Let $K$ be a number field. For any $\alpha, \beta \in \mathcal{O}_K$ such that $N_{K/\mathbb{Q}}(\alpha) | N_{K/\mathbb{Q}}(\beta)$, is there a $\gamma \in \mathcal{O}_K$ such that $N_{K/\mathbb{Q}}(\gamma) = N_{K/\mathbb{Q}}(\beta)/N_{K/\mathbb{Q}}(\alpha)$?

Obviously, we have $N_{K/\mathbb{Q}}(\beta/\alpha) = N_{K/\mathbb{Q}}(\beta)/N_{K/\mathbb{Q}}(\alpha)$, but it is not true in general that $\beta/\alpha \in \mathcal{O}_K$. For example, take $K = \mathbb{Q}(i)$ and $\alpha = 5, \beta = 6 + 8i$. Then $N_{K/\mathbb{Q}}(\alpha) = 25 | 100 = N_{K/\mathbb{Q}}(\beta)$, but $\beta/\alpha = \frac{6 + 8i}{5} \not\in \mathbb{Z}[i] = \mathcal{O}_K$. Of course we know that $N_{K/\mathbb{Q}}(2) = 4 = \frac{100}{25}$ so this is not a counterexample to the question.

But perhaps there is a chance that given $\beta/\alpha$, we could find some element $\mu \in K$ such that $N_{K/\mathbb{Q}}(\mu) = 1$ and such that $\mu\beta/\alpha \in \mathcal{O}_K$. If $\beta/\alpha \not\in \mathcal{O}_K$ then we cannot have $\mu \in \mathcal{O}_K$, but there exist in general plenty of elements of unit norm in $K$ that are not algebraic integers, so the limitation is not as stringent as that given by the structure of the unit group of $\mathcal{O}_K$. For an example of this, see again $K = \mathbb{Q}(i)$ and the element $\frac{3+4i}{5}$ which has norm $1$ but is not in $\mathbb{Z}[i]$ (this is also what I used to produce the example in the paragraph above).

I know that in the case of quadratic fields $K = \mathbb{Q}(\sqrt{d})$, we can at least parametrize the set $\{x + y\sqrt{d}: x^2 - dy^2 = 1 \text{ and } x, y \in \mathbb{Q}\}$ using the method of choosing a starting point like $(1, 0)$ and constructing the intersection of lines of rational slopes passing through this point with the curve defined by $x^2 - dy^2 = 1$. But I don't know if that is of much help.

  • $\begingroup$ Choosing the best answer was somewhat difficult because Wojowu provided a simple and explicit counterexample, but Lord Shark the Unknown provided a method of generating many counterexamples (without explicitly providing one). I chose Lord Shark the Unknown's answer due to its insight. $\endgroup$ – Tob Ernack May 26 at 16:35

I suspect counterexamples do exist; certainly there are counterexamples to the corresponding statement for ideals in $\mathcal{O}_K$.

To be precise, there is a number field $K$, a rational prime $p$, and ideals $I$ and $J$ of $\mathcal{O}_K$ of norms $p^2$ and $p^3$, but there are no ideals of norm $p$. To construct $K$ take any prime $p$, and irreducible polynomials $f_1$ and $f_2$ of degrees $2$ and $3$ over $\Bbb F_p$. Let $f$ be a quintic polynomial over $\Bbb Z$ reducing to $f_1f_2$ modulo $p$, and irreducible modulo another prime. Then a root of $f$ will generate a field $K$ of degree $5$, in which $p$ splits into prime ideals of norms $p^2$ and $p^3$ in $\mathcal{O}_K$. So $\mathcal{O}_K$ has ideals of norms $p^2$ and $p^3$ but none of norm $p$.

It looks likely to me, that there is an example of this construction in which both $I$ and $J$ are principal ideals (if one is principal, so is the other), but that will involve some actual calculation....

  • $\begingroup$ I like the idea here. I guess you'd have to find quintic fields with class number $1$ where reduction mod $p$ gives the desired factorization structure. I tried $x^5 - x + 1$ and it reduces to $(x^2+x+1)(x^3+x^2+1) \pmod 2$ so this is close. The Minkowski bound calculated using the discriminant of the polynomial is around $3.3$, but this might be an overestimate. $\endgroup$ – Tob Ernack May 24 at 6:41
  • $\begingroup$ Hm actually yes, I think this works. Using this bound we only need to check for ideals of norm $2$ and $3$. But these correspond to primes above $(2)$ and $(3)$ respectively. But using the factorizations in $\mathbb{F}_2$ and $\mathbb{F}_3$ we get that the primes above $2$ have norm $2^2$ and $2^3$, and the prime above $3$ has norm $3^5$, so no primes of norm $2$ and $3$. $\endgroup$ – Tob Ernack May 24 at 7:11

Let $K=\mathbb Q[\sqrt{-17}],\mathcal O_K=\mathbb Z[\sqrt{-17}].$ The norm of an element $a+b\sqrt{-17}\in\mathcal O_K$ is $a^2+17b^2$, so clearly $2$ is not a norm. However, $2=18/9=N_{K/\mathbb Q}(1+\sqrt{-17})/N_{K/\mathbb Q}(3)$.

  • $\begingroup$ Nice! Was this found through trial and error, or did you have some idea what to look for? $\endgroup$ – Tob Ernack May 24 at 6:47
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    $\begingroup$ Apart from looking for imaginary quadratic fields (where it's easy to check something is a norm), pretty much trial and error. $\endgroup$ – Wojowu May 24 at 6:48

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