Relative density of primes under extension 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?"
Fact: 


*

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

*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. 
 A: 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.)
A: 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. 
