Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $\mathbb{Z}(\sqrt{5})$

share|improve this question
1  
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
add comment

1 Answer 1

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.

share|improve this answer
    
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
show 9 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.