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$\begingroup$

definite ring $\mathbb{Z}(\sqrt{5})=\{a+\sqrt{5}b\,|\,a,b\in \mathbb{Z}\}$

show that $4+\sqrt{5}$ is a prime member of $\mathbb{Z}(\sqrt{5})$

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    note the norm of the number is $4^2 - 5 1^2 = 11$ which is a prime number, so it's probably a prime we can't rule it out yet..2012-10-26
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    Don't you mean $\mathbb{Z}(\sqrt{5}) = \{a+\sqrt{5}b\,\lvert\, a,b\in\mathbb{Z} \}$?2012-10-26
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    @Sh4pe: yes, this is2012-10-26
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    @Muniain: You could edit your question then... :)2012-10-26
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    This ring is normally called $\Bbb{Z}[\sqrt{5}]$. $\Bbb{Z}(\sqrt{5})$ is its field of fractions.2012-10-26

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