What is the probability that a rational prime remains prime in $\mathbb Z[i,\sqrt{-3}]$?

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

If you're referring to $\mathcal{O}_{\mathbb{Q}(\zeta_{12})}$, the ring of algebraic integers of $\mathbb{Q}(\zeta_{12})$ (see its page in the LMFDB), then the answer is that no prime from $\mathbb{Z}$ remains prime in $\mathcal{O}_{\mathbb{Q}(\zeta_{12})}$.

The intermediate fields are $\mathbb{Q}(i)$, $\mathbb{Q}(\omega)$ and $\mathbb{Q}(\sqrt{3})$. For a prime from $\mathbb{Z}$ to be prime in both $\mathbb{Q}(\omega)$ and $\mathbb{Q}(\sqrt{3})$, it must be $p \equiv 5 \bmod 12$. But then this means that $p \equiv 1 \bmod 4$ and you know what that means for $\mathbb{Z}[i]$...

P.S. A Naiade discovered the density theorem twelve centuries before Chebotarev.

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.)

An important property of a ring $R$ is that if $a$ and $b$ are numbers in it, then $a + b$ and $ab$ are also in $R$.

Clearly $i + \sqrt{-3}$ is in this ring that you're looking at. Big deal, no one cares. Multiplication is a bit more productive (pardon the pun) in this case: $$i \sqrt{-3} = \sqrt{-1} \sqrt{-3} = \sqrt{-1 \times -3} = \sqrt 3.$$

If you had the slightest doubt that 3 ramifies in this ring, doubt no more now.

Then 5 splits on account of $(2 - i)(2 + i)$, and 7 splits on account of $(2 - \sqrt{-3})(2 + \sqrt{-3})$, or, if you prefer, $$\left(\frac{5}{2} - \frac{\sqrt{-3}}{2}\right)\left(\frac{5}{2} + \frac{\sqrt{-3}}{2}\right).$$

Next, 11 is a little trickier: it splits on account of $(1 - 2 \sqrt 3)(1 + 2 \sqrt 3) = -11$. Or how about $(8 - 5 \sqrt 3)(8 + 5 \sqrt 3)$? That's also equal to $-11$. If that's acceptable to you, we can move on.

Actually, that's where I'm going to have to leave it at for tonight...