Re: JSH: Fool all of the people, all of the time?

From: *** T. Winter (***.Winter_at_cwi.nl)
Date: 12/06/04 

Date: Mon, 6 Dec 2004 03:50:11 GMT

In article <3c65f87.0412051913.333cd232@posting.google.com> jstevh@msn.com (James Harris) writes:
...
> That is, given an integer x, either x = 0 mod 2, or x = 1 mod 2.

Yup.

> So trivially you can put up something like (x+1)(x+2) and know it must
> be even.

Indeed.

> However, in the ring of algebraic integers, no such relations exist.

Eh?

> That is, there is no non-unit in the ring of algebraic integers such
> that EVERY algebraic integer is either divisible by that number or has
> the same residue.

This does not parse. But I would think that in the integers a number
is divisible by 3 or has a residu of either 1 or 2 when divided by 3.
So I see nothing like "the same residu" here. So, mod 3, in the integers,
there are three residue classes. So I am not surprised when the same
happens in a larger ring.

> You don't even have the even case with algebraic integers as in the
> ring of algebraic integers it is NOT true that (x+1)(x+2) must have 2
> as a factor.

Indeed, because there are infinitely many residue classes mod 2 in the
algebraic integers. 

-- 
*** t. winter, cwi, kruislaan 413, 1098 sj  amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn  amsterdam, nederland; http://www.cwi.nl/~***/
Quantcast