9
$\begingroup$

Let $D$ be a square free integer. Let $R_D$ be the integral closure of $\mathbb{Z}$ in the field $\mathbb{Q}(\sqrt{D})$.

For some values $D$, the ring $R_D$ is a $UFD$, but not for all. For example, the Gaussian integers $R_{-1}$ are a $UFD$ whereas the ring $R_{-5}$ is not. There are several ways to show this, including computing the class number of $R_D$. However, all the proofs I've seen feel ad hoc and unintuitive.

According to the Stark-Heegner theorem, for $D<0$, the ring $R_D$ is a $UFD$ if and only if $$D \in \{-1,-2,-3,-7,-11,-19,-43,-67,-163\}.$$ Is there any intuitive reason why this should be a complete list? Ideally there would be a structural reason - coming up with a separate proof for each $D$ in the list is deeply unsatisfying to me.

$\endgroup$
  • $\begingroup$ The contention in your second sentence is not correct; the integral closure of $\mathbb{Z}$ in $\mathbb{Q}(\sqrt{D})$ depends on the value of $D$ mod $4$. (In particular, sometimes, it is larger than $\mathbb{Z}[\sqrt{D}]$.) In any event: I'm not sure whether this qualifies as ''intuitive'', but a Dedekind domain is a UFD if and only if it is a PID, so checking whether $R_{D}$ is a UFD amounts to computing its class number. For this, this Minkowski bound is one useful tool; I'm sure there are others too, though I don't know which (if any) come up in the proof of Stark-Heegner. $\endgroup$ – Alex Wertheim Nov 28 '18 at 3:50
  • $\begingroup$ What are you asking exactly? Why the list is finite? Why this very specific list? Why the $D$ are all $-1$ or $-p$ for $p$ prime? Why the $D$ are all $-3 \pmod 4$ after $-2$ or all $-7 \pmod 8$ after $-11$? $\endgroup$ – Lorem Ipsum Nov 28 '18 at 3:52
  • $\begingroup$ @LoremIpsum Yes why this very specific and finite list $\endgroup$ – leibnewtz Nov 28 '18 at 4:01
  • $\begingroup$ @AlexWertheim Thanks for the correction - edited. I'm not as interested in specific techniques for showing why such rings are UFDs, unless that technique does it all at the same time. For example, an answer I'd be satisfied with would be: "this list is all the roots of a given polynomial," or something like that $\endgroup$ – leibnewtz Nov 28 '18 at 4:05
  • 1
    $\begingroup$ It sounds like you are asking for intuition for why the Stark-Heegner theorem is true for exactly this set of 9 numbers, I'm not sure this is completely possible but you may find uni-math.gwdg.de/tschinkel/gauss-dirichlet/stark.pdf particularly section 4 interesting. $\endgroup$ – Alex J Best Nov 28 '18 at 22:40
8
$\begingroup$

I wouldn't be so quick to discount the value of going through the list one by one. For this post, let's stipulate $D < 0$ throughout.

You've noticed $-2$ is the only even value on there, right? If $D$ is even, then $D = (\sqrt D)^2$, which is obvious enough. But if $D$ is even and composite, it means that $N(z) = 2$ for $z \in \mathcal{O}_{\mathbb{Q}(\sqrt D)}$ is impossible. So then $2$ is irreducible, yet $D = 2 \times x = (\sqrt D)^2$, where $x \in \mathcal{O}_{\mathbb{Q}(\sqrt D)}$ also.

You've also noticed that $-5$ is not on the list either. Neither is $-13$, $-17$, $-29$, etc. What these numbers have in common, besides being odd, is that they are congruent to $3 \bmod 4$ (remember that congruence gets "flipped" for negative numbers, so $-3 \equiv 1 \bmod 4$, not $3 \bmod 4$).

So, if $D \equiv 3 \bmod 4$, then $N(1 + \sqrt D) = -D + 1$, which is even. But in this domain, it turns out that $N(z) = 2$ is also impossible. Which means that $-D + 1$ has at least two distinct factorizations. Thus $D = -5$ gives us the classic example $6 = 2 \times 3 = (1 - \sqrt{-5})(1 + \sqrt{-5})$.

Now let's say $D \equiv 1 \bmod 4$ instead. Then it's still the case that $-D + 1 = (1 - \sqrt{D})(1 + \sqrt{D})$, but... $$\frac{1 - \sqrt{D}}{2}, \frac{1 + \sqrt{D}}{2}$$ are also algebraic integers, both with minimal polynomial $$x^2 - x + \frac{-D + 1}{4},$$ e.g., $$\frac{1 - \sqrt{-43}}{2}, \frac{1 + \sqrt{-43}}{2}$$ both have the polynomial $x^2 - x + 11$.

Therefore, the full factorization of $44$ in this domain is not $(1 - \sqrt{-43})(1 + \sqrt{-43})$ but $$2^2 \left(\frac{1 - \sqrt{-43}}{2}\right) \left(\frac{1 + \sqrt{-43}}{2}\right).$$

Why this doesn't work out for $D \leq -167$ is quite a bit more involved, maybe someone else will address that.

$\endgroup$
6
$\begingroup$

I do think there is an intuitive reason, and it has to do with the number 41. You know that $n^2 + n + 41$ is prime for all $0 \leq n < 41$? Of course it's composite for $n = 41$.

Notice that $$\left(\frac{1}{2} - \frac{\sqrt{-163}}{2}\right)\left(\frac{1}{2} + \frac{\sqrt{-163}}{2}\right) = 41,$$ $$\left(\frac{3}{2} - \frac{\sqrt{-163}}{2}\right)\left(\frac{3}{2} + \frac{\sqrt{-163}}{2}\right) = 43,$$ $$\left(\frac{5}{2} - \frac{\sqrt{-163}}{2}\right)\left(\frac{5}{2} + \frac{\sqrt{-163}}{2}\right) = 47,$$ and so on and so forth to $$\left(\frac{79}{2} - \frac{\sqrt{-163}}{2}\right)\left(\frac{79}{2} + \frac{\sqrt{-163}}{2}\right) = 1601.$$

What are the other numbers $d$ such that $n^2 + n + d$ gives primes for $0 \leq n < d$? 2, 3, 5, 11, 17, see OEIS A014556. And then $-4d + 1$ gives $-7, -11, -19, -43, -67$, and $-163$ corresponds to 41.

No higher number $d$ meets this requirement, and thus does not correspond to a Heegner number.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.