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

Solution 1:

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 $\mathbf{Q}(\sqrt{-1})$.