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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 $\mathbb{Q}(\alpha)$:

  1. $(p)$ is stays inert. Density: 1/3.

  2. $(p)=\mathfrak{p}\mathfrak{q}$. Density: 1/2.

  3. $(p)$ 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?

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Do you know the theorem for explicit factorization of primes? If $A$ is a Dedekind ring with quotient field $K$, $E$ is a finite separable extension of $K$, $B$ is the integral closure of $A$ in $B$, and $B=A[\alpha]$ for some $\alpha$ with irreducible polynomial $f(x)$, then the reduction of $f(x)$ modulo $\mathfrak{p}$ (where $\mathfrak{p}$ is a prime of $A$) give the decomposition of $\mathfrak{p}B$. This does not require the extension to be abelian. –  Arturo Magidin Nov 8 '11 at 18:15
Given a prime, I have no trouble finding it's splitting by factorizing $X^3-X-1$ modulo that prime. The problem is the other way around. I want to know for which primes $p$, $X^3-X-1$ has a particular factorization modulo $p$. For a quadratic extension this is given by the quadratic reciprocity theorem. –  pki Nov 8 '11 at 18:19
I might add that the ring of integers in this case is $\mathbb{Z}[\alpha]$, so Kummer's theorem applies for all primes. –  pki Nov 8 '11 at 18:22
Try calclating the artin conductors and norms for the series of extensions $L/\mathbb{Q}(\sqrt{-23}),$ $\mathbb{Q}(\sqrt{-23})/\mathbb{Q}$ and then piece this information together to calculate how primes factor in $L/\mathbb{Q}$ –  jspecter Nov 8 '11 at 18:30

2 Answers 2

up vote 11 down vote accepted

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.

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Thanks. I still need to digest the splitting of $\mathfrak{p}_1$ in $K$. I've never computed explicit splittings for anything except primes of $\mathbb{Q}$, but doing that should be good practice. –  pki Nov 8 '11 at 18:46
I also like to point out that this is the first time that I've seen quadratic forms come up in questions like this. I've been told that quadratic forms are important in number theory and this gave me one of the first convincing proofs of that! –  pki Nov 8 '11 at 19:17

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.

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