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Decompose number in Gaussian interger field

  1. Mar 20, 2012 #1
    Hi everyone, I always have trouble on decomposing number into irreducible factors inside Gaussian integer field. I keep trying to express number as product as (a+bi)(c+di), and trying to solve a,b,c,d inside of integers (Z), then see if they are irreducibles, which of course end of very messy.

    Is there any techniques or algorithms to do these kind of decomposing numbers? Can anyone show me an example please so I may get good idea? Say take a prime 5, try to decompose it into irreducibles in Z

    Thanks a lot.
  2. jcsd
  3. Mar 20, 2012 #2
    You should use that every gaussian integer has an integer norm, and that the norm is multiplicative. The norm is defined as N(a+bi)=a2+b2.

    So if you like to factor the number 13, you first factor its norm 169=13*13, and then look for gaussian integers of norm 13=9+4, so you only need to consider ±3±2i, and ±2±3i, and indeed 13=(3+2i)(3-2i), and this can not be factored further since the norms of the factors are prime.
  4. Mar 21, 2012 #3
    Hi Norwegian, thank you so much for your reply. It is very clear, thank you.

    I have another question now, that for example 13=(3+2i)(3-2i), but since my goal is decompose 13 into irreducibles, how do I show (3+2i) and (3-2i) are irreducibles, so they cannot be decomposed further? Do I have to repeat the whole process like how I decomposed 13 to try to decompose (3+2i) and (3-2i) again?

    Thanks a lot for your time.
  5. Mar 21, 2012 #4
    As I told you above, 3+2i can not be factored further, since the norm N(3+2i)=13 is a prime in Z, (and any factor of 3+2i would have a norm dividing 13). We dont count the units (norm=1) ±1,±i as factors btw.
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