Prime density race : ring $A = \Bbb Z[\frac{1+\sqrt {-7}}{2}]$ vs ring $B = \Bbb Z[\frac{1+\sqrt {-11}}{2}]$?

Question

Consider the ring $A = \Bbb Z[\frac{1+\sqrt {-7}}{2}]$ with the elements $\frac{a+b\sqrt {-7}}{2}$ where $a,b$ are both even or both odd integers.

This is also known as the Kleinian integers; the ring of integers in the imaginary quadratic field $\Bbb Q[\sqrt {-7}]$.

Consider also the ring $B = \Bbb Z[\frac{1+\sqrt {-11}}{2}]$ with the elements $\frac{c+d\sqrt {-11}}{2}$ where $c,d$ are both even or both odd integers.

This is also known as the ring of integers in the imaginary quadratic field $\Bbb Q[\sqrt {-11}]$.

Both $A$ and $B$ are UFD.

Consider $f_7(r)$ as the number of primes in $A$,such that $1 < a^2 + b^2 < r$ and likewise

Consider $f_{11}(r)$ as the number of primes in $B$,such that $1 < c^2 + d^2 < r$.

How to show that

$$ g(r) = f_7(r) - f_{11}(r)$$

1. changes sign infinitely often

2. is unbounded for both positive and negative output

3. is equal to zero infinitely often

I assume a proof exists by redefining the prime norms $a^2 + 7 b^2$ and $c^2 + 11 d^2$ in some list(s) $v$ :

$$p = {v} \mod w(*)$$

for some $w$.

Comparable by primes $1 \mod 4$ vs $3 \mod 4$ in a prime race. So related to dirichlet's theorem. And by using analogues of https://en.wikipedia.org/wiki/Siegel%E2%80%93Walfisz_theorem and https://en.wikipedia.org/wiki/Chebyshev%27s_bias we might get a proof.

I prefer a proof that does not use Chebotarev's density theorem, admittedly because I do not understand it so well.

• I would like alot a proof that does not use the above strategy.

More than one proof would be nice and insightful.

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some motivation and background

Notice I did not pick the Eisenstein or Gaussian integers because they have more than two units, whereas these $A$ and $B$ have only $-1,1$. I also picked two rings with the same class number (1 = UFD), since the asymptotic density of irreducibles is a function of the class number. Also I picked 2 primes of type $3 \mod 4$. This assures us algebraic integers :

The integers extended with a solution $x$ :

$$x^2 + b x + c = 0$$

for nonzero integers $b,c$

and since $b^2 - 4c < 0$ is always of the form $-3 \mod 4$ the list of candidates follows.

$\Bbb Q[\sqrt {-3}]$ gave us the Eisenstein integers but they have $6$ units as mentioned above.

So the next primes of type $3 \mod 4$ where $7$ and $11$.

I started with the smallest instances because it probably makes everything easier. And I assume understanding the small cases is sufficient for understanding the larger ones.

Also class number $1$ is probably best and easiest to start with. Analogue ideas for two quadratic rings with the same class number $w>1$ are ofcourse possible.

(quite mysterious to me is that the first class number $>1$ for similar cases is for $\Bbb Q[\sqrt {-23}]$ and is actually $3$ and not $2$ !! Although that might be pure coincidence or my lack of of understanding)

How to prove this ?

And as a second question, how often do they change ? Or is that superhard to answer ?

Plots are welcome too !

Answer 1

(By "prime" I'll always mean "pair of primes related by $\pm$", for simplicity of writing.)

Primes in $A$ (other than $\sqrt{-7}$ itself) come in two types:

• If $p\equiv1,2,4\pmod7$ is a rational prime, then $-7$ is a quadratic residue modulo $p$, and so there are two primes in $A$ of norm $p$. (Indeed they are $a\pm b\frac{1+\sqrt{-7}}2$ where $a^2+ab+2b^2=p$.)
• If $p\equiv3,5,6\pmod7$ is a rational prime, then $-7$ is a quadratic nonresidue modulo $p$, and so there is a single prime in $A$ of norm $p$ (namely $p+0\frac{1+\sqrt{-7}}2$ itself).

In other words, the number of primes in $A$ of norm not exceeding $x$ is twice the number of $1,2,4\pmod7$ rational primes up to $x$, plus the number of $3,5,6\pmod7$ primes whose square doesn't exceed $x$, plus $1$ for $\sqrt{-7}$ (assuming $x\ge7$). Let's write this as $$ \pi_A(x) = 2\pi(x;7,\{1,2,4\}) + \pi(\sqrt x;7,\{3,5,6\}) + 1 = 2\pi(x;7,\{1,2,4\}) + \biggl( \frac12 + o(1) \biggr) \frac{\sqrt x}{\log\sqrt x}, $$ where $\pi(x;q,S)$ counts the number of primes up to $x$ that are congruent to some element of $S$ modulo $q$.

By a similar argument, we can see that when $x\ge11$, $$ \pi_B(x) = 2\pi(x;11,\{1,3,4,5,9\}) + \biggl( \frac12 + o(1) \biggr) \frac{\sqrt x}{\log\sqrt x}. $$ Therefore $\pi_A(x) - \pi_B(x)$ has to do with counting primes in arithmetic progressions modulo both $7$ and $11$, which we can recast as counting primes in arithmetic progressions modulo $77$: \begin{align*} \pi_A(x) - \pi_B(x) &= 2 \bigl( \pi(x;7,\{1,2,4\}) - \pi(x;11,\{1,3,4,5,9\}) \bigr) + o\biggl( \frac{\sqrt x}{\log x} \biggr) \\ &= 2 \pi(x;77,\{2, 8, 11, 18, 22, 29, 30, 32, 39, 43, 44, 46, 50, 51, 57, 65, 72, 74\} \bigr) \\ &- 2 \pi(x;77, \{ 3, 5, 12, 14, 20, 26, 27, 31, 34, 38, 42, 45, 47, 48, 49, 56, 59, 69, 70, 75 \}) \\ &+ o\biggl( \frac{\sqrt x}{\log x} \biggr). \end{align*} (This is after cancelling $2\pi(x;77,\{1, 4, 9, 15, 16, 23, 25, 36, 37, 53, 58, 60, 64, 67, 71\})$ from both main terms.)

All this specificity is leading to this assertion: the race between $\pi_A(x)$ and $\pi_B(x)$ is directly analogous to other prime number races like the one between $\pi(x;4,3)$ and $\pi(x;4,1)$. In this case, there is no bias (the cancelled residue classes were precise the quadratic residues modulo $77$, so they're not present here to cause a bias), and so the conjecture would be that $\pi_A(x)>\pi_B(x)$ and $\pi_A(x)<\pi_B(x)$ both occur asymptotically 50% of the time in the appropriate sense (limiting logarithmic density). In particular, the conjecture is that $\pi_A(x)-\pi_B(x)$ changes sign infinitely often.