Splitting of primes in an $S_3$ extension

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.

Depends on what you mean by "explicit." Prime splitting in this case is governed by the coefficients of the modular form

$$f(q) = q \prod_{n=1}^{\infty} (1 - q^n) (1 - q^{23n}).$$

See this MO question for some details.