Let $\mathcal{O}_K$ be the ring of integers of a field $K$. I have learnt about the Baker-Heegner-Stark theorem, which implies that if $K=\mathbb{Q}(\sqrt{-n})$ with $n\in\mathbb{Z}^+$ square-free, then $\mathcal{O}_K$ is a UFD if and only if $n$ is a Heegner number. However, I am interested in the possibility of obtaining an easier result: is it possible to prove that all Heegner numbers greater than $1$ are primes, with undergraduate mathematics?
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Check out this answer by Franz Lemmermeyer. May be that will give an idea? I am also trying to figure out a way to modify my answer from that thread to this case. – Jyrki Lahtonen Sep 09 '16 at 17:39
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@Jyrki: I'll think of it, thank you. I accepted Ted's answer but I'm looking forward to a modification of your answer, if your attempt succeeds. :D – Vincenzo Oliva Sep 11 '16 at 18:19
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If $n = rs$ with integers $r,s>1$, then you have 2 different factorizations for $n$: $$n = -1 \sqrt{-n} \sqrt{-n} = rs.$$
We claim that $\sqrt{-n}$ is irreducible in $\mathcal{O}_{\mathbb{Q}(\sqrt{-n})}$, which will complete the argument. If $\frac{a + b \sqrt{-n}}{2} \in \mathcal{O}_{\mathbb{Q}(\sqrt{-n})}$ is a divisor of $\sqrt{-n}$, where $a, b \in \mathbb{Z}$ and where we may assume $a \ge 0$, then the following must be an element of $\mathcal{O}_{\mathbb{Q}(\sqrt{-n})}$: $$2\frac{\sqrt{-n}}{a + b \sqrt{-n}} = 2\frac{\sqrt{-n}(a - b \sqrt{-n})}{a^2 + n b^2} = 2\frac{bn + a \sqrt{-n}}{a^2 + n b^2}.$$
Let $y := \frac{2a}{a^2 + nb^2}$ be the coefficient of $\sqrt{-n}$ of the last term above. Clearly $0 \le y < 2$ (recall we assumed $a \ge 0$). If $y=0$ then $a=0, b = \pm 1$ so the divisor we have found is trivial. Otherwise $y = k/2$ for one of $k = 1,2,3$. We may write: $$\frac{4}{k} = \frac{2}{y} = a + n \frac{b^2}{a}.$$ Both summands on the right side are positive, and $a$ is an integer. So if $k \ge 2$ then the left side is $\le 2$, we conclude that $a=1$ so the right side is $1 + n b^2 > 6$ (since $n$ is square-free composite) which is a contradiction. If $k=1$, we have to step through a few more cases but we still get a contradiction.
Exercise: For real quadratic fields, the corresponding statement is false, e.g., the ring of integers of $\mathbb{Q}(\sqrt{6})$ is a unique factorization domain. Where does this argument go wrong for real fields?
Ted
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$n=24$ has two different factorizations in $\mathbb Z$: $3\cdot 8,4\cdot 6$. Yet $\mathbb Z$ is a UFD. Your argument doesn't establish anything unless you show two different factorizations into irreducibles. – Wojowu Sep 05 '16 at 17:46
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@Wojowu: Thanks, I hate when I brainfart precisely because I think I have... brainfarted. – Vincenzo Oliva Sep 05 '16 at 19:31
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@Wojowu: I admit this argument was a bit terse, but it's basically the right idea. The only step missing is to show that $\sqrt{-n}$ is irreducible. See edits for details. – Ted Sep 09 '16 at 16:37
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Great, thank you, Ted! Is the problem that $\sqrt{n}\sqrt{n}=\sqrt{r^2}\sqrt{s^2}=rs$ while $-\sqrt{-n}\sqrt{-n}\ne-\sqrt{(-n)^2}$ ? – Vincenzo Oliva Sep 09 '16 at 17:46
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Not sure what you mean; the factorization $n = \sqrt{n}\sqrt{n} = rs$ always works. But when $n>0$, $\sqrt{n}$ may no longer be irreducible; e.g., $\sqrt{6} = (3 + \sqrt{6})(-2 + \sqrt{6})$. The factorizations $\sqrt{6} \sqrt{6}$ and $2 \cdot 3$ have a common refinement up to units of $(-2 + \sqrt{6})(-2 + \sqrt{6})(3 + \sqrt{6})(3 + \sqrt{6})$ (and remember, real quadratic fields have lots of units). – Ted Sep 09 '16 at 23:46
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This is not quite enough. It could be that the two factorizations can each be refined into the same factorization.
Rene Schoof
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Did you mean to answer the question or comment on the other answer? – Vincenzo Oliva Sep 05 '16 at 19:32