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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

 E | \ K  L | / Q 

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.

  • 0
    Possible duplicate of [Find all primes $p$ such that $x^3+x+1\equiv0\pmod p$ has $3$ incongruent solutions.](https://math.stackexchange.com/questions/2233442/find-all-primes-p-such-that-x3x1-equiv0-pmod-p-has-3-incongruent-soluti)2018-06-16

0 Answers 0