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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. – Rankeya Nov 19 '12 at 17:09
I'm curious why the question asks about maximal ideals. Any ideas? – sperners lemma Nov 19 '12 at 17:20
It is a Dedekind domain. So, all nonzero primes are maximal. – Rankeya Nov 19 '12 at 17:24
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 $a+b(1+\sqrt{5})/2$, $a,b\in\mathbb{Z}$. – user26857 Nov 19 '12 at 17:40
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. – susan Nov 19 '12 at 18:39

$\phi=\frac{1+\sqrt(5)}2\\ \\ \phi^n=F_n+F_{n-1}$ 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 $F_i<a$ plus the largest $F_j<(a-F_i)$ etc.), therefore the fraction can be rewritten with the numerator and denominator polynomials in $\phi$ for which all coefficients except the term in $\phi$ are $0$ or $1$. Divide top and bottom through by $\phi$ and rewrite as $\frac{a'+b'\phi}{c'+d'\phi}$. Repeating this process will eventually yield a term in $\mathbb{Z}[\phi]$ plus a remainder.

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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.

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