# Decomposition of prime ideals

First of all, the polynomial $p(x)=x^{3}-x-4$ must have one real root. Let $\theta$ be one of its roots, L=Q($\theta$), $O$ the integral closure of Z in L.
Then Kummer's theorem tells us then and there that for a prime q not to be nonsplit it is necessary and sufficient that $p(x) \bmod q$ is reducible.
Nevertheless, it in general is pretty hard to determine whether or not it is true, e.g. hard enough for the case q=the biggest prime ever known until now.
So, my question is that is there an efficient way determining the solvability of this kind of polynomials, or if I mistakenly thought of something pretty easy as very difficult, please inform me, thanks.
Edit:An efficient way means it might work for cases such as q=109 since for small enough primes, it suffices to check by hands, such as for 29, (29)=$P_1$$P_2$$P_3$ where $P_1$=29$O$+($\theta-5)O$,$P_2$=29$O$+($\theta-19)O$, $P_3$=29$O$+($\theta+15)O$, and some similar result for (5) while (5) is not totally split in $O$.

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By the way, I know that the conductor of L over Q is a divisor of 2 which makes things more easier, and an integral basis of L|Q is ${1,a,(a+a^{2})/2}$ where a =$\theta$. – awllower Feb 13 '11 at 7:51
What exactly is it you are asking? Do you want to know how to decide if a specific cubic has a root modulo a very large prime number? – KCd Feb 13 '11 at 9:17
Yes, the question might be formulated as that, while I was intended to solve it in terms of algebraic number theory. – awllower Feb 13 '11 at 13:35
The polynomial $x^3 -x -4$ has one real root and two complex roots. I don't think this actually affects the question, though. – Matt E Feb 14 '11 at 3:39
Sorry, I will improve it. – awllower Feb 14 '11 at 9:26