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Is there a characterization of the homogeneous forms of degree $n$ in $n$ indeterminates over $\mathbb{Z}$ which occur as the norm of some algebraic number ring with a suitable $\mathbb{Z}$-basis?

Let $K/\mathbb{Q}$ be a number field of degree $n$ with Galois group $G$. Let $\beta = \{y_1, \dots, y_n\}$ be an ordered basis for $\mathcal{O}_K$ over $\mathbb{Z}$. The norm $N_\beta(x_1, \dots, x_n) := N_{K/\mathbb{Q}}(x_1y_1 + \dots + x_ny_n)$ is a homogemeous polynomial in $x_1, \dots, x_n$; it is invariant under the action of $G$, hence its coefficients lie in $\mathbb{Q} \cap \mathcal{O}_K = \mathbb{Z}$. Of course $N_\beta$ depends on $\beta$. For instance, picking the basis $\{1, i\}$ of $\mathbb{Z}[i]$ yields the usual form $x_1^2+x_2^2$, but picking the basis $\{1,1+i\}$ yields the form $(x_1+x_2)^2+x_2^2$. Two forms which come from the same algebraic number ring are related by the action of $\text{SL}_n(\mathbb{Z})$, by $N_\beta(\mathbb{x}) = N_{g\cdot \beta}(g\cdot \mathbb{x})$ where $\mathbb x = (x_1, \dots, x_n)$. Thus, to each number field $K$, we can associate canonically the homogeneous form $N_\beta(\mathbb x)$, up to the action of $\text{SL}_n(\mathbb{Z})$.

The form $N_\beta(\mathbb x)$ satisfies some pretty cool properties:

  1. It splits as a product of linear forms over $K$ (by definition of the norm!), hence the affine variety $V$ defined by $N_\beta=0$ over $\overline{\mathbb{Q}}$ is a union of $n$ hyperplanes.
  2. It is nonzero on ${\mathbb{Q}^*}^n$, because $y_1, \dots, y_n$ are algebraically independent over $\mathbb{Q}$.
  3. It satisfies a "multiplicative identity" such as the Brahmagupta–Fibonacci identity, which reflects in a naive way the multiplicative structure of $K$.

Is there a way to pick out these forms easily? In general, what can be said of the classification of homogemeous forms of degree $n$ in $n$ variables over $\mathbb{Z}$?

Many thanks!

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Unless you order the basis, you need to take the action by GL_n instead, I think. –  franz lemmermeyer Apr 2 '12 at 18:27
Dear @franzlemmermeyer, Any two bases differ by the action of an invertible matrix with integer coefficients. Regards, –  Bruno Joyal Apr 24 '12 at 6:03
Interesting question. I guess the first place to start would be to find a form that doesn't come from the norm of a number field. –  fretty Dec 27 '13 at 12:57
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