# 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?

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Shouldn't this kind of material be well-covered in any text on algebraic number theory? –  Qiaochu Yuan Jan 4 '10 at 7:27

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.

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.

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The special case when K is defined by the polynomial P(t)=t^3+t^2−2t−1 has been discussed recently on this site, cf. A problem of Shimura and its relation to class field theory. –  Chandan Singh Dalawat Jan 4 '10 at 11:25

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 $t^3+t^2−2t−1$

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Thanks, David. I will return the favor by recommending your expository post -- these are just the sort of questions that should be discussed in basic courses on algebra and number theory (but too rarely are). –  Pete L. Clark Jan 4 '10 at 17:04

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.

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