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.


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.