Exercise from Marcus on class groups

I stumbled upon this:

Let $p$ be a prime of $K$ (a given number field) and let $m$ be the order of $[p]$ in $Cl(K)$. Suppose $\mathcal{P}|p$ is a prime of $L$ (here, $L/K$ is an extension of $K$). Show that $m/(m, f(\mathcal{P},p))$ divides the order of the class $[\mathcal{P}]$ in $Cl(L)$.

Any help is welcomed, thanks!

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This is closely related to the class-field-theoretic type argument I mentioned in the other answer. Here's a fast solution (if possibly not the most elementary).

Note that by class field theory, specifically properties of the Artin symbol, the order $m$ of $[p]$ equals the relative inertia degree of $p$ in the Hilbert class field $K^{(1)}$ of $K$, and similarly for the order of $[\mathcal{P}]$ and the residue degree of $\mathcal{}P$ in $L^{(1)}/L$. Now consider the following diagram:


Here the labels on each arrow are the inertia degrees of a prime above $p$ in each extension. Now by multiplicativity of inertia degrees in towers, we have $m\mid m'f$, so $\frac{m}{(m,f)}\mid m',$ as desired.

(Apologies for the sloppy diagram, I've never jury-rigged an xypic on MSE or MO before. Can anyone make those arrows "headless"? And/or move the$f(\mathcal{P}/p)$ to the left of the arrow?)

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Makes sense. Thanks! – Anna Dec 13 '11 at 17:54
Sure thing. Happy number-theoring. – Cam McLeman Dec 13 '11 at 21:09
Excuse me, I am a little confused: As far as I know, the order of $P$ is equal to the order of its Frobenius, by the Artin symbol, and then is equal to the order of the decomposition group. So it is the product of $f$ and $e$, the ramification index. I guess it is assumed that $p$ is unramified, but I find no such indications. Thanks in any case in advance. – awllower May 23 '13 at 12:44

Here is an argument avoiding class field theory:

Let $a$ be the order of $\mathcal P$ in the class group of $L$. Then $\mathcal P^a = (\alpha)$ for some integer $\alpha$ of $L$. Taking norms down to $K$, we find that $p^{f(\mathcal P,p) a}$ is principal. Thus $m$ divides $a f(\mathcal P,p)$, and hence $m/(m,f(\mathcal P,p))$ divides $a$, as claimed.

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This is just a hint, that will hopefully help you realize where this $m/(m,f(\mathcal{P},p))$ comes from.

Let $m>1$ be a fixed integer and suppose that $a\in\mathbb{Z}$ is relatively prime to $m$. Then, for any $d\geq 1$, we have:

$$\operatorname{ord}(a^d) = \frac{\operatorname{ord}(a)}{\gcd(\operatorname{ord}(a),d)}.$$

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Thanks a lot. I actually realized that 5 minutes after posting it :)). – Anna Dec 13 '11 at 17:41