I've already asked this question on Math StackExchange but having gotten no responses this may be more obscure than I had initially believed.

Here $\varphi=\frac{1+\sqrt{5}}{2}$. It's true that $\mathbb{Z}[\varphi]=\mathcal{O}_{\mathbb{Q}(\sqrt{5})}$ is Euclidean since $\mathbb{Q}(\sqrt{5})$ is norm Euclidean, and I've read that $A=\mathbb{Z}\left[\frac{1+i}{\sqrt{2}}\right]$ is Euclidean as well, though I'm not certain what the Euclidean function is there (the reasonable candidate being $x\mapsto N_{K/\mathbb{Q}}(x)$, $K$ being $A$'s fraction field). So, my questions are if anything is known about:

  • Is $R=\mathbb{Z}[i,\varphi]$ Euclidean?
    • All I know is that $R=\mathcal{O}_L$ ($L$ being $R$'s fraction field), by computing its discriminant and comparing with $D_L$ given by LMFDB, so $R$ is a PID.
  • If so, what is the Euclidean function?
    • The reasonable candidate is $x\mapsto N_{L/\mathbb{Q}}(x)$, but this has proved difficult to work with.
  • If not, is $R$ a (finite-index) subring of a (nice) Euclidean domain?
  • My main goal is to compute $\gcd$s in $R$, so if all of the above don't have known/affirmative answers, a Euclidean algorithm in $R$ (sans a Euclidean function) would be just as great instead.

Thanks in advance for any answers.

  • 4
    $\begingroup$ It is conjectured, and known under GRH, that any ring of integers which is a PID, except for 4 imaginary quadratic exceptions, is a Euclidean domain. I have no idea if this is known for this particular field, but in general proving this is a difficult problem. $\endgroup$
    – Wojowu
    Apr 16, 2021 at 21:59
  • 3
    $\begingroup$ The number $(1+i)/\sqrt{2}$ is a primitive 8th root of unity. Hendrik Lenstra wrote some papers on Euclidean number fields for the Mathematical Intelligencer in 1979-1980, titled "Euclidean Number Fields 1" and then "... 2" and "... 3". In the first paper he says (p. 14) that the ring of integers of the 8-th cyclotomic field is among the cyclotomic fields whose ring of integers is norm-Euclidean. $\endgroup$
    – KConrad
    Apr 16, 2021 at 23:35
  • 4
    $\begingroup$ As a follow-up to Wojowu's comment, the conjecture is now known to hold unconditionally (due to work of Harper and Murty) if $K/\mathbb{Q}$ is finite, and $\mathscr{O}_K$ has unit rank $r>3$ (and perhaps this has been improved since then). This doesn't cover the case at hand, but covers many other cases. $\endgroup$ Apr 17, 2021 at 0:58
  • 1
    $\begingroup$ If $\mathcal{O}_L$ is a finite index subring of some domain $R'$, then $R'$ is (up to isomorphism) contained in $L$ and is integral over $\mathcal{O}_L$, and since $\mathcal{O}_L$ is the integral closure of $\mathbb{Z}$ in $L$, we must have $R'=\mathcal{O}_L$. $\endgroup$ Apr 17, 2021 at 9:17
  • 2
    $\begingroup$ @Gro-Tsen Good start is the paper mentioned by Pace Nielsen, which you can find here. This was improved here. The original proof under GRH was presented in Weinberger's article cited in both of these. The use of GRH comes in proving some version of Artin's conjecture, the rest is elementary. $\endgroup$
    – Wojowu
    Apr 17, 2021 at 14:36

1 Answer 1


It turns out that $K=\mathbb{Q}[i,\varphi]$ is norm-Euclidean. The proof of this fact appears as Appendix A in https://arxiv.org/abs/2205.03007. I'll explain the heft of the argument here.

Let $R=\mathbb{Z}[i,\varphi]$ and $N=N_{K/\mathbb{Q}}$. We use this formulation of norm-Euclidean: for all $\alpha\in K$ there exists $\beta\in R$ with $N(\alpha+\beta)<1$.

Model $K\cong\mathbb{Q}^4$ via $w+x\varphi+yi+zi\varphi\mapsto(w,x,y,z)$ whence $R$ identifies with $\mathbb{Z}^4$. If $K$ were "extra nice" we could simply verify that for all $\alpha\in\left[-\frac{1}{2},\frac{1}{2}\right)^4\cap K=C$, we have $N(\alpha)<1$ whence for $\alpha\in K$ we could simply let $\beta=-[\alpha]$ where $[\alpha]$ is $\alpha$ with each coordinate rounded to the nearest integer. (This is how the standard proof goes for the Gaussian number field being norm-Euclidean.) Unfortunately, this approach fails, particularly near the corners where exactly three components have the same sign.

Notice, however, that $\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2}\right)$ has very small norm, so that if we shift only some points in $C$ then we may be okay, since this would resolve e.g. $\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},-\frac{1}{2}\right)$, and all points nearby, since $N$ is a polynomial in the coordinates, so it is definitely continuous. The trick then is to exactly quantify continuity. We will consider $C_n=\left[-\frac{1}{2},\frac{1}{2}\right)^4\cap(\frac{1}{n}\mathbb{Z}^4)$ with $n$ to be decided later. If we can show for each $\alpha\in C_n$ that there exists $\delta\in\mathbb{Z}^4$ such that $N(\alpha+\delta+B_n)<1$ where $B_n$ is the radius-$\frac{1}{n}$ $L^\infty$ ball in $\mathbb{Q}^4$, then we are done.

We consider now finding an explicit bound for $N(\alpha+\beta)$ where $\alpha=w+x\varphi+yi+zi\varphi\in K$ is any point and $\beta=d_1+d_2\varphi+d_3i+d_4i\varphi$ has $\|\beta\|_\infty<\varepsilon$, i.e. $\lvert d_1\rvert<\varepsilon$. $N(\alpha+\beta)$ is quartic in the eight variables, so write it out and group terms by "$d$-type," that is, $N(\alpha+\beta)=\sum_{e\in\mathbb{N}^4}d^eP_e(\alpha)$ for polynomials $P_e\in\mathbb{Q}[w,x,y,z]$. (e.g., $d_1d_2^2x+d_2d_3d_4z+2d_1d_2^2y+8d_4wyz$ would have $P_{1200}=x+2y$, $P_{0111}=z$, $P_{0001}=8wyz$, and $P_e=0$ otherwise.) Then, apply the triangle inequality (and positivity of $N$): $N(\alpha+\beta)=\left\lvert\sum_{e\in\mathbb{N}^4}d^eP_e(\alpha)\right\rvert\le\sum_{e\in\mathbb{N}^4}\lvert d\rvert^e\lvert P_e(\alpha)\rvert\le\sum_{e\in\mathbb{N}^4}\varepsilon^{\sum e}\lvert P_e(\alpha)\rvert$. (For the same example, the bound becomes $8\varepsilon\lvert wyz\rvert+\varepsilon^3(\lvert x+2y\rvert+\lvert z\rvert)$.) This now depends entirely on $\alpha$ and $\varepsilon=\frac{1}{2n}$. Let that bound be $f(\alpha,n)$ (also a polynomial in $\alpha$'s coordinates).

The strategy is then to check with a computer all $n^4$ elements $\alpha\in C_n$, and for each evaluate $f(\alpha,n)$. If $f(\alpha,n)<1$ then we move on, otherwise check $f(\alpha\pm e_j,n)$ for $e_j$ one of the four Kronecker basis vectors. It turns out that when $n=6$ we only ever need to check $\alpha$ and $\alpha\pm e_j$ (i.e. 0 or 1 steps) for this number field.

Curiously, it would seem that this approach could work to show that other biquadratic number fields $\mathbb{Q}\left[i,\sqrt{m}\right]$ of relatively small discriminant are norm-Euclidean, but this approach does not yield any results for $5<m,n<100$.


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