homogeneous forms as norms

Question

Motivation/example. Consider $K = \mathbb{Q}(\sqrt[3]{2})$. This is a number field with ring of integers $O_K = \mathbb{Z}[\sqrt[3]{2}]$. We have a norm map $N_{K/\mathbb{Q}}$ which maps $x + y\sqrt[3]{2} + z\sqrt[3]{4}$ to $x^3 + 2y^3 + 4z^6 - 6xyz$; restricting to $\mathcal{O}_K$ gives of course the same form. Using standard results about factorization of prime ideals, it is not too hard to see which integers are norms of elements in $\mathcal{O}_K$. Therefore we can get the values of $n$ (with some work...) for which $x^3 + 2y^3 + 4z^3 - 6xyz = n$ has integral solutions, and I guess it is also possible to say something about the solutions for a fixed $n$ - although it is not obvious to me how to do this in general.

The same is of course true for many interesting quadratic forms - to cite a famous example: we can get all positive integers $n$ which can be written as the sum of two perfect squares, or more generally as $x^2 + \alpha y^2$ (for some interesting values of $\alpha$).

Questions. Is this a fruitful method to study diophantine equations? Are there interesting "large" classes of higher degree polynomials/diophantine equations which can be treated by this sort of argument? What is known in general about such "norm forms"? How to decide whether a polynomial is a norm form? Et cetera :)

(I know that it is a bit vague... I didn't find any useful references.)

Answer 1

Just a remark : It is certainly interesting to study the equation $N_{L|K}(x)=y$, where $L|K$ is an extension of number fields, and there must be an extensive literature on the subject. Hasse proved that if such an equation is solvable everywhere locally (at every place of $K$), and if $L|K$ is cyclic, then there is a global solution. This is sometimes called the Hasse Norm Principle.

Hasse also found examples where $L|K$ is galoisian and the equation is solvable everywhere locally, but not globally. Even for $K=\mathbb{Q}$, there are infinitely many biquadratic extensions $L$ which provide counter-examples.

Answer 2

A necessary condition: Let $f: \mathbb{Q}^d \to \mathbb{Q}$ be a homogenous map of degree $d$. If $f$ is a norm, then the corresponding map $\mathbb{C}^d \to \mathbb{C}$ will be a product of linear forms. That's always true for $d=2$ (fundamental theorem of algebra), but rare for $d=3$.