Decomposition and inertial fields of primes in $\mathbf{Q}(\sqrt{2},\sqrt{3},\sqrt{5})$ I recently ran into this old number theory prelim problem.

Let $K=\mathbf{Q}(\sqrt{2},\sqrt{3},\sqrt{5})$ and let $\mathcal{O}_K$
  be the ring of integers of $K$. Find the ramification index and
  inertial degree of a prime $P|7$. Find the decomposition field and
  inertial field associated to a prime $P|7$.

I have no idea what result to use here. Since $\mathcal{O}_K$ is pretty complicated, I am guessing you are not supposed to just find $P$ and explicitly do the computation, but I cannot think of anything else.
 A: This isn't as hard as it looks. First of all you note that the dimension of this field is $8= 2^3$ since each field is linearly disjoint from all the others and that it is Galois since it is the splitting field of $(x^2-2)(x^2-3)(x^2-5)$. By noting your field is contained in $\Bbb Q(\zeta_{120})=\Bbb Q(\zeta_8)\Bbb Q(\zeta_5)\Bbb Q(\zeta_3)$ you can see that $7\not\mid \operatorname{disc}(\Bbb Q(\zeta_{120}))$, because this field's discriminant divides $120^{120}$ (see for example any section of an algebraic number theory text on cyclotomic fields, eg Neukirch or Lang or just Wikipedia) and $7$ is not in the prime factorization of $120$. We know if it doesn't ramify in the big field, it cannot in a sub-field.
We conclude there is no ramification, which leaves just inertia and splitting behavior. Now what happens in the quadratic sub-fields? Well this is determined by Quadratic Reciprocity, which says that since
$$\begin{cases} \left({3\over 7}\right) = -1 \\ \left({5\over 7}\right) = -1 \\ \left({2\over 7}\right) = 1 \\ \left({-1\over 7}\right) = -1\end{cases}$$
We conclude form this that $7$ does not split in $\Bbb Q(\sqrt 3)$ or $\Bbb Q(\sqrt 5)$ and does in $\Bbb Q(\sqrt 2)$ which we can easily see by $(7)=(3+\sqrt 2)(3-\sqrt 2)$. So the decomposition field of $(7)$ is $\Bbb Q(\sqrt 2)$ which means $(7)$ looks like $\mathfrak{p}_1\mathfrak{p}_2$ in the big field. Which gives $e=1$, and $r=2$. And since $efr=[K:\Bbb Q]$ for Galois extensions $K/\Bbb Q$, we see $f=4$. This also tells you that $K$ itself is the inertia field.
