# 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:

1. 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 $\left|(-1/2)+(-1/2)i\right|^2=1/2$
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?

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For (1): The norm in number theory is different. For Gaussian integers it is $a^2+b^2$. If one took the square root, norm would be usually irrational, and people in number theory are very rational. –  André Nicolas Jul 21 '11 at 14:22
For(2): If you look at the ring $\mathbf{Z}[\sqrt{-5}]$, then the imaginary parts of the elements of the ring are integer multiples of $\sqrt5$ as opposed to (rational) integers as here. Therefore the `error' term $y_0$ may be as large as $\sqrt{5}/2>1$, and this approach to getting a Euclidean algorithm fails. –  Jyrki Lahtonen Jul 21 '11 at 14:35
I looked at the actual passage! Artin is using $|x+yi|$ in its standard complex variable sense, and there is a typo as pointed out by the OP. There is no problem, we don't need $\lt 1/2$ anyway to push through the Gaussian version of the Division Algorithm, $\lt 1$ is good enough. But earlier comment still (mostly) stands. –  André Nicolas Jul 21 '11 at 15:15
Without saying "because $\mathbf{Z}[\sqrt{-5}]$ does not have unique factorization" it seems like it would be very hard to show that there is no norm. If you just want to show that $N(a + b\sqrt{-5}) = a^2 + 5b^2$ is not a norm, I think we can do something for you. –  Dylan Moreland Jul 21 '11 at 17:42
@Xodarap: Artin does place one important restriction on the norm: the norm should be multiplicative. IOW for all complex numbers $w,z\in\mathbf{C}$ you should have $|zw|=|z|\cdot |w|$. This is used in the equation $|a(x_0+iy_0)|^2=|a|^2|x_0+iy_0|^2\le\frac12|a|^2$. You MUST use the norm $|x+y\sqrt{-5}|=x^2+5y^2$. Otherwise the norm is not multiplicative. I don't have access to the text, but whatever the 'size'-function is called, surely it must be multiplicative. –  Jyrki Lahtonen Jul 21 '11 at 17:46

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.
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"! $|x_0+y_0\sqrt{-5}|^2=x_0^2+5y_0^2$.
Nevermind, I think the problem was that I didn't realize $|ab|$ must be $|a||b|$. –  Xodarap Jul 21 '11 at 18:36