Let $L$ be a Galois extension of $\mathbb{Q}$ and $M$ a finite extension of $\mathbb{Q}$, both of degrees $> 1$. A Theorem of Bauer tells that $Spl_1(M)\subset Spl(L)$ up to a finite number of exceptions is equivalent to $L\subset M$. Here $Spl(L)$ denotes the rational primes which splits in $L$ and $Spl_1(M)$ denotes the rational primes such that there exists a prime of inertial degree $1$ above.

This result can be found in Cox, Primes of the form $x^2+ny^2$ (Prop 8.20) or Neukirch (p. 135) for example.

$\textbf{My question}:$ Suppose that $L\cap M =\mathbb{Q}$ with $L$ Galois, Bauer's result impliesthe existence of infinitely primes in $Spl_1(M)$ which are not splitting in $L$. Can we ask for something stronger, I mean a positive density of such primes (and even quantify that density using the information on Galois groups)?

To be less general, in my case I am interested when $L$ is a quadratic extension.

Thanks in advance!

Edit: (This is in response to KConrad's comment on Chebotarev density.) So the proof I refer for Bauer's Theorem is the following: taking $N$ a Galois extension containing both $M$ and $L$ we can show that $Gal(N/M) \subset Gal(N/L)$ which implies $L \subset M$. For doing this, we take an automorphism $\sigma \in Gal(N/M)$ and wants to show that it is trivial on $L$. It uses Chebotarev to construct a lot of rational primes such that the Artin symbol $((N/\mathbb{Q})/p)$ is in the conjugacy class of $\sigma$, we prove that those primes are in $Spl_1(M)$ and by hypothesis they split in $L$ (finite number exceptions). We conclude by showing that this implies the automorphism to be trivial on $L$.

Conversely it means that if $L\subsetneq M$, we have infinitely many primes in $Spl_1(M)$ which are not splitting in $L$.

So I think that the same result is true up to a set of density $<1/|Gal(N/\mathbb{Q})|$ (minimal size of conjugacy class is $1$). In that case I can perform the same argument as above applying Chebotarev. For each automorphism I produce a lot of primes (density $\geq 1/|Gal(N/\mathbb{Q})|)$ such that the Artin symbol $((N/\mathbb{Q})/p)$ is in the conjugacy class of $\sigma$ and the rest of the proof goes in the same lines.

Can we do better than this? The optimal $N$ above to choose is the normal closure of the compositum of $L$ and $M$. Suppose $M$ is of degree $n$, we have at least a density $1/n$ of primes in $Spl_1(M)$. We could loose a lot by this argument is the normal closure is "large".

Suppose the strongest condition that $L\cap M = \mathbb{Q}$ (this is equivalent to the previous conclusion of non inclusion if $L$ is quadratic), can we hope for better proportion of primes in $Spl_1(M)$ which does not split in $L$?

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    $\begingroup$ Strictly speaking the premise of your question is false: try $L = M = \mathbf Q$. Please fix the hypotheses. And have you tried seeing what the Chebotarev density theorem tells you? $\endgroup$ – KConrad Mar 22 '15 at 5:19

Bauer based his results not on Chebotarev (which came 20 years after Bauer's work) but on Kronecker, who "proved" (real proofs were later provided by Frobenius) that the primes splitting completely in a normal extension N of the rationals have density $1/(N:{\mathbb Q})$. Since primes splitting completely in some extension $M$ coincide with those splitting completely in its normal closure $N$, this result gives the densities of completely splitting primes in arbitrary extensions. Finally, the compositum of $N$ and the normal extension $L$ coincides with the normal extension of $LM$, and so Kronecker's density theorem gives you everything you need.


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