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Let's take a cyclotomic field of the form $K=\mathbb{Q}(\zeta_n)$ where $\zeta_p$ is the $n$th root of unity. Then the ring of integers of $K$ is $\mathcal{O}_K= \mathbb{Z}(\zeta_n)$. Is there a generalisation of the rounding function $\left \lfloor \cdot \right \rceil: \mathbb{Q} \to \mathbb{Z}$ to some rounding function $\left \lfloor \cdot \right \rceil_K : K \to \mathcal{O}_K $ for cyclotomic fields that rounds a cyclotomic number to its "nearest" cyclotomic integer?

EDIT: I found something that might be useful. The following definition comes from https://hal.archives-ouvertes.fr/hal-00632997v1/document:

Definition: For any $\eta \in K$, the real number $m_K(\eta)= \min_{z \in \mathcal{O}_K}|N_{K/\mathbb{Q}}(\eta - z)|$ is the Euclidean minimum of $\eta$.

Does this give us a generalisation of the rounding function, and if so does the "rounding" function only hold for Euclidean domains?

Chris
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  • This seems hard since $K$ isn't an ordered field, just consider $K=\mathbb{Q}(i)$. – cansomeonehelpmeout Jun 13 '18 at 09:52
  • @cansomeonehelpmeout can we use the field norm to order the field and do it that way? (sorry if that's a stupid question) – Chris Jun 13 '18 at 09:55
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    Also I was reading that each ring of integers is isomorphic to the $n$-dimensional integer lattice $\mathbb{Z}^n$ - would finding the closest point on such a lattice correspond to finding the closest algebraic integer? – Chris Jun 13 '18 at 09:56
  • @cansomeonehelpmeout I believe it's possible for $\mathbb{Q}(i)$ - for each $a \in \mathbb{Q}(i)$ we need some $b \in \mathbb{Z}(i)$ such that $a/b = 1+r$ where $Norm(r)<1$, e.g. the closest integer to $3/2-i$ is $2-i$ (though I'm unsure if b is unique in general, and whether the field has to be a Euclidean domain). – Chris Jun 13 '18 at 10:06
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    Consider When is $\mathbb{Z}[\alpha]$ dense in $\mathbb{C}$ and e.g. $\mathbb{Z}[\zeta_8]$. With the usual distance, there is no nearest algebraic integer. – ccorn Jun 13 '18 at 12:18
  • @ccorn Sorry for the late reply. I'm not great at analysis - why does $\mathbb{Z}[\zeta_n]$ have to be dense in $\mathbb{C}$ for there to be a nearest integer? – Chris Jun 14 '18 at 10:22
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    If $\mathbb{Z}[\zeta_n]$ is dense in $\mathbb{C}$, then there are infinitely many integers from $\mathbb{Z}[\zeta_n]$ in every neighborhood of a given non-integer element of $\mathbb{Q}[\zeta_n]$ (with the continuous distance). Therefore there is no nearest integer then. – ccorn Jun 14 '18 at 10:30
  • This is easy to see for $\mathbb{Z}[\zeta_8]$ because it contains $1,\mathrm{i},\sqrt{2}$. This allows you to get arbitrarily close to any point in $\mathbb{C}$. – ccorn Jun 14 '18 at 10:33
  • @ccorn I see - so for roots of unity whose degree is greater than 2, there are infinitely many so-called "closest integers"? Would you care to submit this as a formal answer? I'd be happy to mark it as correct – Chris Jun 14 '18 at 15:05
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    There might be more general answers (considering other distance functions). The paper you have linked to goes in that direction. – ccorn Jun 14 '18 at 15:35
  • Possibly of interest: When is $\mathbb{Z}[\zeta_n]$ a Euclidean Domain? – ccorn Jun 14 '18 at 17:10
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    You can certainly convert this into a reasonably related nearest lattice point prolem. Just use the usual embedding of $K=\Bbb{Q}(\zeta_n)$ into, not just $\Bbb{C}$ but $\Bbb{C}^{\phi(n)/2}$. If $z\in K$, then its image is the vector $(\sigma_i(z))_{1\le i\le\phi(n)/2}$ where $\sigma_i$ ranges over a set of cosets of the subgroup generated by the usual complex conjugation in the Galois group. This is bread-and-butter in ANT. Then $\mathcal{O}_K$ is turned into a discrete lattice, and by blowing up the dimension we avoid the topological difficulties ccorn commented on. – Jyrki Lahtonen Jun 15 '18 at 06:31
  • Essentially, for two elements of $K$ to be close to each other in this space, all their Galois conjugates need to be (pairwise) close to each other in the usual metric. – Jyrki Lahtonen Jun 15 '18 at 06:33
  • Another straightforward way of doing something similar is to write elements of $K$ as $\Bbb{Q}$-linear combinations of some basis, the monomial basis $\zeta_n^j, 0\le j<\phi(n)$, comes to mind. Then you can simply round the rational coefficients of any element $z\in K$ w.r.t. this basis to nearest integers. But, this has the air of arbitrarily favoring one basis over another, and gives a bit distorted view. Using the Galois group gives a better behaved map, but won't do everything you might want it to do (e.g. won't turn the ring into a Euclidean domain in general). – Jyrki Lahtonen Jun 15 '18 at 06:38
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    Related MO thread: Find closest integers in Euclidean rings – ccorn Jun 16 '18 at 07:51

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Short answer: Yes, there is such a function. Just map every element of your number field to $0$. Since you do not specify what properties your rounding function should have, you can't even complain that this answer is trivial.

As suggested by the remarks, you can choose an integral basis and round each element of the number field to the integer that is coordinatewise closest to this number. This also gives you many rounding functions, and you if you choose $1$ as an element of your integral basis then you may even pick a rounding function that restricts to the usual rounding function in the rationals.

The Euclidean minimum has little to do with the question; its image is a real number, most often rational, but rarely integral. You can of course map $\eta$ to some $z$ that minimzes the absolute value of the norm of $\eta - z$, but this is not well defined if your number field has nontrivial units.