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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
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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
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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
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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
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@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

1 Answer 1

up vote 3 down vote accepted

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"! $|x_0+y_0\sqrt{-5}|^2=x_0^2+5y_0^2$.

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Could you look at my edit? I still don't think I understand. –  Xodarap Jul 21 '11 at 17:10
    
Nevermind, I think the problem was that I didn't realize $|ab|$ must be $|a||b|$. –  Xodarap Jul 21 '11 at 18:36
    
Ah. Indeed! ... –  paul garrett Jul 21 '11 at 18:39

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