Elements in $\mathcal{O}_K$ with norm equal to ratio of norms of two elements in $\mathcal{O}_K$.

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.

• 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. – 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....

• 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. – Tob Ernack May 24 at 6:41
• 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 $3$. – 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, $$2=18/9=N_{K/\mathbb Q}(1+\sqrt{-17})/N_{K/\mathbb Q}(3)$$.

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