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Let $\zeta \in \overline{\mathbb{Q}}$ be a primitive 8th root of unity and let $K = \mathbb{Q}(\zeta)$. Let $N_{K/\mathbb{Q}}$ be the corresponding norm.

Consider the three-dimensional $\mathbb{Q}$-torus defined by $N_{K/\mathbb{Q}}(x) = 1$ for $x \in K^\times$. Is it rational over $\mathbb{Q}$?

I think the answer is no, but I don't have a proof.

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What do you mean by rational over $\mathbb{Q}$? do you mean "split"? –  Joël Sep 30 '11 at 20:26
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I want the underlying variety to be birational to $\mathbb{P}^3$ over $\mathbb{Q}$. –  Wanderer Sep 30 '11 at 20:29
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I deleted my post because ${\bf Q}(\zeta)$ has is not cyclic over ${\bf Q}$ (the Galois group is elementary abelian of order 4), so the result quoted by Serre does not apply. I still expect that the answer is Yes, though. –  Noam D. Elkies Sep 30 '11 at 20:51
    
...but apparently I was wrong there (expecting a positive answer) as well... –  Noam D. Elkies Sep 30 '11 at 22:50

1 Answer 1

up vote 11 down vote accepted

The torus is not rational.

This is due to B.E. Kunyavskii, On Tori with a biquadratic splitting field, Izv. Akad. Nauk SSSR Ser. Mat 42 (1978), 580-587; English translation Math USSR-IZV. 12(1978), 536-542.

An easier to find reference is V.E. Voskresenskii's book Algebraic Groups and their Birational Invariants. See Section 4.8, concluding with Example 4.8.1, at the top of page 54. Also, see the beginning of Section 4.10.

The argument given in that book actually shows something more general: the Norm 1 torus is non-rational, if the Galois group of your extension is $\mathbb{Z}/p\mathbb{Z} \times \mathbb{Z}/p\mathbb{Z}$, for any prime $p$. This is what is done in Section 4.8.

In Section 4.10, Voskresenskii recalls the work of Kunyavskii, where it is shown that the Norm 1 torus for a biquadratic extension is not stably rational.

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OK, I'll check the reference. Is the argument difficult? :) –  Wanderer Oct 1 '11 at 6:57
    
Basically, by taking character lattice of a torus, you are able to move back and forth between the world of algebraic tori and lattices of finite rank on which the Galois group acts. There is a theorem that says that a torus $T$ is stably rational if and only if the the character module $\hat{T}$ is stably permutation,, i.e. there exists a short exact sequence \begin{equation} 1 \rightarrow \hat{T} \rightarrow P_1 \rightarrow P_2 \rightarrow 1, \end{equation} with $P_1$, $P_2$ permutation modules. The book proves that such a sequence does not exist, so the torus is not stably rational. –  mark Oct 3 '11 at 18:16

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