Let $\mathbb{P}_{\mathbb{C}}$ be the set of Gaussian primes and $\mathbb{P}_{\mathbb{N}}$ the set of primes in $\mathbb{N}$.

Let $\pi_{\mathbf{C}}(\sqrt{n})$ be the number of Gaussian primes with norm $\leq \sqrt{n}$ and $\pi_{\mathbf{N}}(n)$ be, as usual, the number of primes $\leq n$ in $\mathbb{N}$. Recall that norm($x+iy$)=N($x+iy$)=$x^2 +y^2$; hence, my taking of a square root above.

I am interested to know what the order of magnitude is for $$\frac{\pi_{\mathbf{N}}(n)}{\pi_{\mathbf{C}}(\sqrt{n})}$$i.e. Has the extension of the definition of primes increased/decreased the relative density of primes with respect to their set of definition? A rather quixotic question could be " Is there a general asymptotic for the number of primes in an arbitrary infinite field with the definition of being a prime as usual?"


  1. Prime numbers of the form $4n + 3$ are also Gaussian primes.

  2. Gauss's circle problem which asks for the number of Gaussian integers with norm less than a given value is presently unresolved. I think this is tangentially related to the asymptotic I am looking for.

  • $\begingroup$ The asymptotic solution to the circle problem is just $\pi r^2$. The problem itself is about the exact value. $\endgroup$ – Qiaochu Yuan Mar 15 '11 at 19:24
  • $\begingroup$ In dillema of which answer to accept. How can I distribute the incentive I have in hand? To Akhil for exposition or to Qiaochu for mentioning the last theorem. $\endgroup$ – Chulumba Mar 15 '11 at 19:25
  • $\begingroup$ @Qiaochu, that is true and that was why I said it only relates to my case tangentially. $\endgroup$ – Chulumba Mar 15 '11 at 19:34
  • $\begingroup$ I don't understand why you say "recall that norm($x+iy$)=$x^2+y^2$"... this is not the usual definition, and as indicated by the answers, leads to some confusion. The norm, or absolute value, of a complex number $x+iy$ is $\sqrt{x^2+y^2}$. $\endgroup$ – mjqxxxx Mar 15 '11 at 20:41
  • $\begingroup$ @mjqxxxx, that is a standard definition for norm of Gaussian integers as also stated in wikipedia. $\endgroup$ – Chulumba Mar 15 '11 at 21:03

I'm going to answer a different question, when $\sqrt{n}$ is replaced by $n$. The answer is that it goes to one!

We can see this by noting that rational primes are evenly split (in terms of asymptotic density) between $4n+1$ and $4n+3$; this is a slight strengthening of the usual Dirichlet density theorem. Primes of the former type split into two primes in the Gaussian integers, while primes of the latter remain prime. The norms of the former are the same as the prime $4n+1$; the norms of the latter are $(4n+3)^2$.

When you take asymptotics by considering Gaussian primes whose norm is less than $N$ for $N$ large, only the former case matters ($(4n+3)^2 \gg 4n+3$ for $n$ large), and since there are roughly $N/(2 \log N)$ rational primes of the form $4n+1$ in this range, each of which splits into two distinct Gaussian primes, we get that the number of Gaussian primes of norm less than $N$ is roughly $N/\log N$.

In fact, though, this is true for general number fields. This is an extension of the usual prime number theorem, and follows by essentially similar arguments: the point is that one can define the Dedekind zeta function which has the analogous properties of the Riemann-zeta function, and following the same proof of the prime number theorem, one can show that in a number field, the number of prime ideals of norm at most $N$ is asymptotically $N/\log N$. (Note that the primes that completely split over $\mathbb{Q}$ are the only ones that count asymptotically in the zeta-function; this is the generalization of the statement I made about the primes $4n+3$ not contributing much.)

  • $\begingroup$ Dear Akhil, Regarding your last paragraph: you could add that (as you noted in the Gaussian case) the bulk of the contribution comes from prime ideals that are totally split over $\mathbb Q$. Regards, $\endgroup$ – Matt E Mar 15 '11 at 19:22
  • $\begingroup$ @Akhil: thanks Akhil. Your reformulation is agreeable. $\endgroup$ – Chulumba Mar 15 '11 at 19:24
  • $\begingroup$ Landau prime ideal theorem is really the answer here. $\endgroup$ – Chulumba Mar 15 '11 at 19:45
  • $\begingroup$ @Matt: Dear Matt, thanks for suggesting that; I've added it. $\endgroup$ – Akhil Mathew Mar 15 '11 at 20:46
  • 1
    $\begingroup$ @Akhil: Dear Akhil, You're welcome. Note that this is an important fact in applications. (See e.g. Serre's article on CM in Cassels--Frolich for one application: in verifying the main theorems on CM, he gets to restrict attention to split primes. Or, in the theory of automorphic forms on unitary groups, one can typically restrict attention to Hecke operators at primes that split in the quadratic imaginary field giving rise to the unitary group, where everything is much simpler (because locally at these primes the unitary group is just $GL_n$).) Regards, $\endgroup$ – Matt E Mar 15 '11 at 23:26

I don't think you're asking the question you wanted to ask. First, if you're using $\sqrt{n}$ then it seems more natural to define the norm as the square root of what you've defined it. Second, either way the contribution from primes congruent to $3 \bmod 4$ is negligible, so you're basically only counting the contributions from the primes congruent to $1 \bmod 4$.

What you should be asking for is the relative density of primes congruent to $1 \bmod 4$ and primes congruent to $3 \bmod 4$, and it is known that both of these are $\frac{1}{2}$ by Dirichlet's theorem.

A very general statement which might answer your follow-up question (in what sense is this quixotic? This is an extremely natural question to ask) is the Chebotarev density theorem.

  • $\begingroup$ @Qiaochu: Thanks Qiaochu. I feel you missed my first point as I understand from your first paragraph. But that is not a problem as it is very generally answered in the theorem mentioned. $\endgroup$ – Chulumba Mar 15 '11 at 19:22
  • $\begingroup$ @Chulumba: what is your first point? As written the limit is $\infty$ and this does not tell you very much. $\endgroup$ – Qiaochu Yuan Mar 15 '11 at 19:24
  • $\begingroup$ @Qiaochu, I think my first point was to not only to count the density of primes $\equiv 3(mod4)$ but also to compare this with the density of Gaussian primes. Sorry for confusing you. $\endgroup$ – Chulumba Mar 15 '11 at 19:31
  • $\begingroup$ @Qiaochu, you have indicated me what I have to learn and the rest shall not worry you. I may come back if I find difficulties. Thanks. $\endgroup$ – Chulumba Mar 15 '11 at 19:32
  • $\begingroup$ @Chulumba: I think you're missing my point. $x + iy$ is a Gaussian prime with $N(x + iy) = x^2 + y^2$ a rational prime if and only if $x^2 + y^2 = 2$ or $x^2 + y^2 \equiv 1 \bmod 4$. If you're only counting by the square root of the norm then there are less than $\sqrt{n}$ of these with norm at most $\sqrt{n}$, and there are less than $\sqrt[4]{n}$ primes $p$ with $p \equiv 3 \bmod 4$. The norm is not really the right way to compare these to regular primes; you get strictly more information out of knowing how common each of them are separately. $\endgroup$ – Qiaochu Yuan Mar 15 '11 at 19:36

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