# Special cases of the Stark-Heegner theorem with simple proofs

The Stark-Heegner theorem states that the ring of integers of the quadratic number field $\mathbb Q(\sqrt{m})$, where $m$ is a squarefree negative integer, is a principal ideal domain, iff $$m\in\{-1,-2,-3,-7,-11,-19,-43,-67,-163\}$$ I am looking for special cases of this theorem (including references where I can find them), which can be proven with elementary number theory or not-too-complicated algebraic number theory.

This is what I have so far:

• $\mathbb Q(\sqrt{m})$ is a PID for $m\in\{-1,-2,-3,-7,-11\}$ (It is easy to prove, that this rings are norm-Euclidean and that Euclidean rings are PIDs)
• $\mathbb Q(\sqrt{m})$ is a PID for $m\in\{-19,-43,-67,-163\}$ (Either prove, hat the norm is a Dedekind–Hasse norm for these rings, or that, if Euler's prime generating polynomial $x^2+x+q$ generates primes for $x=0,\dots q-1$, then $\mathbb Q(\sqrt{1-4q})$ is a PID)
• If $m\not\equiv 1 \bmod 4$ and $m<-2$, then $\mathbb Q(\sqrt{m})$ is not a PID (Again, there are several proofs, one of them is this one)

Does anyone know any other simple result in the above spirit, particulary for proving that some rings are not PIDs? Maybe something like a proof, that there are only finitely many PIDs, maybe even an upper bound for $m$, or the fact that $m\equiv 1\bmod{8}$ does not work, or maybe some other restriction on $m$?

Thanks!

EDIT: All Heegner numbers are prime. Maybe, there is some argument, why it won't work for composite numbers?

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I think you are aware of it, but I'm not sure how to interpret your last sentence, so to be sure: the case $m \equiv 1 \pmod 8$ is settled by the fact that $2$ is not a prime then, and for $m < -7$, it is irreducible. – Daniel Fischer Jun 28 '13 at 17:38
Well, this was actually meant as a question, as I saw this in an other more difficult context. I didn't think about this easy solution. Thank you! – Tomas Jun 28 '13 at 18:17
Happy to help. Don't forget to ping me when you find something simple for the $m \equiv 5 \pmod 8$ case. – Daniel Fischer Jun 28 '13 at 18:20
Haha, well that would be nice. – Tomas Jun 28 '13 at 18:24
Well, I should have written any $m \equiv 5 \pmod 8$ case. Some may be simple. – Daniel Fischer Jun 28 '13 at 18:26

There are tons of easy cases. If $m$ is even, or if $m \equiv -1 \bmod 4$, then $2$ ramifies in $\mathbb{Q}(\sqrt{m})$. But $2$ is not of the form $a^2+|m| b^2$, except for $m=-1$ and $m=-2$. So we can limit ourselves to $m \equiv 1 \bmod 4$. As pointed out above, we can also assume $m \not \equiv 1 \bmod 8$. So we are down to $m \equiv -5 \bmod 8$. This means that the ring of integers in $\mathbb{Q}(\sqrt{m})$ is $a+b(1+\sqrt{m})/2$, with norm function $a^2 + ab + \frac{|m|+1}{4} b^2$. (Of course, $m$ is negative, but I think a lot of these formulas are more readable with $|m|$ instead of $-m$ because it makes the sign immediately visually clear.)

In particular, if $b \neq 0$, the norm $N(a+b(1+\sqrt{m})/2) \geq \frac{|m|+1}{4}$.

First of all, this explains why $m$ must be prime. If $m$ is an odd, square-free composite, then there is a prime $p$ dividing $m$ with $p \leq |m|/5$. This prime $p$ ramifies, so there is a prime ideal $\pi$ with $N(\pi) = p$. Since $p \leq m/5 < \frac{|m|+1}{4}$, the ideal $\pi$ can't be principal.

Also, suppose that $p$ is an odd prime with $\left( \frac{m}{p} \right) =1$ and $p < (|m|+1)/4$. Then, $p$ splits into $\pi \bar{\pi}$ and the same argument as the above paragraph shows that $\pi$ can't be principal.

Summary Any counterexample must have $|m|$ a prime, $m \equiv 5 \bmod 8$, and must have $\left( \frac{m}{p} \right) = -1$ for $p < \frac{|m|+1}{4}$.

I can't find any $m$ more negative than $-163$ which passes this test, searching through the first $50,000$ primes. Nothing even comes close. Define $r(m)$ to be the least odd prime $p$ for which $\left( \frac{m}{p} \right) = 1$. A PID would have $r(m)/|m| > 0.25$; the largest value I can find after $163$ is $r(-193)/193 = 11/193 \approx 0.057$. In case you want to do some computations yourself, here is a bit of Mathematica code:

FirstSplit[m_] :=
(i = 2; While[KroneckerSymbol[m, Prime[i]] == -1, i++]; Prime[i])

SplitRatio[m_] := FirstSplit[-m]/m

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Quite impressive, thank you! – Tomas Jun 28 '13 at 20:24
Since it seems to be possible to find an arbitrary big lower bound for $m$ with your computational method (as long as there are no counterexamples to $r(m)/|m|>0.25$), any upper bound would complete the proof of the theorem, so this suggests, that there is no simple proof for an upper bound for $m$. – Tomas Jun 28 '13 at 20:38
The problem of the first quadratic residue is famously hard but does have some good bounds; see terrytao.wordpress.com/2009/08/18/… . Regarding the first prime residue, all I could find quickly was this unanswered question: math.stackexchange.com/questions/421188/… – David Speyer Jun 28 '13 at 20:43
Here is a bound, but it's derived from Heegner's theorem. – Tomas Jun 28 '13 at 21:19
More precise: The paper shows, that for $p>163$, the least prime quadratic residue modulo $p$ is bounded by $\sqrt{p}$. The result is derived from Heegner's theorem, but nevertheless it shows, that there is no counterexample to $r(m)/|m|>0.25$ for $|m|>163$, am I right? – Tomas Jun 30 '13 at 19:54