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> I've already asked this question [on Math StackExchange](https://math.stackexchange.com/questions/3981370/is-mathbbzi-varphi-a-euclidean-domain) but having gotten no responses this may be more obscure than I had initially believed.
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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](https://www.lmfdb.org/NumberField/4.0.400.1), 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.



Is $\mathbb{Z}[i,\varphi]$ a Euclidean domain?
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.




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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 $5<m,n<100$.


