Is $(4+\sqrt{5})$ a prime ideal of $\mathbb{Z} \left[ \frac{1+\sqrt{5}}{2} \right]$

abstract-algebraalgebraic-number-theorymaximal-and-prime-idealsring-theory

Consider the integral domain $\mathbb{Z} \left[ \frac{1+\sqrt{5}}{2} \right]$. Is $(4+\sqrt{5})$ a prime ideal of $\mathbb{Z} \left[ \frac{1+\sqrt{5}}{2} \right]$?

I know the following elementary facts. We have
\begin{equation}
\mathbb{Z} \left[ \frac{1+\sqrt{5}}{2} \right] = \left\{ \frac{m + n \sqrt{5}}{2} : m, n \in \mathbb{Z} \text{ are both even or both odd} \right\}.
\end{equation}

For every $\frac{m + n \sqrt{5}}{2} \in \mathbb{Z} \left[ \frac{1+\sqrt{5}}{2} \right]$, define its norm as usual:
\begin{equation}
N\left(\frac{m + n \sqrt{5}}{2}\right)=\frac{m^2-5n^2}{4}.
\end{equation}
Since $m, n$ are both even or both odd, it is easy to see that the norm is an integer. From this fact it is easily seen that $\frac{m + n \sqrt{5}}{2}$ is a unit of $\mathbb{Z} \left[ \frac{1+\sqrt{5}}{2} \right]$ if and only if $m^2 – 5n^2=4$ or $m^2 – 5n^2=-4$. Now since $N(4+\sqrt{5})=11$ we easily get that $4+\sqrt{5}$ is an irreducible element of $\mathbb{Z} \left[ \frac{1+\sqrt{5}}{2} \right]$. If $\mathbb{Z} \left[ \frac{1+\sqrt{5}}{2} \right]$ were a unique factorization domain, we could conclude that $(4+\sqrt{5})$ a prime ideal of $\mathbb{Z} \left[ \frac{1+\sqrt{5}}{2} \right]$. But I do not know if $\mathbb{Z} \left[ \frac{1+\sqrt{5}}{2} \right]$ is a unique factorization domain. Does someone know if it is?

Thank you very much in advance for your attention.

Best Answer

Call $A = \mathbb Z \left[ \frac{1 + \sqrt 5}2\right]$. We can show that $A / (4+\sqrt 5) \cong \mathbb Z/11 \mathbb Z$, so that the ideal $(4 + \sqrt 5)$ is maximal.

  1. As $N(4 + \sqrt 5) = 11$, it is clear that the elements $0, 1, \ldots, 10$ are pairwise incongruent modulo $4 + \sqrt 5$.

  2. Every element of $A$ is congruent to an integer modulo $4 + \sqrt 5$: indeed, if it is of the form $a + b \sqrt 5$ with $a, b \in \mathbb Z$ we can subtract a suitable integer multiple of $4 + \sqrt5$ to land in $\mathbb Z$. If it is of the form $(a+b\sqrt5)/2$ with $a, b$ odd, we can subtract $$\frac{1 + \sqrt5}2 \cdot (4 + \sqrt5) = \frac{9 + 5\sqrt5}2$$ to land in $\mathbb Z + \mathbb Z\sqrt5$.

Consider the ring homomorphism $$\mathbb Z / (11) \to A / (4+\sqrt5) \,.$$ By the first observation, it is injective. By the second, it is surjective.

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