Classifying splittings of primes? I was wondering what general strategies are available to figure out if a prime splits? I know for quadratic extensions there aren't too many possibilities for how a prime can split, so we essentially only need to check that $X^2-d$ has a root modulo $p$ and that $p$ does not divide the discriminant. This can be one using quadratic reciprocity.
For e.g. an extension $\mathbb{Q}(\alpha)/\mathbb{Q}$, where $\alpha^3-\alpha-1=0$ there are a few more options. In this particular example computing the discriminant is easy and in general there are algorithms for it. The question is that for a prime $\mathfrak{P}\mid p$ how do we know that $f(\mathfrak{P}/p)=1$? Is there any "simple" algorithm that works for the polynomials $X^3+aX+b$ and $X^5+aX+b$?
EDIT: Per instructions below I'm editing the question. I was more interested in actually classifying and finding the density of the set of primes with a particular factorization. Given a particular prime, its factorization is much easier to find. In the quadratic case this is done by using quadratic reciprocity as I described above. What I don't know is how to work with cubics. I've been told that this can be done for cubics $X^3+aX+b$ and for simplicity I'm interested in the case $X^3-X-1$. Apparently another set of "easier" equations is $X^5+aX+b$.
 A: For an extension $\mathbb{Q}(\alpha)/\mathbb{Q}$, where $\alpha$ has minimal polynomial $P$ which we can assume has all its coefficients in $\mathbb{Z}$, for every prime $p$ which does not divide the discriminant of $P$, there is a $\mathfrak{P}$ above $p$ with $f(\mathfrak{P}/p)=1$ iff $P \mod p$ has a root in $\mathbb{F}_p$ (which by Hensel's lemma can be lifted to a root of $P$ in $\mathbb{Z}_p$).
But note that $p$ can split even if there is no such root: a general version of Hensel's lemma tells us that the factorization of $P \mod p$ into irreducible (and coprime, since $p$ does not divide the discriminant) can be lifted to a factorization of $P$ (this gives you the factors of $P$ as an element of $\mathbb{Q}_p[X]$).
Each factor corresponds to a place $\mathfrak{P}$ above $p$, and $f(\mathfrak{P}/p)$ is equal to the degree of the factor.
However in degree $\leq 3$, a polynomial is irreducible iff it has no root.
Note that nothing is said about primes dividing the discriminant. They can be unramified or not. It is a bit harder (but there is an algorithm) to compute the decomposition in this case.
EDIT: of course, everything remains true if we take an arbitrary number field instead of $\mathbb{Q}$
A: For a given prime $p$, the structure of the prime factorization of $p$ in the (ring of integers of the) field $\mathbb{Q}(\alpha)$ mirrors the factorization of the generating polynomial (say, $X^3-X-1$) over the finite field $\mathbb{Z}/p\mathbb{Z}$. This factorization can be algorithmically determined quite easily. 
If, however, your task is to determine the entire set of primes which split, then this is essentially a higher reciprocity law and becomes quite difficult when the Galois group is not Abelian. I'm not sure what the general algorithm is, or even if there is one.
