gaussian prime factorisation

if you look at the wolframalpha calculation page:
-13-26i - Wolfram|Alpha

it factorises -13-26i as

i×(1+2 i)×(2+3 i)×(3+2 i)

my question is why is the i a necessary factor to take out? are the rules regarding this?

Thanks

Re: gaussian prime factorisation

Hint: (1-i) is NOT a Gaussian Prime. It can be factored: i(1+i)

Re: gaussian prime factorisation

Quote:

Originally Posted by TKHunny

Hint: (1-i) is NOT a Gaussian Prime. It can be factored: i(1+i)

I don't think that this is correct. There is a characterisation of Gaussian primes that says that a Gaussian integer is prime if and only if one of the conditions is true:

Its norm is prime (as an integer)
It is a prime (as an integer) of the form $4k+3$

In this case, the norm of $1-i$ is $1^2+(-1)^2=2$ , which is prime. So $1-i$ is prime.

The issue that should be brought up is that unique factorisation only holds up to a unit (i.e. invertible number). For example, normal integer factorisation follows this idea: $25=5\times 5$ or $25=(-5)\times (-5)$ . In the second example, we have multiplied the factors by $-1$ (which is invertible), but we do not consider it to be inherently different from the first. So in this case, $1-i$ only differs from $1+i$ by multiplying by the unit $i$ , which is no difference at all, as far as unique factorisation goes.

So to address original question, throwing in $i$ is analogous to repeatedly multiplying by $-1$ with the usual integer factorisation. If it bothers you, you may let one of the terms absorb it.

Re: gaussian prime factorisation

I'm making stuff up again?! Only explanation I could think of for the question presented. It's -i, anyway. Wow. Taking a break...

Re: gaussian prime factorisation

No worries! Mistakes happen.