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Show that the ring of integers of $\mathbb{Q}(i+\sqrt{2} )$ is a Euclidean domain.

Not sure how to go about doing this. I've tried showing that $\mathbb{Q}(i+\sqrt{2} )$'s norm is a Euclidean function on its ring of integers, as is often the case in number fields whose integers form a Euclidean domain, but I haven't had any luck (if I am being unclear, this norm of an element $x$ of $\mathbb{Q}(i+\sqrt{2} )$ is the product of all the distinct monomorphisms from $\mathbb{Q}(i+\sqrt{2} )$ to $\mathbb{C}$ evaluated at $x$). Any help would be appreciated.

Petros
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    @Mark Saving: the ring of integers of a number field is not a field. – Thomas Preu Aug 05 '22 at 00:12
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    Realizing that the field is actually $\mathbb Q(\zeta_8)$ with a primitive $8$th root of unity, and that the ring of integers is known to be $\mathbb Z[\zeta_8]$, may facilitate checking that Euclidean-ness... – paul garrett Aug 05 '22 at 00:40
  • If you want to circumvent general theory: use that $o=o_{\mathbb{Q}(i+\sqrt 2)}$ is Euclidean with the norm function. A. Compute the norm function: $N(a_0+a_1\zeta+a_2\zeta^2+a_3\zeta^3)=a_1^4+4a_1^2a_2a_4+2a_1^2a_3^2-4a_1a_2^2a_3+4a_1a_3a_4^2+a_2^4+2a_2^2a_4^2-4a_2a_3^2a_4+a_3^4+a_4^4$, B. Show that the $N$ satisfies $N\left([-\frac12,\frac12]^4\right)<1$ (in fact $\leq\frac12$ should be true), C. Show that for $0\neq b,a\in o$: If you have $a/b=c_0+c_1\zeta+c_2\zeta^2+c_3\zeta^3$ and set "$q=\text{rd}(a/b)$" where you round the $c_i$ to nearest integer, you satisfy the Euklidean property. – Thomas Preu Aug 06 '22 at 11:04
  • The advantage of showing this explicitly is that you get an algorithm to compute in $o$. Sometimes that makes a big difference. Showing B. can be done by usual real multivariate analysis: Find all stationary points of $N$ in the bulk and the boundary facets of the hypercube $\left[-\frac12,\frac12\right]^4$ and show that the values at these points is $\leq\frac12$. – Thomas Preu Aug 06 '22 at 11:12
  • B. is actually a bit more involved, since $N=0$ does not only have the single solution $(a_i)_{i=1}^4=(0,0,0,0)$, but e.g. any point on the line $P\in{(0,x,\sqrt{2}x,x):x\in\mathbb{R}}$ also satisfies $N(P)=0$. There are even more strata with this property. However this does not invalidate the claim, that $o$ is norm-Euklidean, since we need $N(P)=0\Leftrightarrow P=0$ only for $P\in\mathbb{Q}^4$ and the coefficient $\sqrt{2}$ interferes with that. It only makes it harder to show part B. explicitly. – Thomas Preu Aug 06 '22 at 11:55

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