3

Moreover, how can I prove that $2+i$ and $1+i$ are irreducibles?

mechanodroid
  • 46,490
Lola
  • 81

2 Answers2

4

Why don't you start by factoring it in $\mathbb{Z}$, and then check if those divisors can be further reduced?

For your second question, recall that for any complex numbers $a$ and $b$, $\|ab\|^2=\|a\|^2\|b\|^2$. So since $\|2+i\|^2 = 5$ is prime, if $2+i=ab$ then... can you take it from there?

user7530
  • 49,280
  • Yes, the (field-theoretic) norm, $z\mapsto z\overline z$ in this case, is the key to all questions of this sort. – Lubin Feb 14 '13 at 18:03
0

i have noticed that $6=2*3 $. now $2=(1+i)(1-i)$ and 3 must be irreducible because cannot be written as product of two squares. but again I have tried to prove that 1+i is irreducible and I end up in a vicious circle, ie that it can be written as a product of a unit times either $(+1+i),(1-i) , (-1-i),(i-1)$.

Lola
  • 81
  • Can it be written as the product of two non-units, though? (See my answer for a hint on how to prove the answer). – user7530 Feb 14 '13 at 18:44
  • See this answer http://math.stackexchange.com/questions/302502/show-if-these-gaussian-integers-are-irreducible-or-not/302537#302537 – Andreas Caranti Feb 14 '13 at 18:45