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 $L$?