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Let $K$ be a number field with ring of integers $O_K$. It is well known that for almost all prime $p\in\mathbb{Z}$, the prime $p$ is unramified in $K$, that is, if $pO_K=\mathfrak{p}_1^{e_1}\ldots \mathfrak{p}_r^{e_r}$, where $\mathfrak{p}_1,\ldots,\mathfrak{p}_r$ are the different primes in $O_K$ which lie above $p$, then $e_i=1$ for $i=1,\ldots,r$.

1-I want some reference where I can read a proof of this fact.

But I am mainly interested in the following question: Is it true that for almost all primes $p\in\mathbb{Z}$ we have that $N\mathfrak{p}_1=N\mathfrak{p}_2$ for all primes $\mathfrak{p_1}$ and $\mathfrak{p}_2$ which lie above $p$? (here $N\mathfrak{p}_i$ means the cardinality of $O_K/\mathfrak{p}_i$). This is certainly true when the extension is Galois, but I am intersted in the general case. That's why the expression "for almost all primes".


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3 Answers

IIRC, your first question isn't hard to work out yourself: the key point is that if a prime over $p$ is ramified, then any monic defining polynomial for the number field must have a double root modulo $p$, which is detected by the discriminant.

For your second question, it is surely not true in the general case. e.g. a "random" degree 3 polynomial will, over a "random" prime, have a degree 1 and a degree 2 factor with probability $1/2 + O(p^{-2})$ -- this can be seen by simply counting all factorizations of degree 3 polynomials over $\mathbb{F}_p$ (I think that's the right error term). Correspondingly, in the number field defined by that polynomial, asymptotically half of the integer primes ought to split into a degree 1 and degree 2 factor.

I don't recall how to convert this from a heuristic argument to a rigorous one. It probably has something to do with Chebotarev's density theorem.

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No, it's definitely not true, nowhere near. Take your favorite nonGalois extension, like $\mathbb{Q}(\root3\of5\,)$. Then the unramified primes are those other than $3$ and $5$. Look at the cubic $X^3-5$ over $\mathbb{Z}/(p)$, for such a prime. It may (a) factor into three linear factors; (b) factor into a linear times a quadratic; or (c) remain irreducible. In case (b), you'll get one prime above $p$ of norm $p$, one prime of norm $p^2$. This happens! Look at $p=11$, $p=17$, $p=23$. See a pattern?

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Are there infinitely many primes which satisfy (b)? –  Makoto Kato Sep 21 '12 at 21:41
The “pattern” is that if $p$ is congruent to $-1$ modulo $3$, case (b) obtains. –  Lubin Sep 22 '12 at 5:25
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Concerning your reference request, the proof of the first statement can be found in "Algebraic Number Theory" by Jürgen Neukirch (I think it should be in section 1.8 but I am not sure; I don't have my copy at hand, unfortunately).

You can also check out Milne's notes on algebraic number theory. In those notes, it is Theorem 3.35 on page 54. Note that Milne proves a somewhat stronger result, and the proof is a little more complicated, in my opinion. So maybe you should check out Neukirch first.

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