how to prove that the ring of algebraic integers of $\mathbb Q(\sqrt {69})$ is a PID or UFD

Question

I am now reading an article about Euclidean ring and one part is about the integers in $\mathbb Q(\sqrt{69})$. One step needed is to prove that the ring is a UFD (or to prove PID). Since in the article this will be used before the definition of the Euclidean function, the fact need to be proven without using that it is Euclidean.

Answer 1

Suppose that the ring is in fact not a principal ideal domain. That would mean we can find a number with at least two distinct factorizations, and the principal ideal generated by that number is the product of ideals that are themselves not principal.

Let's look at $\textbf Z[\sqrt{-69}]$ for just a brief moment. We readily find that $$70 = 2 \times 5 \times 7 = (1 - \sqrt{-69})(1 + \sqrt{-69})$$ and so $\langle 70 \rangle = \langle 2, 1 + \sqrt{-69} \rangle^2 \langle 5, 1 - \ldots$ you get the idea. The class number is $8$, if I'm not mistaken, but since I just wanted to make the point that this ring is not UFD, we're done.

Turning our attention back to $\mathcal O_{\textbf Q(\sqrt{69})}.$ Thanks to Minkowski, it is enough to prove that $\langle 2 \rangle$ and $\langle 3 \rangle$ can be factorized into products of principal ideals. (I was having some connectivity problems yesterday, so I apologize I did not post a complete answer yesterday.)

Anyway, the Minkowski bound for $\mathcal O_{\textbf Q(\sqrt{69})}$ is roughly $4.1533119$, so we just need to look at $\langle 2 \rangle, \langle 3 \rangle, \langle 4 \rangle$.

Since $69 \equiv 5 \pmod 8$, then, by Theorem $10.2.1$ in Alaca & Williams Introductory Algebraic Number Theory, $\langle 2 \rangle$ is prime, and therefore $\langle 4 \rangle = \langle 2 \rangle^2$.

That just leaves us $\langle 3 \rangle$ to concern ourselves with. Since $69 = 3 \times 23$, it's a ramifying ideal: $\langle 3 \rangle = \langle 3, \sqrt{69} \rangle^2$.

However, since $$\left(\frac{9}{2} - \frac{\sqrt{69}}{2}\right) \left(\frac{9}{2} + \frac{\sqrt{69}}{2}\right) = 3$$ and $$\left\langle \frac{9}{2} - \frac{\sqrt{69}}{2} \right\rangle = \left\langle \frac{9}{2} + \frac{\sqrt{69}}{2} \right\rangle$$ it follows that $$\langle 3 \rangle = \left\langle \frac{9}{2} + \frac{\sqrt{69}}{2} \right\rangle^2.$$

We have thus verified that the principal ideals generated by positive integers below the Minkowski bound are either themselves prime ideals, or the squares of prime principal ideals. Therefore $\mathcal O_{\textbf Q(\sqrt{69})}$ is a principal ideal domain, and its numbers factorize uniquely, disregarding multiplication by units.