Find irreducible but not prime element in $\mathbb{Z}[\sqrt{5}]$

Question

I have tried various numbers of the form $a+b\sqrt{5},\ a,b \in \mathbb{Z}$, but cannot find the one needed.

I would appreciate any help.

Update: I have found that $q=1+\sqrt{5}$ is irreducible. Now if I show that 2 is not divisible by $q$ in $\mathbb{Z}[\sqrt{5}]$ then $2\cdot2 = (\sqrt{5}-1)(\sqrt{5}+1)$ and I'm done. Can it be shown without use of norm ?

The last update: I have written the equation $(\sqrt{5}+1)(x\sqrt{5}+y)=2$ and have deduced that the equation has no integer solutions. Thanks to all who helped.

Indeed. Especially since "using the norm" is a bit of an overstatement -- all you're doing is multiplying by $\overline{q}$. — Cam McLeman, Feb 07 '12 at 15:13
@AlexBecker because then I need to prove that there is a norm on the ring and the ring is integral. — Sergey Filkin, Feb 07 '12 at 15:15
Try to divide $2$ by $\sqrt{5}+1$. So write $\frac{2}{\sqrt{5}+1}$, and rationalize the denominator. — André Nicolas, Feb 07 '12 at 15:19
@CamMcLeman question is about $\mathbb{Z}[\sqrt{5}]$, not $\mathbb{Z}[\sqrt{-5}]$. — Sergey Filkin, Feb 07 '12 at 15:19
Let's assume that bar means Galois conjugate :). But I think André's suggestion is simpler. — Dylan Moreland, Feb 07 '12 at 15:21
It is actually easier to show that $2$ is irreducible, and that it does not divide $1+\sqrt{5}$ (or $1-\sqrt{5}$)... — Arturo Magidin, Feb 07 '12 at 15:59

André Nicolas · Accepted Answer · 2012-02-07T15:24:05.780

5

Hint: $(\sqrt{5}+1)(\sqrt{5}-1)=4$

Added: (after the OP's edit) When it comes to showing that $\sqrt{5}+1$ does not divide $2$, note that $$\frac{2}{\sqrt{5}+1}=\frac{2(\sqrt{5}-1)}{(\sqrt{5}+1)(\sqrt{5}-1)}.$$

edited Feb 07 '12 at 15:24

answered Feb 07 '12 at 15:10

André Nicolas

507,029

I have shown that by another method. Maybe It's the same as rationalization of denominator. – Sergey Filkin Feb 07 '12 at 15:28
@Sergey Filkin: Your method works just fine. Both your method and rationalizing solve the problem of finding rationals $x$ and $y$ such that $2/(\sqrt{5}+1)=x+y\sqrt{5}$. – André Nicolas Feb 07 '12 at 15:43

score 3 · Answer 2 · edited Apr 13 '17 at 12:21

3

Hint: the norm map is absolutely crucial here. Because it is multiplicative, taking norms preserves the multiplicative structure and transfers divisibility problems from quadratic number fields to simpler integer divisibility problems (in a multiplicative submonoid of $\mathbb Z$). In this case the transfer is faithful, i.e. an element of your quadratic number ring is irreducible iff its norm is irreducible in the monoid of norms. Similarly, in many favorable cases, a quadratic number ring will have unique factorization iff its monoid of norms does.

For much more on this conceptual viewpoint see this answer. It is crucial to understand this conceptual viewpoint in order to master (algebraic) number theory. Do not settle for ad-hoc proofs when much more enlightening conceptual proofs are easily within grasp.

edited Apr 13 '17 at 12:21

Community

1

answered Feb 07 '12 at 16:14

Math Gems

19,574

I have solved the problem without use of norm. See my edited post. – Sergey Filkin Feb 07 '12 at 18:28
Also it's not true that if an element of $\mathbb{Z}[\sqrt{5}]$ is irreducible, then it's norm is irreducible. Take $1+3\sqrt{5}$, it's irreducible, but its norm is not. – Sergey Filkin Feb 07 '12 at 18:33
But I agree with you that it's enlightening to apply to conceptual proofs instead of low level proofs. – Sergey Filkin Feb 07 '12 at 18:41
@Sergey The irreducibility of the norm occurs in the monoid of all norms of elements of $\mathbb{Z}[\sqrt{5}]$. Since this is a proper multiplicative submonoid of $\mathbb Z$, the meaning of "irreducible" may differ from that in $\mathbb Z$. Please see the linked post and its references,. – Math Gems Feb 07 '12 at 18:42
this topic is deeper than I thought. Thank you for the link. – Sergey Filkin Feb 07 '12 at 18:58

Find irreducible but not prime element in $\mathbb{Z}[\sqrt{5}]$

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