How do we determine the decomposition of $p\mathcal{O}_K$ in $K = \mathbb{Q}(\sqrt[3]{5})$?

Question

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!

Answer 1

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.