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May 10th, 2014, 03:51 AM  #1 
Member Joined: Sep 2013 Posts: 93 Thanks: 2  Show that 2+i and 2i is irreducible in Z[i]
Hello I need to give the irreducible factorization of 5 in Z[i],which is (2i)*(2+i).But how can i show that it is really the irreducible factorization? 
May 10th, 2014, 06:20 AM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,700 Thanks: 2682 Math Focus: Mainly analysis and algebra 
Can you not prove it by contradiction? An alternative approach that suggests. Itself to me is that $2 \pm I = \sqrt{5}e^{\pm i\theta}$. Can you prove that the magnitude $\sqrt{5}$ precludes a factorisation in $\mathbb{Z}[i]$? 
May 10th, 2014, 07:12 AM  #3  
Senior Member Joined: Apr 2014 From: Greater London, England, UK Posts: 320 Thanks: 156 Math Focus: Abstract algebra  Quote:
Taking conjugates gives $(abi)(cdi)=2\mp i$. Multiplying yields $(a^2+b^2)(c^2+d^2)=5$ $\implies$ (as $5$ is prime) $a^2+b^2=1$ or $c^2+d^2=1$ $\implies$ $a+bi$ or $c+di$ is a unit in $\mathbb Z[i]$. Therefore $2+i$ and $2i$ are irreducible in $\mathbb Z[i]$. PS: In general, if $p$ is prime and $p=(m+ni)(mni)$ then $m+ni$ and $mni$ are irreducible in $\mathbb Z[i]$. Last edited by Olinguito; May 10th, 2014 at 07:27 AM.  
May 12th, 2014, 05:44 PM  #4 
Senior Member Joined: Mar 2012 Posts: 294 Thanks: 88 
This is really Olinguito's approach "in disguise", but it is worth mentioning as it can be applied to other rings: Note that for any $u = a + bi \in \Bbb Z[i]$, we can define: $N(u) = a^2 + b^2$ (this is called the NORM of $u$). One properties of the map $N: \Bbb Z[i] \to \Bbb N$ is that it is multiplicative: $N(uv) = N(u)N(v)$ (I urge you to prove this, it is enlightening!). Note that $N(2+i) = 5$ which is prime, so if we did have a factorization of $2+i$, say: $2 + i = uv$, we would have $5 = N(u)N(v)$, and since $N(u),N(v)$ are natural numbers, one of them must be 1. It is also easy to prove that $N(u) = 1$ if and only if $u$ is a unit (note that you probably meant there is no factorization of $2+i,2i$ into two NONunits). 

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