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

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 $$$$\mathbb{Z} \left[ \frac{1+\sqrt{5}}{2} \right] = \left\{ \frac{m + n \sqrt{5}}{2} : m, n \in \mathbb{Z} \text{ are both even or both odd} \right\}.$$$$

For every $$\frac{m + n \sqrt{5}}{2} \in \mathbb{Z} \left[ \frac{1+\sqrt{5}}{2} \right]$$, define its norm as usual: $$$$N\left(\frac{m + n \sqrt{5}}{2}\right)=\frac{m^2-5n^2}{4}.$$$$ Since $$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?

• It is a unique factorization domain! This post and the linked posts there may be helpful: math.stackexchange.com/questions/18589/golden-number-theory – Bart Michels Nov 26 '20 at 16:23
• You don't need to know whether the ring is a UFD or not. Because the norm is equal to $11$, the quotient ring $\Bbb{Z}[(1+\sqrt5)/2]/\langle 4+\sqrt5\rangle$ is isomorphic to $\Bbb{Z}_{11}$, which is a field and an integral domain. Therefore.... – Jyrki Lahtonen Nov 26 '20 at 16:27
• @JyrkiLahtonen Dear Jyrki, how could you guess it?! I feel pure wonder! Thanks very very ... much for your incredible insights! – Maurizio Barbato Nov 26 '20 at 17:12
• Thanks, you see that Bart detailed this same idea. How? Experience and practice! And a strong tendency to try and find ways to prove things with as little technology as possible. The last point may be related to having a background in contest math (but also to relative lack of success in research). – Jyrki Lahtonen Nov 26 '20 at 21:16

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.

• Dear Bart, this is an incredibly beautiful and remarkably elegant solution!!! It is a pure joy to grasp it! Thank you very very ... much for having carefully written down it in detail! – Maurizio Barbato Nov 26 '20 at 17:07
• You are welcome! – Bart Michels Nov 26 '20 at 17:10

The number field $$K=\Bbb Q(\sqrt{5})$$ has class number one because its Minkowski bound satisfies $$B_K<2$$ . Hence its ring of integers $$\mathcal{O}_K=\mathbb{Z} \left[ \frac{1+\sqrt{5}}{2} \right]$$ is even a PID and hence a UFD.

On the other hand, it is enough to see that $$\mathcal{O}_K/(4+\sqrt{5})$$ is a field, so that the ideal $$(4+\sqrt{5})$$ is prime.

Yes, $$\mathbb{Z}\left[\frac{1+\sqrt{5}}{2}\right]$$ is a UFD because it is norm-Euclidean.