5
$\begingroup$

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.

$\endgroup$
6
  • $\begingroup$ I don't have an answer for you, but you might find interesting to peak at the class group and Minkowski's "geometry of numbers" theory. He embedds a number field $K$ in a finite dimensional $\mathbb{R}$-vector space $V_{K}$ and proves some results concerning lattices in $V_{K}$. When you put an ideal of $\mathcal{O}_{K}$ inside $V$ through this map, it becomes a lattice and you can apply the theorems. This allows you to prove that in each class of the class group, there is an ideal $I\subseteq\mathcal{O}_{K}$ with bounded index $(\mathcal{O}_{K}\colon I)$, making things a lot easier! $\endgroup$
    – Shoutre
    Commented May 9, 2017 at 13:26
  • $\begingroup$ Once the bound of $(\mathcal O:I)$ can be 1, it is finished. $\endgroup$ Commented May 9, 2017 at 13:44
  • 1
    $\begingroup$ Suppose for the sake of the example (i guess won't work there) that we are dealing with $\mathbb{Q}(\alpha)$, where $\alpha$ satisfies $x^{5}-x+1=0$, instead, and that we have Minkowski bound 4. Then supposing it's ring of integers is not a PID, every non-trivial class has an ideal $I$ with $\mathcal{O}_{K}/I\cong 2$ or $\mathcal{O}_{K}/I\cong 3$. But then we can reduce the polynomial $x^{5}-x+1$ mod $2$ and $3$ and see that there are actually no solutions, a contradiction. This is to show that even bounds $>1$ can be useful $\endgroup$
    – Shoutre
    Commented May 9, 2017 at 13:51
  • 3
    $\begingroup$ Fun fact: this domain is Euclidean but the absolute value of the norm is not the Euclidean function. That's getting a little bit ahead of things, but I'm telling you just in case you wonder why they picked this one for this. $\endgroup$ Commented May 9, 2017 at 17:51
  • 1
    $\begingroup$ For information, see the survey The Euclidian Algorithm in Algebraic Number Fields by Franz Lemmermeyer . $\endgroup$ Commented May 10, 2017 at 6:27

1 Answer 1

7
$\begingroup$

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.

$\endgroup$
1
  • 2
    $\begingroup$ You might also want to mention that $$\frac{\frac{9}{2} + \frac{\sqrt{69}}{2}}{ \frac{9}{2} - \frac{\sqrt{69}}{2}} = \frac{25}{2} + \frac{3 \sqrt{69}}{2},$$ which is the fundamental unit. $\endgroup$
    – Mr. Brooks
    Commented May 11, 2017 at 20:46

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .