Re: how to factor gaussian integers ?

Arturo Magidin <magidin@xxxxxxxxxxxxxxxxx> wrote:

There are very efficient algorithms for factoring in integral domains
with only finitely many primes.

Most certainly not. If that were true then one could factor
any integer N "efficiently" in the localization of Z whose
primes are precisely those integer primes smaller than N.

Let D be the set of all rational numbers that can be written as a/b,
with a and b integers, and b relatively prime to, say, 2, 3, and 5.

This is an integral domain. The only primes in this domains are 2, 3,
and 5.

Given any element x/y of D, with gcd(x,y)=1, find the highest power of
2, of 3, and of 5 that divide x (which can be done both easily and
efficiently). Write x = 2^a*3^b*5^c*e, with gcd(e,30)=1. Then x/y
factors in D as (e/y)*2^a*3^b*5^b. Both easy and efficient.

Same idea if you have an integral domain with only finitely many
primes.

That's simply trial division. Certainly easy, but hardly efficient.

--Bill Dubuque
.

Relevant Pages

  • Re: how to factor gaussian integers ?
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  • Re: how to factor gaussian integers ?
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  • Re: how to factor gaussian integers ?
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  • Re: how to factor gaussian integers ?
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