2
$\begingroup$

I would appreciate a reference to the following statement, which, I was having an impression, is known:

Let $L, M$ be field extensions of finite degree of a number field $K$, such that $L \cap M = K$. Then probabilities of any two chosen splitting types in $L$ and $M$ respectively, of a randomly chosen prime ideal in $\mathcal{O}_K$, are independend.

Thank you.

$\endgroup$

1 Answer 1

6
$\begingroup$

As stated, this is false: take $L$ and $M$ to be two distinct intermediate cubic subfields of a Galois $S_3$-extension. In particular, they are isomorphic. If a prime of $K$ splits completely in one, then it also splits completely in the other.

Perhaps you wanted to assume that both $L$ and $M$ are Galois and disjoint, or at least that their Galois closures are disjoint over $K$, rather than just the fields themselves. Then what you are saying is true, and is simply Chebotarev over the compositum of the Galois closures. That's because the Galois group of this compositum will be the direct product of the two separate Galois groups, so all the relevant probabilities of being a Frobenius, etc. will multiply.

$\endgroup$
1
  • $\begingroup$ Thanks! (yes, I meant Galois extensions). I did not see that it actually follows from Chebotarev's theorem. $\endgroup$
    – Albertas
    Commented Jul 4, 2013 at 12:11

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .