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Let $L$ be a Galois number field over $\mathbb{Q}$. Based on classical algebraic number theory (specifically, the Chebotarev density theorem), I can answer lots of numerical questions about how primes $p\in\mathbb{Z}$ split in $\mathfrak{O}_L$. For example the probability that $p$ splits completely is $1/[L:\mathbb{Q}]$, and the average number of factors is the sum over the Galois group of the reciprocals of the orders (so, for a quadratic extension, we get $1+1/2=3/2$ since half the primes split, and half don't.

My question: Another question one can ask is if the factors of $p$ are principal (since $L$ is Galois, they are all conjugate, and thus all principal or all not). Are there any quantitative results about this?

One might optimistically hope that the probability of this is the reciprocal of the class number, but I have no idea if this is true.

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As Pete explains below you probably meant "splits completely" rather than "is still prime" in your example in the first paragraph. –  Noah Snyder Jan 6 '10 at 18:34
    
There is an ambiguity here, it seems to me, which is I think making people give answers which are at cross-purposes. Is Ben asking for the probability that a prime that splits completely, splits into principal factors? Or is he asking what the chances are that all the factors of a random prime of Q are principal? The two answers are different, as Pete's answer indicates. I think that Felipe might have correctly answered the question about primes which split completely, because the factors of the split primes have density 1 in the top field. The factors of the inert primes have density 0. –  Kevin Buzzard Jan 6 '10 at 18:56
    
I definitely did a double-take in reading the question myself; in the end I decided on my interpretation. Is the point that Felipe's answer is measuring the density in L rather than in Q? –  Pete L. Clark Jan 6 '10 at 18:58
    
Right! So your inert primes, which for a quadratic extension have density 1/2 in Q, become a set of density zero in the extension itself (because you're evaluating the zeta function at 2 not 1, and it converges at 2). So Ben has to be clear about what he's asking. –  Kevin Buzzard Jan 6 '10 at 19:03
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@Ben: upon further reflection...you win. –  Pete L. Clark Jan 6 '10 at 23:06
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2 Answers

Yes, the density is 1/h. Apply Chebotarev to the Hilbert class field of L.

Edit: As Pete points out, this is not right if I am counting primes in Q. It is only correct if I count prime ideals of L and, as Kevin points out, this only sees the primes in Q splitting completely in L.

Edit again: Isn't the Hilbert class field of L galois over Q with the Galois group being a semi-direct product of Gal(L/Q) and the class group? If that's the case, then we apply Chebotarev to the Hilbert class field of L as an extension of Q and the density is whatever comes out. So for L/Q quadratic, for example, the density is 1/2 + 1/2h. The 1/2 comes from the inert primes and the 1/2h from the split principal primes.

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I don't think Gal(H/Q) is always a semidirect product. But having said that, of course the technique you outline works fine anyway. The crucial point (as you know) is that a prime of L splits completely in H/L iff it's principal, and splitting completely in H/L is something detectable using Frob_p in Gal(H/Q), whose behaviour we know by Cebotarev. This reduces the question to a counting argument. –  Kevin Buzzard Jan 6 '10 at 20:16
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I believe that Felipe's answer is not correct. [Edit: rather, it is correct according to a different interpretation of the question. But my interpretation is also natural, I think.]

Say a prime $p$ is inert in $K$ if $p \mathbb{Z}_K$ remains prime. In particular, inert primes have all of their factors principal.

If $L/\mathbb{Q}$ is not cyclic, then there are no inert primes. However, if it is cyclic, say of degree $d$, then the density of inert primes is $\frac{\varphi(d)}{d}$, which gives a lower bound on the answer to Ben's question. This lower bound can certainly be greater than $\frac{1}{h(L)}$: take for instance imaginary quadratic fields with sufficiently large discriminant.

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I've edited my answer accordingly. –  Felipe Voloch Jan 6 '10 at 19:45
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