at 16:26
    
Plugging your answer into wolframalpha was discouraging: wolframalpha.com/input/… –  Hurkyl Oct 1 '11 at 18:17
1  
But it can do the linear algebra for you, after replacing $\sqrt[3]{2}$ with $u$ and $\sqrt[3]{4}$ with $v$: wolframalpha.com/input/… (you want the second solution, since that's the one that doesn't assert any equality involving $y$ and $z$) –  Hurkyl Oct 1 '11 at 18:17

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