prime ideal factorization in an extension field Let $p$ be a rational prime and $K$ a number field.
Dedekind's discriminant theorem tells us that
$p$ ramifies in $K$ $\iff$ $p$ divides the discriminant of $K$.
Hence if $p$ does not divide discriminant of $K$,
$(p)$ either splits, i.e.,   
(i) $(p)=P_1 \cdots P_g$ for $P_i \neq P_j$ and $g \geq 2$
or  
(ii) $(p)$ remains prime.  
Now, my question is: what are some criteria which can tell if $p$ will split or remain prime?
 A: Pete's answer is very good. Some self-promotion:
An expository post on the relationship between factoring polynomials and factoring primes.
The previous answer regarding the polynomial $t^3+t^2−2t−1$
A: The best explicit criterion that I know is the criterion of Kummer-Dedekind, which involves writing $K = \mathbb{Q}[t]/(P(t))$ and factoring $P(t)$ modulo the prime $p$.  Then the factorization of $(p)$ in $\mathbb{Z}_K$ "has the same shape" as the factorization of $P(t)$ in $(\mathbb{Z}/p\mathbb{Z})[t]$: see e.g.
http://www.math.uconn.edu/~kconrad/blurbs/gradnumthy/dedekindf.pdf
This criterion does not apply to primes dividing the discriminant of the polynomial $P(t)$ but not the number field $K$.  
At a more theoretical level, the Chebotarev density theorem gives some powerful asymptotic information: for instance, it says that the density of the set of primes which split completely in $K$ is $\frac{1}{[M:K]}$, where $M$ is the Galois closure of $K/\mathbb{Q}$.  Also class field theory has things to say in the special case when $K/\mathbb{Q}$ is abelian.  
In some sense, the general problem is unsolved: it is one of the things that we imagine we might know better if we knew a "non-abelian class field theory".
Addendum: After copyediting your question, I interpret is as being especially interested in determining which primes $p$ remain prime in $K$ (or are $\textbf{inert}$).  Again the Chebotarev Density Theorem is helpful for this: suppose for simplicity that $K/\mathbb{Q}$ is Galois.  Then there exist primes $p$ which remain inert in $K$ iff the Galois group of $K/\mathbb{Q}$ is cyclic, in which case the density of such primes is $\frac{\varphi([K:\mathbb{Q}])}{[K:\mathbb{Q}]}$.  So for example, if $\ell_1$ and $\ell_2$ are distinct prime numbers and $K = \mathbb{Q}(\sqrt{\ell_1}, \sqrt{\ell_2})$, the Galois group of $K/\mathbb{Q}$ is isomorphic to $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$, so there are no inert primes. 
A: Adding a few more words to Pete's answer...In the Galois case, at least, the inertia degrees $f$ and ramifications indices $e$ of each prime $P_i$ lying over $p$ are the same. If $p$ is unramified then $e = 1$ and so the only question is what is $f$ and what is $g$? The common degree $f$ is the order of any (and each) of the Frobenius elements $(P_i | K/\mathbb Q)$ which map to the generator of the local Galois group $G(k(P_i)/\mathbb F_p)$. I always recommend Milne's online notes http://www.jmilne.org/math/CourseNotes, both the algebraic number theory and class field theory, for this kind of question. I also remember well reading this material in the Janusz book "Algebraic Number Fields". In the latter, I believe Chapter 1 and Chapter 3 are most relevant, but I do not have a copy with me.
