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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
up vote 2 down vote accepted

For a particular prime, you can use Cantor-Zassenhaus or another factoring algorithm to factor the polynomial modulo the prime.

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Thanks, it is refreshing to me. – awllower Feb 14 '11 at 9:36
Thanks, @Qiaochu Yuan Although I am seeking for further ideal-theoretical methods or valuation-theoretical ones, thanks, in any case. – awllower Feb 15 '11 at 9:00
I am not really sure what you are looking for. – Qiaochu Yuan Feb 15 '11 at 9:03
I know there is a well-developed theory of ramification in the Galois case, nonetheless, this case is not Galois, and hence I would like to know if there is any result on the direction. – awllower Feb 16 '11 at 5:02

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