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Euclidean Domain

Date: 08/12/2003 at 23:16:57
From: Mauro
Subject: Quadratic Fields and the Division Algorithm

How can I prove that there does not exist a division algorithm in 
any quadratic field K = Q(sqrt(D)), where D <= -15?  Basically, I 
need to prove that, for a and b integers in K such that the norm of 
a is greater than or equal to the norm of b, there exists an integer 
c such that the norm of a - bc is less than the norm of b - unless 
I have this wrong.

A previous question asked me to prove that such an algorithm exists 
for the Gaussian integers, but I did that geometrically, so I have no 
idea where to go on this one - there are too many cases and variables 
for me to do it algebraically, I think.

Date: 08/13/2003 at 02:27:36
From: Doctor Jacques
Subject: Re: Quadratic Fields and the Division Algorithm

Hi Mauro,

Actually, what you have to prove is the opposite of what you say:

In Q(sqrt(d)), with d <= -15, for some a and b, there is no c such 
that N(a-bc) < N(b) (otherwise, there would be a division algorithm).

An integral domain with a division algorithm is called a Euclidean 

I answered a similar question some time ago; see:

   D is not Euclidean 

However, a few remarks are in order.

In the above article, the question was only about rings of the form Z
[sqrt(d)]. As you should know, not all rings of quadratic integers 
are of that form; for example, for d = -7, the integers are of the 

  x + yw

with w = (1+sqrt(-7)) / 2

The argument must be expanded to account fot those cases. In 
particular, it is stated (at the end) that Z[sqrt(-3)] is not 
Euclidean. This is correct, but that ring is not the largest ring in 
its field of fractions. The ring of all integers of Q[sqrt(-3)] also 
includes (1+sqrt(-3))/2, and _that_ ring is Euclidean.

If you assume that the Euclidean function is the norm (i.e. the 
product of the number by its complex conjugate), the reasoning would 
be based on the same geometric arguments as for the Gaussian 
integers: you have a division algorithm based on the complex norm if 
you can cover the plane with open discs of radius 1 centered at the 
lattice points, as explained in the first part.

But there could still be the case that you would have a division 
algorithm based on another Euclidean function g, defined on the non-
zero ring elements, ans satisfying:

1.  g(x) is a positive integer
2.  If a | b, then g(a) <= g(b)
3.  For all a, b, with b <> 0, there exist q and r such that either
    r = 0 or g(r) < g(b)

To show that this is not the case, you have to use the second part of 
the article. You will also need to fill in what is necessary to cover 
the cases where d = 1 (mod 4) and the ring contains elements 
with "half-integer" coordinates. The rings of all integers in
Q[sqrt(d)], d < 0,  are Euclidean for d = -1, -2, -3, -7, and -11.

Does this help?  Write back if you'd like to talk about this 
some more, or if you have any other questions.

- Doctor Jacques, The Math Forum 
Associated Topics:
College Modern Algebra

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