Artin's proof of the order of $\mathbb Z[i]/(a+bi)$ As Ben suggested in my earlier question on the subject, I looked at Artin's proof that $\left|\cdot\right|^2$ is a "size function" which makes $\mathbb Z [i]$ into a Euclidean domain. To quote page 398:

We divide the complex number b by a: $b=aw$, where $w=x+yi$ a complex number, not necessarily a Gauss integer. The we choose the nearest Gauss integer point $(m,n)$ to $(x,y)$, writing $x=m+x_0,y=n+y_0$, where m,n are integers and $x_0,y_0$ real numbers such that $-1/2\leq x_0,y_0<1/2$. Then $(m+ni)a$ is the required point of $Ra$. For, $\left|x_0 + y_0i\right|^2<1/2$ and $|b-(m+ni)a|^2=|a(x_0+y_0i)|^2<\frac{1}{2}|a|^2$.

I have two questions:


*

*I assume he's using the notation $\left|a+bi\right|=\sqrt{a^2 + b^2}$. If so, it seems like $\left|x_0 + y_0i\right|^2<1/2$ is not always true since $\left|(-1/2)+(-1/2)i\right|^2=1/2$

*He never uses the identity $i^2=-1$, so it seems like this proof could be expanded to all rings $\mathbb Z[x]/(x^2 + a)$, or indeed anything which has a vectorspace-like structure like $\mathbb Z^2$. But I remember hearing that $\mathbb Z[\sqrt{-5}]$ is not Euclidean - why does this proof fail for $x^2 = -5$?



EDIT: $\sqrt{-5}\approx 2.2i$ so we can write for example $3i \approx 1.3\sqrt{-5}$. By my understanding, $y_0=.3$ here and the norm $|0+.3|=0^2+.3^2$ is less than one. Furthermore, every $x_0,y_0$ is less than $1/2$, so this norm will always be less than 1, which is all we need.
Why doesn't this show that $\mathbb Z[\sqrt{-5}]$ is Euclidean?
 A: Whatever the normalizations, the requirement of "Euclidean" is that the remainder is strictly smaller than the divisor. Thus, in the highlighted paragraph above, the necessary conclusion is simply that the norm of the leftoever be strictly smaller than the divisor, not than half the size of the divisor.
In the case of $\mathbb Z[\sqrt{5}]$, the analogous leftover $a+b\sqrt{5}$ again has $|a|,|b|\le 1/2$, but the usual complex absolute value, squared or not, is $(1/2)^2+5(1/2)^2=6/4>1$. Taking a square root or squaring or not... does not affect the crucial issue of whether it's $<1$ or not.
Some examples of computations about Euclidean-ness of rings of alg integers are in  my notes , and many other places on the internet, I'm sure.
Edit: in response to further query... in $\mathbb Z[\sqrt{-5}]$, the leftovers are $a+b\sqrt{-5}$ with $|a|,|b|\le 1/2$. The norm (-squared) is
$a^2+5b^2\le (1/4)+5(1/4)=6/4>1$. In the "EDIT" in the query, yes, there is a particular example where the norm is $<1$. Ok, but that doesn't imply that all leftovers are $<1$. Yes, every $x_0,y_0$ are at most $1/2$ in size, but that's without the "5"! $|x_0+y_0\sqrt{-5}|^2=x_0^2+5y_0^2$.
