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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

2 Answers 2

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$\phi=\frac{1+\sqrt(5)}2\\ \\ \phi^n=F_n+F_{n-1}$ where $F_n$ is the $n^{th}$ fibonacci number.

For any $\frac{a+b\phi}{c+d\phi}$ with $a$ and $c$ positive (if either is negative take out the factor of $-1$), $a$ and $c$ can be written uniquely as the sum of distinct fibonaccci numbers (take the largest $F_i

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Here's a sketch --- see how it goes.

Let $\alpha,\beta$ be in $B$, $\beta\ne0$. First show that $${\alpha\over\beta}=\gamma+\delta,\quad\delta=p+q\sqrt5$$ for some $\gamma$ in $B$ and some rationals $p$ and $q$, $0\le p\lt1$, $0\le q\lt1$. Consider $\delta-\epsilon$ for the following five values of $\epsilon$, all of which are in $B$: $0,1,\sqrt5,1+\sqrt5,(1+\sqrt5)/2$. Show that for at least one of these five values of $\epsilon$ the norm of $\delta-\epsilon$ is less than $1$ in absolute value (the norm of $r+s\sqrt5$ is $r^2-5s^2$). Then we have $$\alpha=(\gamma+\epsilon)\beta+(\delta-\epsilon)\beta$$ and the norm of $(\delta-\epsilon)\beta$ is the norm of $(\delta-\epsilon)$ times the norm of $\beta$, so it's less, in absolute value, than the absolute value of the norm of $\beta$.

EDIT: It's done nicely in Cohn, Advanced Number Theory, pp 108-109 (with some references to earlier pages). I'll summarize.

With $\alpha,\beta$ as above, rationalize the denominator and write $${\alpha\over\beta}={A_1+A_2\omega\over C}$$ with $A_1,A_2,C$ integers and $\omega=(1+\sqrt5)/2$. We want to find $\gamma=a+b\omega$ with $a,b$ integers such that $$|N((\alpha/\beta)-\gamma)|\lt1$$ which is to say we want $$|N((A_1/C)-a+((A_2/C)-b)\omega)|\lt1$$ Computing this norm, it's $$((A_1/C)-a)^2+((A_1/C)-a)((A_2/C)-b)-((A_2/C)-b)^2$$ Choose $a,b$, respectively, as the integers closest to $A_1/C,A_2/C$, respectively, and write $$P=(A_1/C)-a,\qquad Q=(A_2/C)-b$$ Then $$-1/2\le P\le1/2,\qquad-1/2\le Q\le1/2$$ and we are looking at $$f(P,Q)=P^2+PQ-Q^2$$ Now you can use calculus to show that $$\max|f(P,Q)|=5/16$$ given the restriction on $P,Q$, and you're done.