Let $K = \mathbb{Q}(\sqrt[3]{5})$, and $\mathcal{O}_K$ be its ring of integers.

In general, how do we decide the decomposition of $p\mathcal{O}_K$, for an odd prime $p$?

I know that by Kummer's theorem, one can decide the decomposition based on factorization of $X^3 - 5 \pmod{p}$.

For quadratic polynomials, this is easily decided by the reciprocity law.

  1. How can one factorize $X^3 - 5 \pmod{p}$ ?

  2. Also, given another cubefree $d$, is it possible to (easily) decide the decomposition of $p\mathcal{O}_L$ in $L = \mathbb{Q}(\sqrt[3]{d})$ ?

Thank you!


Part $1$ is easy, because a polynomial of degree $3$ over a field is irreducible, if and only if it has a zero. For example, $x^3-5$ is irreducible over $\mathbb{F}_7$, whereas $x^3-5=(x^2 + 3x + 9)(x + 8)$ over $\mathbb{F}_{11}$. If there is a zero, we obtain a quadratic polynomial, and this we know how to factorize further. For part $2$ see here.

  • $\begingroup$ Hi, thank you sir! How may one decide given an arbitrary prime $p$ ? I was told that the factorization occurs differently when $p\equiv 1 \pmod{3}$ and $p\equiv -1 \pmod{3}$ $\endgroup$ – user147794 Apr 23 '15 at 10:29
  • $\begingroup$ Yes, you are right. We have $7\equiv 1 \bmod 3$, and $11\equiv -1\bmod 3$, in the examples above. $\endgroup$ – Dietrich Burde Apr 23 '15 at 19:16
  • $\begingroup$ But what if $p$ is arbitrary ? Will a frobenius map sending $x$ to $x^3$ help? $\endgroup$ – user147794 Apr 24 '15 at 17:51

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