Let $K$ and $F$ be two number fields of degree $n$ and $m$ and $K\cap F = \mathbb{Q}$. Then we consider the number field $KF$ generated by $K$ and $F$. It is known that we have the relation between corresponding discriminant, i.e, $$ d_{KF}\ \Big|\ d_K^m\cdot d_F^n. $$ If $p$ is ramified in $KF$ and umramified in $F$. Can we get the following identity for $p$-part of discriminants? $$d_{KF,p} = d_{K,p}^m.$$
Any help is welcome and appreciated!
Consider the fields $K=\mathbb{Q}(\sqrt[6]{2})$ and $F=\mathbb{Q}(\zeta_8\sqrt[4]{2})$, where $\zeta_8$ is the fourth root of unity. It's not hard to see that $\mathbb{Q}$ is the only real subfield of $F$ and so we have that $K \cap F = \mathbb{Q}$, as $K$ is a real subfield.
Now using PARI/GP one can calculate that $d_F=2^{11}$, while $d_K=2^{11}3^{6}$. On the otherside $KF = \mathbb{Q}(\sqrt[6]{2}+\zeta_8\sqrt[4]{2})$. The minimal polynomial of the primitive element according to Wolfram Alpha is $x^{12} - 6x^8 + 48x^7 - 4x^6 + 12x^4 + 160x^3 + 168x^2 + 48x - 4$. Again by PARI/GP we get that $d_{KF} = -2^{35}3^{12}$ and hence we have that the $p$-parts aren't the same.
The reason why this happens is that $K \cap F = \mathbb{Q}$ isn't enough to conclude that $KF$ is an extension of degree $mn$. Indeed if one of the fields is Galois then this would be true. In fact it's enough for $K$ to have a trivial intersection with the Galois closure of $F$ or vice versa. Note that this isn't the case, as $K \cap F_1 = \mathbb{Q}(\sqrt{2})$, while $K_1 \cap F = \mathbb{Q}(i\sqrt{2})$, where $K_1$ and $F_1$ are the respective Galois closures.
If the condition from above is satisfied we have that the claim is true. You can use the theorem in this answer. It's not hard to notice that $d_{KF/K} \mid d_F$ (as ideals in $\mathcal{O}_K$) and then we have that $$N(d_{KF/F}) \text{ }\Big | \text{ }N(d_F) = d_F^n$$
The last equality follows, as $d_F$ is an element of $\mathbb{Q}$. Hence $N(d_{KF/F})$ doesn't contribute with any $p$-factor and so we have that the $p$-part of $d_{KF}$ is same as the one of $d_{K}^m$
[UPDATE]: In fact we can do most of the proof without those computer aided calculations. Consider $K,F$ as above and let $L=\mathbb{Q}(i\sqrt{2})$. Then it's not hard to conclude that $KF = KL$. Indeed $KL \subseteq KF$ and the equality holds as they have the same extension degree over $\mathbb{Q}$, namely $12$. Now we have that $d_{KL} \mid d_{K}^2d_{L}^6$. It's easy to see that $3 \mid d_K$, while not being a factor of $d_L$
Similarly $d_{KL} = d_{KF} \mid d_{K}^4d_{F}^6$ and $3 \nmid d_F$. Finally from above we have that $d_{KF}$ and $d_K^4$ can't have the same $p$-parts.
