# Homogeneous forms of degree $n$ in $n$ indeterminates over $\mathbb{Z}$: which ones come from the norm of a number field?

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

Lemma: Suppose you have a homogenous form $F$ in $n$ variables and degree $n$ over $\mathbf{Q}$ with the following properties:

1. The form $F$ splits into linear factors,
2. $F$ is non-zero on $\mathbf{Q}^n \setminus (0,0,\ldots,0)$.

Then $F$ is a scalar multiple of a norm form.

I presume this is what you intended by your second condition anyway --- note that $x^2_2$ is non-zero on $\mathbf{Q}^{\times 2}$, but it is not a norm form. A small piece of notation; for a polynomial $P$, let $Z(P)$ be the zero set of $P$.

Proof: By property one, $F$ contains a linear factor

$$H:=x_1 y_1 + x_2 y_2 + \ldots + x_n y_n.$$

The absolute Galois group $G_{\mathbf{Q}}$ acts on the component group of any algebraic variety over $\mathbf{Q}$. It follows that any irreducible component is defined over a number field, and so the $y_i$ must be algebraic.

The variety $Z(H)$, which is just a hyperplane, is determined precisely by the corresponding coordinates up to scalar multiple, or equivalently, by the corresponding point in projective space. Let

$$P = [y_1:y_2: \ldots:y_n] \in \mathbf{P}^{n-1}(\overline{\mathbf{Q}}).$$

The action of $G_{\mathbf{Q}}$ on $P$ corresponds to action of this group on the geometrically irreducible components of $Z(F)$ which are conjugate to $H$. Recall that $F$ has degree $n$. It follows that the orbit of $P$ has order at most $n$. Hence the stabilizer of $P$ has index at most $n$, and so, by Galois theory, the point $P$ lies in $\mathbf{P}^{n-1}(K)$ for a field $K$ of degree at most $n$. We may multiply $H$ by a scalar so that the coefficients actually lie in $K$. If $[K:\mathbf{Q}] = n$, then the orbit fills out all of $Z(F)$, and we see that $F$ is a linear multiple of the norm form associated to $H$.

On the other hand, suppose that $[K:\mathbf{Q}] < n$. Then the $n$ elements $y_i$ are linearly dependent over $\mathbf{Q}$, and hence $F$ will vanish no a non-zero rational point, violating (2'). $\square$

To see that the adjective "scalar multiple" is required, note that $3 x^2_1 + 3 x^2_2$ is not literally a norm form, because $3$ is not a norm in $\mathbf{Q}(\sqrt{-1})$.

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