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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.. – sperners lemma Oct 26 '12 at 12:48
Don't you mean $\mathbb{Z}(\sqrt{5}) = \{a+\sqrt{5}b\,\lvert\, a,b\in\mathbb{Z} \}$? – Sh4pe Oct 26 '12 at 12:53
@Sh4pe: yes, this is – Muniain Oct 26 '12 at 12:56
@Muniain: You could edit your question then... :) – Sh4pe Oct 26 '12 at 12:58
This ring is normally called $\Bbb{Z}[\sqrt{5}]$. $\Bbb{Z}(\sqrt{5})$ is its field of fractions. – Chris Eagle Oct 26 '12 at 15:56

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? – Chris Eagle Oct 26 '12 at 15:53
All you seem to be showing is that $4+\sqrt{5}$ is irreducible. How do you then conclude that it is prime? – Chris Eagle Oct 26 '12 at 15:53
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.) – Lukas Geyer Oct 26 '12 at 16:00
If $|x|$ is absolute value, then $N(4+\sqrt{5})=(4+\sqrt{5})^2=21+8\sqrt{5}$. – Chris Eagle Oct 26 '12 at 16:08
Ah, I guess it is too early for me to post answers here... – Lukas Geyer Oct 26 '12 at 16:26

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