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May 10th, 2014, 03:51 AM   #1
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Show that 2+i and 2-i is irreducible in Z[i]

Hello
I need to give the irreducible factorization of 5 in Z[i],which is (2-i)*(2+i).But how can i show that it is really the irreducible factorization?
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May 10th, 2014, 06:20 AM   #2
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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 $\mathbb{Z}[i]$?
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May 10th, 2014, 07:12 AM   #3
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Quote:
Originally Posted by ricsi046 View Post
Hello
I need to give the irreducible factorization of 5 in Z[i],which is (2-i)*(2+i).But how can i show that it is really the irreducible factorization?
Suppose $(a+bi)(c+di)=2\pm i$.

Taking conjugates gives $(a-bi)(c-di)=2\mp i$.

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 $2-i$ are irreducible in $\mathbb Z[i]$.

PS: In general, if $p$ is prime and $p=(m+ni)(m-ni)$ then $m+ni$ and $m-ni$ are irreducible in $\mathbb Z[i]$.
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Last edited by Olinguito; May 10th, 2014 at 07:27 AM.
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May 12th, 2014, 05:44 PM   #4
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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 $2+i,2-i$ into two NON-units).
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