TITLE: What is the probability that a rational prime remains prime in $\mathbb Z[i,\sqrt{-3}]$?
QUESTION [11 upvotes]: Using Chebotarev's density theorem, asymptotically, what is the probability that a rational prime remains prime in $\mathbb Z[i, \sqrt{-3}]$?

REPLY [6 votes]: The simplest answer, which is already implicit in several of the other answers is as follows: Let $p \geq 5$ be prime. Then at least one of $-1$, $-3$ and $3$ are square modulo $p$. 
If $a$ is a square root of $-1$ modulo $p$ then I claim that $\langle p \rangle = \langle p, a - i \rangle \langle p, a + i \rangle$. To see this, note that $\langle p, a - i \rangle \langle p, a + i \rangle = \langle p^2, p(a - i), p(a+i), a^2 + 1 \langle$. Each term on the right is clearly divisible by $p$, so $\langle p \rangle \supseteq \langle p, a - i \rangle \langle p, a + i \rangle$. Conversely, $p(a + i) + p(a - i) = 2 ap$ and $\gcd(2a, p) = 1$, so $p$ is in the ideal generated by $2ap$ and $p^2$. Similarly, if $b$ is a square root of $-3$, then $\langle p \rangle = \langle p, b + \sqrt{-3} \rangle \langle p, b - \sqrt{-3} \rangle$ and, if $c$ is a square root of $3$, then $\langle p \rangle = \langle p, c + \sqrt{3} \rangle \langle p, c - \sqrt{3} \rangle$. 

Having given the simple answer, let me address two issues raised in the comments and other answers: Is $\mathbb{Z}[i, \sqrt{-3}] = \mathbb{Z}[\zeta_{12}]$? Should we care? And where does Chebotaryov come in?
In fact, we have $\mathbb{Z}[i, \sqrt{-3}] \subsetneq \mathbb{Z}[\zeta_{12}]$. To prove the containment, note that $i = {\zeta_{12}}^3$ and $\sqrt{-3} = 2 {\zeta_{12}}^2 -1$. However, we do not have equality. We have 
$$\mathbb{Z}[i, \sqrt{-3}] = \{ a+bi+c\sqrt{-3} + d \sqrt{3} : a,b,c,d \in \mathbb{Z} \}$$ and each element of $\mathbb{Z}[i, \sqrt{-3}]$ has a unique expression of this form. But $$\zeta_{12} = \tfrac{1}{2} i + \tfrac{1}{2} \sqrt{3}$$ so it is not in $\mathbb{Z}[i, \sqrt{-3}]$. In fact, working a little harder, we have $$\mathbb{Z}[\zeta_{12}] = \left\{ \tfrac{1}{2} (a+bi+c\sqrt{-3} + d \sqrt{3}) : a,b,c,d \in \mathbb{Z},\ a \equiv c \bmod 2,\ b \equiv d \bmod 2 \right\}.$$
So $\mathbb{Z}[i, \sqrt{-3}]$ is an order in $\mathbb{Z}[\zeta_{12}]$ of index $4$.
Should we care? For some purposes, this is important. The ring $\mathbb{Z}[i, \sqrt{-3}]$ is not a Dedekind domain and the semigroup of fractional ideals is not a group. This leads to problems such as that $\langle 2, 1 + \sqrt{-3} \rangle^2 = \langle 2 \rangle \langle 2, 1 + \sqrt{-3} \rangle$ (exercise!) but $\langle 2, 1 + \sqrt{-3} \rangle \neq \langle 2 \rangle$.
However, to answer the question about asymptotic density, we don't care. The rings $\mathbb{Z}[i, \sqrt{-3}]$ and $\mathbb{Z}[\zeta_{12}]$ become equal after inverting $2$. So all odd primes factor the same way in both rings, and we can discard the prime $2$ for asymptotic purposes.
Where does Chebotaryov come in? We have the quality of fields $\mathbb{Q}(i, \sqrt{-3}) = \mathbb{Q}(\zeta_{12})$. The Galois group over $\mathbb{Q}$ is $C_2 \times C_2$, where $C_k$ is the cyclic group of order $k$. We can think of this group as either switching the signs in $\pm i$, $\pm \sqrt{-3}$, or as the group of units in $\mathbb{Z}/12$. Either way, a prime $p$ remains inert in a Galois extension if and only if the Frobenius at that prime generates the Galois group; since this Galois group is not cyclic, no primes remain inert.
The deep part of Chebotaryov is that the $4$ elements of $\mathrm{Gal}(\mathbb{Q}(i, \sqrt{-3})/\mathbb{Q}))$ occur equally often as Frobenius elements, but, since none of them correspond to remaining inert, we don't actually need this. (Also, in this case, we can explicitly write down the Frobenius at $p$ as a function of $p$ modulo $12$, so Chebotaryov density is simply Dirichlet's theorem on density of primes in arithmetic progressions in this case.)