Primes in composite lying over given primes Let $L_1,L_2$ be finite linearly disjoint extensions of $\Bbb Q$ and ${\mathfrak p_i}\subset{\mathcal O} _ {L_i}$ prime ideals lying over the same prime $p\in\Bbb Z$.
Question:

Is there a prime $\mathfrak p$ of ${\cal O}_L$, where $L:=L_1L_2$, such that $\mathfrak p$ lies over both $\mathfrak p_1$ and $\mathfrak p_2$?

Some thoughts:
Assume that the discriminants of $L_1,L_2$ are coprime. In that case we have a canonical isomorphism ${\cal O}_{L_1}\otimes_{\Bbb Z}{\cal O}_{L_2}\cong {\cal O}_{L}$ and by elementary properties of the fiber product we see that there is an element of $\operatorname{Spec} {\cal O}_{L_1}\times\operatorname{Spec} {\cal O}_{L_2}\cong\operatorname{Spec} {\cal O}_{L}$ mapping to $\mathfrak p_1$ and $\mathfrak p_2$ respectively.
I think by localization this argument also works for the case when $p$ is not a common divisor of the the discriminants of $L_1,L_2$.
But what about the case when $p$ is ramified in both fields?
Maybe another approach would be to count the number of primes in $L$ lying above $p$ and use some kind of pigeonhole argument, but I think this won't work because there might be several primes of $L$ lying over both ${\mathfrak p}_i$.
Edit: While writing this question I thought of the following: As ${\cal O}_L$ is integral over ${\cal O}_{L_1}{\cal O}_{L_2}$ the map $\operatorname{Spec} {\cal O}_{L}\to\operatorname{Spec} {\cal O}_{L_1}{\cal O}_{L_2}$ is surjective. Now consider the map ${\cal O}_{L_1}\otimes{\cal O}_{L_2}\to{\cal O}_{L_1}{\cal O}_{L_2}$ given by $a\otimes b\mapsto ab$. By writing out bases of ${\cal O}_{L_i}$ we get that this map is an isomorphism, hence we get a canonical surjective map $\operatorname{Spec} {\cal O}_{L}\to\operatorname{Spec} {\cal O}_{L_1}\otimes{\cal O}_{L_2}$ and the same argument as above should work. Is this correct?? Are there other arguments, maybe by directly considering ${\cal O}_{L}$ instead of the (potentially) non-maximal order ${\cal O}_{L_1}{\cal O}_{L_2}$? Also I am primarily interested in the case where the $L_i$ are Galois over $\Bbb Q$, so an argument in this case would be sufficient for me.
 A: Maybe not the way you want, as it uses $p$-adic numbers, but the answer is yes:
We have $L_1\otimes_{\mathbb Q} L_2=L_1L_2=L$ (linearly disjoint).
We also have (for $i=1,2$) $L_i\otimes_{\mathbb Q} \mathbb Q_p= \bigoplus_{\mathfrak q_i}(L_i)_{\mathfrak q_i}$, where $\mathfrak q_i$ runs over all the prime ideals of $\mathcal O_{L_i}$ containing $p$.
Combining these, we have
$$A:=L\otimes_{\mathbb Q} \mathbb Q_p = \bigoplus_{\mathfrak q_1,\mathfrak q_2}
(L_1)_{\mathfrak q_1}\otimes_{\mathbb Q_p}(L_2)_{\mathfrak q_2}. \qquad(*)$$
Again this $\mathbb Q_p$-algebra splits to a sum of fields,
$$A=\bigoplus_\mathfrak r L_\mathfrak r\qquad(+)$$
where $\mathfrak r$ runs over all the prime ideals of $\mathcal O_L$ containing $p$.
The direct sum $(*)$ contains, in particular $B:=(L_1)_{\mathfrak p_1}\otimes_{\mathbb Q_p}(L_2)_{\mathfrak p_2}$. If you split $B$ to a sum of fields (a sub-sum of $(+)$)
$$B=\bigoplus_\mathfrak s L_\mathfrak s\subset \bigoplus_\mathfrak r L_\mathfrak r = A$$
then every $\mathfrak s$ can serve as a $\mathfrak p$ you're looking for.
