NetNews Usenet Archive 1992 #26/NN_1992_26.iso/spool/sci/math/14693

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From: [e-mail] (G.J. McCaughan)
Newsgroups: sci.math
Subject: Re: Generalizing Prime Numbers
Message-ID: <[e-mail]>
Date: 10 Nov 92 00:26:22 GMT
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In article <[e-mail]>, [e-mail] (Jonathan Katz) writes:

> I was wondering if anyone knows about any method of generalizing the concept of
> prime numbers to the complex plane.
> Is there more than one way to generalize it?
> Any help on the would be appreciated.

Yeah. Prime numbers live in the integers, which sit inside the rationals.
If you take any finite-degree field extension of the rationals (translation:
write down finitely many algebraic numbers, like "root(3)" or "the smallest
positive real root of x^5+x+1=0" and consider the set of linear combinations
of them with rational coefficients), then there is a corresponding "ring of
integers" inside it (namely the set of elements of the field which are roots
of polynomials with integral coefficients, whose "top" coefficient is 1).

An element of this ring is a "unit" if it has an inverse in the ring (note
that the units in the usual ring of integers are precisely 1 and -1). We say
that an element of the ring is "irreducible" (which is probably what you mean;
"prime" is in fact a different concept, for nasty technical reasons) if it
can't be written as the product of two elements neither of which is a unit.

Just about the simplest example is in the case of the so-called Gaussian
integers, namely the integers in the smallest field containing the rationals
and i. (i^2=-1). The integers here turn out to be exactly the numbers of the
form a+bi with a,b both integers (warning: things don't always work out in
quite such an obvious way). If you do a bit of work you can prove that the
(usual) primes which are of the form 4m+3 are still irreducible in this ring,
and that the ones of the form 4m+1 factor into two irreducible factors, and
that 2 (the only prime not of form 4m+1 or 4m+3 !) factors as (1+i)(1-i).

I said something about primes and irreducibles being different; I might as
well explain very briefly and handwavingly. It so happens that "irreducible"
isn't really quite the right concept in general; what we really need for
most purposes about prime numbers is not that they don't factor, but that
if a prime divides the product of two numbers then it divides one or other
of the two numbers. By a happy coincidence prime=irreducible in the usual
integers, but this doesn't always hold. (Example: if you throw in root(-5)
instead of root(-1), the integers are again the numbers of form a+bw where
a and b are integers (and w=root(-5)); now 6 is both 2x3 and (1+w)(1-w),
and 2,3,1+w,1-w are irreducible.) So in fact it's no longer the case that
every number factors in a unique way into irreducibles, which is a pain.
The way out involves things called "ideals", which you can think of as
non-existent divisors of numbers, sort of. (Very much sort of.)

I hope this makes some kind of sense to you...

--
Gareth McCaughan     Dept. of Pure Mathematics & Mathematical Statistics,
[e-mail]  Cambridge University, England.    [Research student]