List of prime numbers in imaginary quadratic fields with UFD

Question

I am interested in a list of natural prime numbers in the ring of integers of imaginary quadratic fields with UFD e.g. for $\mathbb{Q}[\sqrt{-7}]$ or $\mathbb{Q}[\sqrt{-11}]$. Especially, I want to know the smallest number that is prime in all of these integer rings that is also prime in $\mathbb{Z}$. The possible values for the discriminant $d$ of $\mathbb{Q}[\sqrt{d}]$ to fulfill these requirements are the Heeger numbers $$-1,-2,-3,-7,-11,-19,-43,-67,-163.$$ So far, I only managed to find construction tools for the cases $$d = -1: p \textrm{ is prime and } p \equiv 3 \textrm{ mod } 4 \, ,$$ $$d = -2: p \textrm{ is prime and } p \equiv \{0,2,4,5,6,7\} \textrm{ mod } 8 \, , $$ $$d = -3: p \textrm{ is prime and } p \equiv 2 \textrm{ mod } 3 \, .$$ I browsed the internet for many hours but could not find restrictions for integer primes for the case $d = -7$ (Kleinian integers). I read that there are no prime elements in the integer ring of Hurwitz quaternions so does this transfer to the Kleinian integers, too ?

Thanks in advance, Levigeddon.

Answer 1

If I'm understanding correctly, your question is whether it's possible for $p$ to be a positive prime number in $\mathbb Z$ and also prime in each of $\mathbb Z[i]$, $\mathbb Z[\sqrt{-2}]$, $\mathbb Z[\omega]$, $\mathcal O_{\mathbb Q(\sqrt{-7})}$, $\mathcal O_{\mathbb Q(\sqrt{-11})}$, $\mathcal O_{\mathbb Q(\sqrt{-19})}$, $\mathcal O_{\mathbb Q(\sqrt{-43})}$, $\mathcal O_{\mathbb Q(\sqrt{-67})}$ and $\mathcal O_{\mathbb Q(\sqrt{-163})}$? And if so, what is the smallest such $p$?

In a comment yesterday, I wrote I didn't think such a prime exists. I had failed to notice your earlier comment about 3167. This suggests that these primes do exist but are spaced far apart.

Obviously $p$ must not be one of 2, 3, 7, 11, 19, 43, 67 or 163, which still leaves an infinitude of primes.

But the criterion for $\mathbb Z[i]$ gives us a way to discard "half" the primes: we're looking for $p \equiv 3 \pmod 4$, though obviously $p \neq 3$ itself.

I'm not sure why you have 0, 2, 4, 6 in the criterion for $\mathbb Z[\sqrt{-2}]$, there should be no even values in that one. This leaves $p \equiv 5, 7 \pmod 8$, but we can cross off 5 since $5 \pmod 8 \equiv 1 \pmod 4$, further narrowing it down to $7 \pmod 8$.

In order to combine this with the criterion for $\mathbb Z[\omega]$, we need to broaden our modulus to 24. Hence $p \equiv 7, 15, 23 \pmod{24}$. But clearly $15 \pmod{24}$ can't be prime. And $7 \pmod{24}$ can't be prime in $\mathbb Z[\omega]$ since it's equivalent to $1 \pmod 3$. That leaves us $p \equiv 23 \pmod{24}$.

Next we move to the Kleinian integers — hmm, I don't think I had ever come across that term before seeing your question. We need to broaden our modulus out again, to 168 this time, so we have $p \equiv 23, 47, 71, 95, 119, 143, 167 \pmod{168}$.

It shouldn't bother us too much at this point that $95 = 5 \times 19$, since $\gcd(95, 168) = 1$. Likewise with $143 = 11 \times 13$. It is far more important that $119 = 7 \times 17$, so for that reason we discard 119. So our possibilities are $p \equiv 23, 47, 71, 95, 143, 167 \pmod{168}$.

However, $$\left(\frac{-7}{23}\right) = (-7)^{11} \pmod{23} = 1$$ and indeed $(4 - \sqrt{-7})(4 + \sqrt{-7}) = 23$. $p = 47$ is still in the running since $$\left(\frac{-7}{47}\right) = (-7)^{23} \pmod{47} = -1.$$ But 71 is out, with $(8 - \sqrt{-7})(8 + \sqrt{-7}) = 71$.

For the sake of the Legendre symbol, let's substitute 95 with 263. We quickly find that $(16 - \sqrt{-7})(16 + \sqrt{-7}) = 263$. Replacing 143 with 311, we find it's still in the running. So is 167 (as a residue class, not as a number itself).

So our list of possibilities is now $p \equiv 47, 143, 167 \pmod{168}$, and we have to broaden our modulus out once again to 1848. Taking a shortcut is looking very good right about now. But let's soldier on with 1848.

The possibilites explode, so instead of going through all of them one by one, I'm gonna try to take more of these on with Mathematica's help (you can use Wolfram Alpha, to some extent, if you don't have Mathematica on your computer). So JacobiSymbol[-11, 47 + 168Range[0, 10]] gives us the sequence $1, -1, 1, 1, 1, -1, -1, -1, 1, -1, 0$.

But that second $-1$ corresponds to 215, so we substitute 2063 and see it's still in the running...

That's where I left it last night. This morning, I decided it would make more sense to just do a brute force search in Mathematica and see if anything comes up.

And then there was your gentle reminder about 3167, which checks out with the Legendre symbol in Mathematica (directly, or indirectly through Wolfram Alpha) as well as on the command line with a little library I got off GitHub. So I ran the the brute force search in Mathematica among the first two thousand primes and it gave me 3167, 8543, 14423, which are all $143 \pmod{168}$.

But there are plenty of other primes congruent to $143 \pmod{168}$, why don't they also stay inert in the nine quadratic imaginary UFDs? 311, for example. Indeed,

• 311 is prime in $\mathbb Z[i]$
• 311 is prime in $\mathbb Z[\sqrt{-2}]$
• 311 is prime in $\mathbb Z[\omega]$
• 311 is prime in $\mathcal O_{\mathbb Q(\sqrt{-7})}$, but...
• $311 = (6 - 5 \sqrt{-11})(6 + 5 \sqrt{-11})$, and
• $$311 = \left(\frac{35}{2} - \frac{\sqrt{-19}}{2}\right) \left(\frac{35}{2} + \frac{\sqrt{-19}}{2}\right)$$
• $$311 = \left(\frac{13}{2} - \frac{5 \sqrt{-43}}{2}\right) \left(\frac{13}{2} + \frac{5 \sqrt{-43}}{2}\right).$$ By the way...
• 311 is prime in $\mathcal O_{\mathbb Q(\sqrt{-67})}$
• 311 is prime in $\mathcal O_{\mathbb Q(\sqrt{-163})}$

Maybe if we had broadened our modulus all the way out to 16488700536, we might find something along the lines of that only primes congruent to 3167 and maybe a few other values modulo 16488700536 can also be prime in all nine quadratic imaginary UFDs.

---

P.S. If you're interested, 15073 is composite in all these rings.

• $15073 = (113 - 48i)(113 + 48i)$
• $15073 = (31 - 84 \sqrt{-2})(31 + 84 \sqrt{-2})$
• $15073 = (121 - 12 \sqrt{-3})(121 + 12 \sqrt{-3})$
• $15073 = (39 - 44 \sqrt{-7})(39 + 44 \sqrt{-7})$
• $$15073 = \left(\frac{89}{2} - \frac{69 \sqrt{-11}}{2}\right) \left(\frac{89}{2} + \frac{69 \sqrt{-11}}{2}\right)$$
• $$15073 = \left(\frac{199}{2} - \frac{33 \sqrt{-19}}{2}\right) \left(\frac{199}{2} + \frac{33 \sqrt{-19}}{2}\right)$$
• $15073 = 111 - 8 \sqrt{-43}, 111 + 8 \sqrt{-43}$
• $$15073 = \left(\frac{107}{2} - \frac{27 \sqrt{-67}}{2}\right) \left(\frac{107}{2} + \frac{27 \sqrt{-67}}{2}\right)$$
• $$15073 = \left(\frac{217}{2} - \frac{9 \sqrt{-163}}{2}\right) \left(\frac{217}{2} + \frac{9 \sqrt{-163}}{2}\right)$$

Answer 2

For the most part, the problem is already solved by Robert Soupe. Here I make some additional observations. To sum up, we should have expected just under 0.2% of primes to be irreducible over all nine imaginary quadratic UFDs (IQUFDs) -- and that is what we get.

• Once the condition $p\equiv-1\bmod 24$ is established, it can be combined with any nonquadratic residue $\bmod 7$, then any nonquadratic residue $\bmod 11$, etc up to $\bmod 163$. Every such combination produces a unique combination $\bmod(8×3×7×...×163)=\bmod16488700536$.

• Numerically this corresponds to three nonquadratic residues $\bmod7$, times five residues$\bmod 11$, times nine residues $\bmod19$, etc up to 81 residues$\bmod163$, for a total of $7577955$ residues $\bmod16488700536$. Thus more than "a few" residue classes are allowed when we combine all the relevant moduli. By Dirichlet's Theorem all such residues allow infinitely many primes, so $3167$ is just the smallest of infinitely many solutions. If we number the residue classes from $0$ to $16488700535$, then the $3167$ also turns out to be the smallest candidate residue.

• The Euler totient function of $16488700536$ is $3879912960=7577955×512$. Thus $1/512=1/2^9$ of all natural primes are expected to be irreducible over all nine IQUFDs. This goes along with each domain allowing half the natural primes to be irreducible; the halves corresponding to the nine IQUFDs are statistically uncorrelated with each other.

• We would have thus expected the first successful prime to lie somewhere around the 512th overall prime. The latter is $3671$ compared with the actual answer $3167$, so the statistical prediction is "in the ballpark".

• Looking out to the first billion numbers, $50847534$ are prime. Dividing this by $512$ gives (to the nearest whole number) $99312$ versus the actual count (revealed in the comments) of $99192$ irreducible primes over all nine IQUFDs. The occurrence of such primes is therefore matching well with what we should have been expecting all along.

Answer 3

OEIS has a list of rational primes that are inert in the intersection of Heegner UFDs: https://oeis.org/A309024