Given a finite abelian extension of number fields $L/K$, the prime ideals $\mathfrak{p}$ in $O_K$ split into primes $\mathfrak{P}$ in $O_L$. The number of primes $\mathfrak{p}$ splits into is necessarily a divisor $d$ of $[L:K]$. One can ask:

**Question (1):** Which primes in $O_K$ split completely in $O_L$?

**Question (2):** Given a divisor $d$ of $[L :K]$, which primes $\mathfrak{p}$ split into $d$ primes in $O_L$?

For $K = \mathbb{Q}$ and $L$ a cyclotomic extension of $\mathbb{Q}$, Question (2) seems genuinely more difficult than Question (1). A friend pointed this out to me in response to some notes that I put together titled A Prelude to the Study of Reciprocity Laws.

Let $p$ be a prime and let $$\Phi_{p}(x) = x^{p-1} + x^{p-2} + \ldots + x^2 + x + 1.$$

As I discuss in the notes, a prime $\ell$ divides a number of the form $\Phi_{p}(n)$ if and only if $\ell = p$ or $\ell \equiv 1 \pmod p$. In Section 7 of the notes, I show how this follows immediately from the basic theorems of modular arithmetic.

Let $\zeta_p$ be a root of the polynomial $\Phi_{p}(x)$. Then a prime $\ell \in \mathbb{Z}$, $\ell \neq p$ divides a number of the form $\Phi_{p}(n)$ if and only if $\ell$ splits completely in $\mathbb{Z}[\zeta_p]$. So the above theorem answers Question (1) in this case where $K = \mathbb{Q}$ and $L = \mathbb{Q}(\zeta_p)$ .

In my notes I (apparently) erroneously remark that the answer to Question (1) in this case implies quadratic reciprocity. In fact, quadratic reciprocity follows the answer to Question (2). I knew this, but thought that the answer to Question (2) follows formally from the answer to Question (1). My friend questioned whether this is true, and after spending a few hours on it, I realized that I can't see a way to use the answer to Question (1) to derive the answer to Question (2).

Is there a way to do this?

The proofs of the theorems of class field theory proceed by answering Question (2) directly, which is consistent with the above paragraph, but I always thought that the reason for this was that the only way to answer Question (1) in general was to answer Question (2). For this reason, I saw the questions as being of comparable difficulty. So I was surprised to see a case in which this doesn't seem to be true.

Aside from cyclotomic extensions of $\mathbb{Q}$, are there finite extensions of number fields $L/K$ for which it's easier to answer Question (1) than Question (2)?

A few potentially relevant thoughts:

The imaginary quadratic case is similar to the cyclotomic case in the sense that one has an explicit construction of the number fields in question via complex multiplication, so it's natural to try to answer the latter question by looking at, e.g., the polynomials over $\mathbb{Z}[i]$ that generate the ray class fields of $\mathbb{Q}(i)$ and try to see if the prime divisors of their values at the Gaussian integers can be determined in a direct way analogous to as in the cyclotomic case.

In the case where $[L : K]$ is a prime, Questions (1) and (2) are the same, because a prime splits completely if and only if it splits at all.

The set of primes that splits completely in a number field uniquely determines the number field and so in principle determines the splitting behavior of the other primes.

abelianwas not mentioned in the earlier version of your question. The answer for abelian extensions (of global fields) is contained in class field theory. – Chandan Singh Dalawat Dec 27 '12 at 5:41