Irreducible factorization in number ring.

**featherbox** · April 24th 2010, 11:47 AM

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

**NonCommAlg** · April 24th 2010, 01:43 PM

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.

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