When is the norm of a ring $\Bbb Z[\sqrt{-p}]$ multiplicative?

Question

I got inspired by quadratic rings and zeta functions. I know for instance that the norm $a^2 + 17 b^2$ for the ring $\Bbb Z[\sqrt{-17}]$ is multiplicative yet the ring $\Bbb Z[\sqrt{-17}]$ is not a UFD !!

So I wonder, for an integer $n>0$ :

When is the norm of a ring $\Bbb Z[\sqrt{-n}]$ multiplicative ?

And in particular when $p$ is a prime :

When is the norm of a ring $\Bbb Z[\sqrt{-p}]$ multiplicative ?

Im not sure what the best way is to test it for a given $n$ or $p$ but I can imagine it is doable with basic modular arithmetic ( quadratic residue , generator , fermats little etc ) and/or testing some small cases.

But I am not (only) interested in a specific case, I wonder about the set of them. Are there infinitely many and do they have a closed form or a good asymptotic ? Do they have number theoretical properties apart from satisfying the above ?

Answer 1

For every nonsquare integer $d$ (positive or negative), the norm mapping ${\rm N} : \mathbf Z[\sqrt{d}] \to \mathbf Z$ where ${\rm N}(a+b\sqrt{d}) = a^2 - db^2$ (equivalently, ${\rm N}(\alpha) = \alpha\overline{\alpha}$, where $\overline{\alpha}$ is the conjugate of $\alpha$) is multiplicative. This has nothing at all to do with modular arithmetic or with whether or not $\mathbf Z[\sqrt{d}]$ is a UFD.

Answer 2

It turns out the norm is always multiplicative.

$$(a^2 + n b^2)(c^2 + n d^2) = (ac - nbd)^2 + n(ad + bc)^2 = (ac + nbd)^2 + n (ad - bc)^2 $$

The bigger picture is that norms are multiplicative because norms are special cases of determinants of matrices.

Notice a ring maps $\Bbb N^m$ to $\Bbb N^m$ hence the matrix representation must be a square. . And determinants of square matrices are indeed always multiplicative.

A proof of that is given here

https://en.wikipedia.org/wiki/Determinant

It is natural to define norms as determinants of square matrices for the same reasons as working with determinant for square matrices in linear algebra is logical.

Im talking here in the context of integral domains, not sure how to see it in more general cases but it probably is the same. Afterall a zero-divisor is a determinant zero in its matrix representation.