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I'm trying to figure out as precisely as possible how primes split in a cubic extension $\mathbb{Q}(x)$, where $x^3-x-1=0$. This extension has discriminant $-23$. Write $K=\mathbb{Q}(x)$, $L=\mathbb{Q}(\sqrt{-23})$ and $E$ for the splitting field of $X^3-X-1$. Then we have the lattice

| \
K  L
| /

The only thing I can figure out precisely is that $23\mathcal{O}_K=\mathfrak{p}^2\mathfrak{q}$, because $$X^3-X-1\equiv (X-a)^3\,(\textrm{mod }p)$$ does not have a solution which is easy to see by expanding the cubic.

Is there any approach for finding out more things explicitly? I know at least the following:

  1. Possible splittings of an unramified prime in $K$ are $$p=\mathfrak{p},\; p=\mathfrak{p}\mathfrak{q},\; p=\mathfrak{p}_1\mathfrak{p}_2\mathfrak{p}_3.$$

  2. A prime splits completely in $K$ if and only if it splits completely in $E$, so this would take care of the 3rd type.

Is it even reasonable to expect that there's some nice formula in terms of congruences to figure out the possibly splittings? At least we know that splittings of the first and second type above can both result in a similar splitting in $E$.

Does anyone have any ideas what I could try? The fact that $E/\mathbb{Q}$ is not abelian, seems to make this far more annoying.

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Another thing I might add is that I've also figured out that there are exactly 3 subfields $K$ s.t. $[K:\mathbb{Q}]=3$ and these are all isomorphic. Simiarly, there is one subfield $L$ s.t. $[L:\mathbb{Q}]=2$ and this is the quadratic field above. It follows that if a prime in $E$ has non-trivial decomposition group, then the decomposition field must be one of these fields. I haven't figured out if this can be used in any useful way though. – dstt May 1 '11 at 20:01
The discriminant is $-23$. There is a modular form written down whose coefficients describe the way primes split at this MO question:… – Qiaochu Yuan May 1 '11 at 20:11
Haha, you're right... I made a stupid sign error while computing the discriminant. I'll fix that. – dstt May 1 '11 at 20:35

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