TITLE: Splitting of primes in an $S_3$ extension
QUESTION [10 upvotes]: Let $\alpha$ be a root of the polynomial $X^3-X-1$. The polynomial has discriminant $-23$ which is not a square, so the splitting field must have Galois group $S_3$. I would like to figure out the splitting of primes in $\mathbb{Q}(\alpha)$. The ramified case is trivial, so assume that $p$ is not ramified. By looking at possible splittings of a prime $p$ of $\mathbb{Q}$ in the Galois closure and then looking at which splittings in $\mathbb{Q}(\alpha)$ give rise to which splittings in the Galois closure, we get the following options for splittings in $\mathbb{Q}(\alpha)$:

$(p)$ is stays inert. Density: 1/3.
$(p)=\mathfrak{p}\mathfrak{q}$. Density: 1/2.
$(p)$ splits. Density: 1/6.

An old qual question I found online asks for which primes $p$ in $\mathbb{Q}$ give rise to which splitting, so to me this seems like the question asks for explicit conditions on $p$ which would tell us how it splits. This is just from a transcript written by the student, so I have no idea whether or not an answer was actually expected or if this was a trick question.
The way I've understood the norm limitation theorem of class field theory is that we can only expect to give congruence conditions in $\mathbb{Q}$ for how primes split in an abelian extension. Thus, for any nonabelian extension there is no way to express the condition in the form
$$p\equiv a_1,\ldots,a_k\,(\textrm{mod } n)$$
since any congruences would only give us information about splittings in the maximal abelian subextension. In this case knowing splitings in $\mathbb{Q}(\sqrt{-23})$ does not let us distinguish between splittings in$\mathbb{Q}(\alpha)$. Can anyone tell me if my intuition is right or if there actually are ways to write down explicit conditions on the primes?

REPLY [14 votes]: As Qiaochu says, it depends what you mean by explicit. Here is another sort-of-explicit answer.
Let $K$ be the splitting field of $x^3-x-1$. As you already realized, this is an $S_3$-extension of $\mathbb{Q}$, and the quadratic subfield is $\mathbb{Q}(\sqrt{-23})$. Let $p$ be a rational prime other than $23$. As I think you realize, the following are equivalent: 

$p$ factors in $\mathbb{Q}(\alpha)$ as $\mathfrak{p} \mathfrak{q}$ 
The Frobenius of $p$ in $S_3$ is a two-cycle 
$p$ is inert in $\mathbb{Q}(\sqrt{-23})$
$-23$ is not a quadratic residue modulo $p$
$p$ is congruent to $5$, $7$, $10$, $11$, $14$, $15$, $17$, $19$, $20$, $21$ or $22 \mod 23$.

Now for the interesting case. Suppose that the above conditions are not true. So $p$ factors as $\mathfrak{p}_1 \mathfrak{p}_2$ in $\mathbb{Q}(\sqrt{-23})$. Then the following are equivalent

$p$ splits completely in $\mathbb{Q}(\alpha)$
The Frobenius of $p$ in $S_3$ is the identity
The prime $\mathfrak{p}_1$ of $\mathbb{Q}(\sqrt{-23})$ splits in $K$.

Now, $K/\mathbb{Q}(\sqrt{-23})$ is abelian, so there should be a congruence condition for when $\mathfrak{p}$ splits in $K$. Moreover, a direct computation will show you that $K/\mathbb{Q}(\sqrt{-23})$ is unramified, so the congruence condition must depend only on the ideal class of the ideal $\mathfrak{p}_1$. Working it out, the following are equivalent:

The prime $\mathfrak{p}_1$ of $\mathbb{Q}(\sqrt{-23})$ splits in $K$ 
The prime ideal $\mathfrak{p}_1$ is principal.
There are some integers $x$ and $y$ such that $\mathfrak{p}_1 = \langle x+y \theta \rangle$, where $\theta = (1+\sqrt{-23})/2$.
The prime $p$ is of the form $x^2 - xy + 6y^2$.

The last condition is the norm of the previous condition; this is the standard trick for going between ideals in quadratic number fields and quadratic forms.
Therefore, my most explicit answer is 

The prime $p$ splits completely in $\mathbb{Q}(\alpha)$ if and only if $p$ if of the form $x^2-xy+6 y^2$.

Remark $K$ is not just an unramified abelian extension of $\mathbb{Q}(\sqrt{-23})$, it is the maximal such extension, also known as the class field.