Theorem 24 of Marcus' Number Fields states the following:
Let $p$ be a prime number and assume that $p$ is ramified in a number ring $\mathcal{O}_K$. Then $p\mid disc(\mathcal{O}_K)$.
There is a specific step in the proof that I don't undestand. So here is how the proof goes:
Of course, $p$ ramified in $\mathcal{O}_K$ impies that $\exists \mathfrak p \in Spec(\mathcal{O}_K)$ (by $Spec(A)$ I denote the set of prime ideals of a ring A) such that $e(\mathfrak p \mid p) > 1$ , where by $e(\mathfrak p \mid p)$ I denote the ramification index of $\mathfrak p$ over $p$. Then $p \mathcal{O}_K = \mathfrak p I $ , where $I$ is divisible by all primes of $\mathcal{O}_K$ lying over $p$ (so $\mathfrak p$ divides $I$).
If $\sigma_1,...,\sigma_n$ denote the embeddings of K in $\mathbb C$,extend them all to automorphisms of some extension $L$ of $K$ that is normal over $\mathbb Q$.
Now, $p \mathcal{O}_K = \mathfrak p I $ of course implies $p \mathcal{O}_K \subseteq I$. If equality was true, then (by unique factorization) $\mathfrak p $ would be equal to $\mathcal{O}_K$, which is absurd. So we have that $p \mathcal{O}_K \subsetneq I$. Let $\alpha \in I- p \mathcal{O}_K$. So $\alpha$ is in every prime of $\mathcal{O}_K$ that lies over $p$. It follows, that $\alpha$ is in every prime of $\mathcal{O}_L$ that lies over $p$ (this is not exactly immediate but repeated use of Theorem 19 of Marcus (characterization of when a prime $\mathfrak q$ lies over a prime $\mathfrak p$) proves it).
Now we are close to the obsticle. We fix any prime $\mathfrak q$ of $\mathcal{O}_L$ lying over $p$. Marcus then claims that $\sigma(\alpha) \in \mathfrak q$ for every automorphism $\sigma$ of $L$. To see this, he continues, we must notice that " $\sigma^{-1}(\alpha)$ is a prime of $\sigma^{-1}(\mathcal{O}_L) =\mathcal{O}_L $ lying over $p$, and hence it contains $\alpha$. My problem is with the phrase "$\sigma^{-1}(\alpha)$ is a prime of $\sigma^{-1}(\mathcal{O}_L) =\mathcal{O}_L $ lying over $p$". So let me make some comments:
- The fact that $\sigma^{-1}(\mathcal{O}_L) = \mathcal{O}_L$ is OK.
- I assume when he writes $\sigma^{-1}(\alpha)$ what he really means is $\sigma^{-1}(\alpha) \mathcal{O}_L$ (notice how $\sigma^{-1}(\alpha) \in \mathcal{O}_L$)
- As far as showing that $\sigma^{-1}(\alpha) \mathcal{O}_L$ lies over $p$: At first, I thought we should take advantage of the fact that $\mathfrak q$ lies over $p$, so $p \mathcal{O}_L = \mathfrak q I$, for some ideal $I$ of $\mathcal{O}_L$, and "apply" $\sigma^{-1}$. Of course maybe that could work if $\alpha \mathcal{O}_L$ showed up in the right hand side of the last equation (before applying $\sigma$). But $\alpha \in \mathfrak q$ implies $\alpha \mathcal{O}_L \subseteq \mathfrak q$, so in fact $\mathfrak q \mid \alpha \mathcal{O}_L$ (not the other way around)
- I don't really why $\sigma^{-1}(\alpha) \mathcal{O}_L$ is a prime ideal of $\mathcal{O}_L$.
Any suggestions or thoughts would be much appreciated. Thank you in advance.
EDIT: As it turns out, $\sigma^{-1}(\alpha)$ is a typo! It should be $\sigma^{-1}(\mathfrak q)$. Then we just apply $\sigma^{-1}$ to the equation $p \mathcal{O}_L = \mathfrak q I$ and we are done because $\sigma^{-1} (p \mathcal{O}_L)= p \mathcal{O}_L$ and automorphisms carry prime ideals to prime ideals.