complex factorization of rational primes over the norm-Euclidean imaginary quadratic fields

Question

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!

Answer 1

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...

Answer 2

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 $p$