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Let $K$ be a number field with integral basis $\{\omega_1,\ldots,\omega_n\}$. Then $$ \Phi(X_1, \ldots, X_n) = N_{K/{\mathbb Q}}(\omega_1 X_1 + \ldots + \omega_n X_n) $$ is a homogeneous polynomial of degree $n$ with integral coefficients, and the integral points on the affine variety $$ \Phi(X_1,\ldots,X_n) = 1 $$ correspond to units with norm $+1$ in the ring of integers of $K$.

For quadratic extensions, this "unit variety" is defined by $X_1^2 - mX_2^2 = 1$ (a Pell conic) whenever $m \equiv 2, 3 \bmod 4$ is squarefree; for other extensions of small degree it is similarly easy to write down explicit equations.

It is well known that the rational points on the Pell conic can be parametrized. The same thing holds for general cyclic extensions: the proof via Hilbert 90 that Pell conics can be parametrized generalizes easily. This suggests the following question:

Can unit varieties of number fields be parametrized by rational functions with rational coefficients?

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Is each "unit variety" an algebraic torus, or is this only the case in dimension one? Certainly they are all algebraic groups over Q – Daniel Loughran Apr 2 '11 at 14:56
@Daniel: The unit equations are the kernel of the Norm map from the $R_{K/\mathbb{Q}}\mathbb{G}_m$ to $\mathbb{G}_m$. Indeed these are algebraic groups, and also algebraic tori, of rank one less than that of the field extension. I think that there are cases where these are not rational. If I recall correctly, a non-cyclic cubic extension of the rationals should give a non-rational surface, but I can't seem to find a reference for this (maybe it's false??). – Dror Speiser Apr 2 '11 at 15:01
Thanks Dror for clarifying! I think I have found a reference... – Daniel Loughran Apr 2 '11 at 15:04
Right, I'm wrong about the cubic remark. We get a parametrisation by going up to the quadratic field and back down. – Dror Speiser Apr 2 '11 at 15:24
up vote 8 down vote accepted

Thanks to Dror for pointing out a nice geometrical way to think of such varieties. This paper shows that such varieties are not rational in general:

However, any algebraic torus over a number field $K$ of dimension one or two is rational over $K$. Indeed, the case of dimension one is easy, and the case of dimension two is a theorem of Voskresenskiĭ, but algebraic tori of dimension three and above do not have to be rational over the ground field in general.

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