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Only nine imaginary quadratic fields are unique factorization domains: $\mathbb Q (\sqrt{-1})$, $\mathbb Q (\sqrt{-2})$, $\mathbb Q (\sqrt{-3})$, $\mathbb Q (\sqrt{-7})$, $\mathbb Q (\sqrt{-11})$, $\mathbb Q (\sqrt{-19})$, $\mathbb Q (\sqrt{-43})$, $\mathbb Q (\sqrt{-67})$, and $\mathbb Q (\sqrt{-163})$. I call these the Heegner domains. Only the first five of these fields are norm-Euclidean. Integers in some of these fields have names or nicknames: the Gaussian, Hippasus, Eisenstein, and Kleinian integers. I wish I knew what to call the $\mathbb Q (\sqrt{-11})$ integers.

Correct me if I'm wrong, but I think I already know how to calculate whether a given rational prime decomposes in each of these fields. One simply finds the rational prime's residue, modulo the field's fundamental discriminant. Its congruence or non-congruence with any residue of a small enough perfect square, modulo the same discriminant, indicates whether the rational prime decomposes in the field.

Yet there could be a problem even with what I think I know. I'm not sure how to handle the case of a rational prime that is ramified in a given field. The online sources that explain such cases typically assume prior knowledge well in advance of my own.

What I mainly wish to ask, though, is how to find the complex factorizations of rational primes over all norm-Euclidean imaginary quadratic fields where they decompose. Does a rational prime's residue, modulo a field's fundamental discriminant, help to determine this? Is there a form of notation for each field, like the $\omega$ notation for Eisenstein integers, that can help to make the complex factorization of rational primes within the field more straightforward? Or is polar form adequate?

Should I also be asking how to find the complex factorization of a rational prime that's ramified in a given imaginary quadratic field? Does the procedure that I seek for finding complex factorizations also apply to the $\mathbb Q (\sqrt{-19})$, $\mathbb Q (\sqrt{-43})$, $\mathbb Q (\sqrt{-67})$, and $\mathbb Q (\sqrt{-163})$ domains, even though they are not norm-Euclidean?

I'm asking because I want to list, for some relatively small rational primes, all the complex factors they have in the Heegner domains, or at least the norm-Euclidean Heegner domains. If this can be done with ordinary spreadsheet formulas, then so much the better!

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  • $\begingroup$ I recommend Lehman's recent book, he does binary quadratic forms and quadratic fields together. Also click the links on the right under "Related." bookstore.ams.org/dol-52 $\endgroup$
    – Will Jagy
    Mar 3, 2020 at 2:16
  • $\begingroup$ Thank you for this recommendation! $\endgroup$ Mar 3, 2020 at 11:33
  • $\begingroup$ Better for self-study, Weissman bookstore.ams.org/mbk-105 although I don't believe he goes that far in quadratic fields. They make a point on the back cover "Requiring only high school algebra and geometry" As an option, if you go through either book, you could still post questions here. $\endgroup$
    – Will Jagy
    Mar 4, 2020 at 16:42
  • $\begingroup$ Thanks again! I will def look into both of these in the next few months. $\endgroup$ Mar 5, 2020 at 10:48

2 Answers 2

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I'm not sure how to handle the case of a rational prime that is ramified in a given field.

That's the easiest case, I think. Notice that, aside from $-1$, these Heegner numbers are all primes. If $p$ is an odd prime that happens to match one of these Heegner numbers, then $p = (\sqrt p)^2$ and $-p = (-1)(\sqrt p)^2$. For example, $-7 = (\sqrt{-7})^2$ and $7 = (-1)(\sqrt{-7})^2$. For the rest of this answer, $p$ refers to a positive odd prime.

For $\mathbb Z[i]$, it suffices to note whether or not $p$ is the sum of two squares in $\mathbb Z$. As that is not the case with $7$, we conclude that it is prime in $\mathbb Z[i]$. In $\mathbb Z[\sqrt{-2}]$, we need to solve $p = a^2 + 2b^2$ in integers. There is no such solution for $7$, so we conclude it is also prime in this ring.

For the rest of these, use the Legendre symbol to see whether $4p = a^2 + db^2$ has solutions, where $d$ is the pertinent Heegner number. If $$\left(\frac{d}{p}\right) = -1,$$ then $p$ is prime in $\mathcal O_{\mathbb Q(\sqrt d)}$, while $1$ means it does split. As for $0$, that's ramification. Thus,

  • Since $28 = 5^2 + 3 \times 1^2$, $$\left(\frac{5}{2} - \frac{\sqrt{-3}}{2}\right) \left(\frac{5}{2} + \frac{\sqrt{-3}}{2}\right) = 7$$
  • There are no solutions in integers to $28 = a^2 + 11b^2$, so $7$ is prime in $\mathcal O_{\mathbb Q(\sqrt{-11})}$
  • Since $28 = 3^2 + 19 \times 1^2$, $$\left(\frac{3}{2} - \frac{\sqrt{-19}}{2}\right) \left(\frac{3}{2} + \frac{\sqrt{-19}}{2}\right) = 7$$
  • There are no solutions in integers to $28 = a^2 + 43b^2$, so $7$ is prime in $\mathcal O_{\mathbb Q(\sqrt{-43})}$. At this point, it is clear that $db^2 > 28$. Hence $7$ is also prime in $\mathcal O_{\mathbb Q(\sqrt{-67})}$ and $\mathcal O_{\mathbb Q(\sqrt{-163})}$.

I'm asking because I want to list, for some relatively small rational primes, all the complex factors they have in the Heegner domains, or at least the norm-Euclidean Heegner domains.

That doesn't really enter into it. The Legendre symbol will mislead you in non-UFDs, but not UFDs.

If this can be done with ordinary spreadsheet formulas, then so much the better!

Maybe if you can use Visual Basic for Applications, or something like that...

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  • $\begingroup$ Thanks, that is abundantly clear! $\endgroup$ Mar 5, 2020 at 11:52
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ummm You want to express $p = x^2 + xy + k y^2$ where $1-4k = \delta$ is your discriminant. This requires Legendre symbol $(\delta|p) = 1,$ unless $\delta = p.$ First, solve $u^2 \equiv \delta \pmod p.$ There are algorithms for this, Cipolla, others. Now, if $u$ is even, replace it by $p - u,$ arriving at an odd $w$ with $$ w^2 \equiv \delta \pmod {4p}. $$ That is, $$ w^2 = \delta + 4 p t $$ for some integer $t,$ thus $w^2 - 4pt = \delta.$ Using notation $\langle a,b,c \rangle$ for binary quadratic form $f(x,y) = a x^2 + b x y + c y^2,$ we have constructed form $$ \langle p, w, t \rangle $$ with discriminant $\delta.$ Use Gauss reduction to find the particular equivalence matrix that takes this to $\langle 1,1,k \rangle.$ Meaning an integer matrix of determinant 1 where $P^T G P = H,$ with $G$ the Hessian of $ \langle p, w, t \rangle $ and $H$ the Hessian of $\langle 1,1,k \rangle.$ Then, taking $Q = P^{-1}$ leads to $Q^THQ = G$ provides the requested expression of $p$

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  • $\begingroup$ Thank you for your willingness to guide me on this topic. The last course I took in math was in high school, circa 1977, so I hope you'll be patient with me as I work to understand your guidance. For starters, the vertical bar or pipe notation in (𝛿|𝑝)=1 is unfamiliar to me. Does this refer to divisibility? $\endgroup$ Mar 3, 2020 at 11:46

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