# Irreducible factorization in number ring.

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• April 24th 2010, 11:47 AM
featherbox
Irreducible factorization in number ring.
In the quadratic number ring $Z[\sqrt{2}]$ find an irreducible factorization of the elements $4+\sqrt{2}$ and $6+\sqrt{2}$.
• April 24th 2010, 01:43 PM
NonCommAlg
Quote:

Originally Posted by featherbox
In the quadratic number ring $Z[\sqrt{2}]$ find an irreducible factorization of the elements $4+\sqrt{2}$ and $6+\sqrt{2}$.

$4+\sqrt{2}=(2+\sqrt{2})(3-\sqrt{2})$ and $6+\sqrt{2}=(2+\sqrt{2})(5-2\sqrt{2}).$ note that, since the norm of the factors are prime numbers, they are irreducible.