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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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OK, after giving a wrong solution this morning (I blame it on the lack of coffee...), here is another try:

The standard field norm on $\mathbb{Q}[\sqrt{5}]$ is $N(a+b\sqrt{5}) = |a^2 - 5b^2|$. It is multiplicative, and $N(4+\sqrt{5}) = |4^2 - 5\cdot 1^2| = 11$ is prime, so $4+\sqrt{5}$ is irreducible in $\mathbb{Z}[\sqrt{5}]$, and more generally in the ring of integers of $\mathbb{Q} [\sqrt{5}]$, which is $\mathbb{Z}\left[\frac{1+\sqrt{5}}2\right]$.

Now irreducibility in general does not imply primality, but it does in Euclidean domains. (More general, this statement is true in Unique Factorization Domains, and every Euclidean domain is a UFD.) It is known that $\mathbb{Q}[\sqrt{5}]$ is norm-Euclidean, i.e., that the standard field norm $N$ is Euclidean on the ring of integers $\mathbb{Z}\left[\frac{1+\sqrt{5}}2\right]$. The standard reference for the discussion of the question for which integers $d$ the domain $\mathbb{Z}[\sqrt{d}]$ is Euclidean seems to be the book of Hardy and Wright.

Now if $4+\sqrt{5}$ divides some $a+b\sqrt{5}$ in $\mathbb{Z}\left[\frac{1+\sqrt{5}}2\right]$, then $$ \begin{split} a+b\sqrt{5} &= (4+\sqrt{5})\left(c+d\frac{1+\sqrt{5}}2\right) \\ & = 4c+2d+\frac{5}2d + \left( c+\frac{d}2+ 2d)\right)\sqrt{5}. \end{split} $$ Since the coefficients are integers $a$ and $b$, we get that $d$ is even, and so $c+ d \frac{1+\sqrt{5}}2 \in \mathbb{Z}[\sqrt{5}]$. This implies that $4+\sqrt{5}$ actually divides $a+b\sqrt{5}$ in $\mathbb{Z}[\sqrt{5}]$. Now if $4+\sqrt{5}$ divides a product $xy$ with $x,y \in \mathbb{Z}[\sqrt{5}]$, then it divides $x$ or $y$ in $\mathbb{Z}[\frac{1+\sqrt{5}}2]$, so by the above argument it divides $x$ or $y$ in $\mathbb{Z}[\sqrt{5}]$, showing that it is a prime element of this ring.

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    What are you using $|x|$ to mean?2012-10-26
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    All you seem to be showing is that $4+\sqrt{5}$ is irreducible. How do you then conclude that it is prime?2012-10-26
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    Chris, $|x|$ just means the absolute value in $\mathbb{C}$, and you are right, I missed that irreducibility does not imply primality (unless we know that $\mathbb{Z}[\sqrt{5}]$ is a unique factorization domain, which I don't.)2012-10-26
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    If $|x|$ is absolute value, then $N(4+\sqrt{5})=(4+\sqrt{5})^2=21+8\sqrt{5}$.2012-10-26
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    Ah, I guess it is too early for me to post answers here...2012-10-26
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    $\Bbb{Z}[\sqrt{5}]$ is not Euclidean, or even a UFD. Perhaps you're confusing it with the ring of integers of $\Bbb{Q}[\sqrt{5}]$, which is $\Bbb{Z}\left[\frac{1+\sqrt{5}}{2}\right]$?2012-10-26
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    True, this is what I get as an analyst wading into algebraic number theory... But anyway, this still implies that $4+\sqrt{5}$ is prime, since $\mathbb{Z}[\sqrt{5}]$ is a subring of the ring of integers.2012-10-26
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    Why? $2$ is prime in the ring of integers, but not in this subring.2012-10-26
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    Chris, you are right, this needed another argument, but I think it works now.2012-10-26
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    I don't understand what did you do?. By definition, $\mathbb{Z}[\sqrt{5}]$ is integer domain, member $a$ is prime if $a\notin \mathbb{Z}*$ and from condition $a|bc$ implies $a|b$ or $a|c$.2012-10-29
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    It doesn't matter if $a \in \mathbb{Z}$ or not, the condition is $a|bc$ implies $a|b$ or $a|c$. And I think this is exactly what I do in the last paragraph, after having shown that $4+\sqrt{5}$ is prime in $\mathbb{Z}\left[\frac{1+\sqrt{5}}2\right]$ first.2012-10-29
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    so, $\mathbb{Z}[\frac{1+\sqrt{5}}{2}]$ is Euclide domain?. Can you give me a material say about this problem?. Thankful2012-11-01
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    The reference is linked to in the answer above, it is the book of Hardy and Wright.2012-11-01
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    Can you give me page number in that book?2012-11-02