# unramified extensions

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".

Thanks!

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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