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How can I prove that every maximal ideal of $B= \mathbb{Z} [(1+\sqrt{5})/2] $ is a principal?

I know if I show that B has division with remainder, that means it is a Euclidean domain. It follows that B is PID, and then every maximal ideal is principal ideal in PID.

However, I haven't been able to show that $B$ has division with remainder.

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    Do you know about Minkowski bound on ideal norms? If so, you can use that to show that the ring is a PID.2012-11-19
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    I'm curious why the question asks about maximal ideals. Any ideas?2012-11-19
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    It is a Dedekind domain. So, all nonzero primes are maximal.2012-11-19
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    Well, I guess it's easier to prove that $B$ is factorial. Its primes must be the following: $0$, $\sqrt{5}$, $p\in\mathbb{Z}$ prime with $p\equiv \pm 2\pmod 5$ and the divisors of primes $p\in\mathbb{Z}$ with $p\equiv \pm 1\pmod 5$ which have the form $a+b(1+\sqrt{5})/2$, $a,b\in\mathbb{Z}$.2012-11-19
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    I'm sorry but i am just undergraduate student. if there is anyone to show that B has division with remainder i will be happy.because it seems only way that i can understand.2012-11-19
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    Do you know the proof that $\mathbb{Z}[i]$ is a Euclidean domain? The only thing I can think of is trying to modify that proof for the case above. I have not tried to see if the proof works, but may be you can give it a shot?2012-11-19
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    I know that proof of the ℤ[i] is Euclidean domain.At first i tried to prove the same way but i stuck on somewhere and then i confused.forexample ℤ[√5] is not Euclidean domain.what's the difference? it seems the same when i try to same proof? what's the point i missed?2012-11-19
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    @susan When you show that $\mathbb{Z}[i]$ is a Euclidean domain, you're actually using the norm on $\mathbb{C}$ and the fact that $\mathbb{Z}[i]$ is a lattice in $\mathbb{C}$; if you have $d\in \mathbb{Z}[i]$ nonzero, then for all $a\in\mathbb{Z}[i]$, $a/d$ is some element of $\mathbb{C}$ and sits in a square of this lattice, so there's an element $q$ of the lattice which is closer than $1$ to this element; $a = qd + d(a/d-q)$ gives your division with remainder. The problem with $\mathbb{Z}[\sqrt{5}]$ is that this set isn't a good lattice in $\mathbb{C}$, so the same argument won't work.2012-11-19

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