# Ramification of a prime ideal in the inertia field

At the momement, I'm trying to understand ramification in the context of Galois extensions and wanted to draw a connection between the different results I've found. However, somehow there must be some mistake/misconception in the following argument but I'm not able to find it.

Let $A$ be a Dedekind domain, $K$ its quotient field, $L/K$ a finite Galois extension and $\mathcal{O}_L$ the integral closure of $A$ in $L$. Moreover let $T_p$ be the inertia field of some non-zero prime ideal $p$ in $\mathcal{O}_L$ and $\mathcal{O}_T$ the integral closure of $T_p$ in $L$ and $p_T$ a non zero prime ideal in $\mathcal{O}_T$ such that $p_T=p \cap \mathcal{O}_T$. We then have that $p$ is the only prime lying above $p_T$.

From this I wanted to conclude: As $p$ is the only prime lying above $p_T$, we have $p_T\mathcal{O}_L=p$ as the decomposition of $p_T$ in $\mathcal{O}_L$, which would imply that $p_T$ is inert in $\mathcal{O}_L$ which again implies that the inertia degree $f(p/p_T)$ is $1$. This gives $1= f(p/p_T)=[\mathcal{K}(p) : \mathcal{K}(p_T)]$.

On the other hand, I have that $[\mathcal{K}(p) : \mathcal{K}(p_T)]= [L:T_p]=|I_p|$ with $I_p$ being the inertia group.

I'm sure that there must be some mistake as $|I_p|$ is for sure not always $1$.

Thank you very much for your help!

When you say "As $p$ is the only prime lying above $p_T$, we have $p_T\mathcal{O}_L=p$", this is not correct.
Actually, we have $p_T\mathcal{O}_L=p^{e(p / p \cap K)}$. In other words, $p_T$ is totally ramified in $\mathcal O_L$.
Then $[K(p) : K(p_T)] = \sum_{i=1}^r e_i f_i = e(P/p) \cdot f(p/p_T) = e(P/p)$.
• Here is what happens in general for abelian extensions: $p \cap K = p \cap O_K$ is totally split in the decomposition field, into primes which remain inert in the inertia field, and all these primes finally are totally ramified in $L$. – Watson Feb 7 '17 at 19:17
• Thank you. I somehow forgot about the ramification degree. Just one more question to the definition of totally ramified: $p_T$ is totally ramified because I know that $f(p/p_T)$ is one and that $e(p/p_T)=e(p/p_0)$ for $p_0$ the prime ideal in K st $p \cap \mathcal{O}_K = p_0$? So the case that $e(p/p_T)$ respectively $e(p/p \cap K)$ is one could not happen? – SallyOwens Feb 8 '17 at 7:32
• @SallyOwens : If $L/K$ is an extension of number fields, then a prime $p$ of $\mathcal O_K$ is totally ramified in $\mathcal O_K$ if $p\mathcal O_L = P^e$ where $P$ is a prime of $\mathcal O_L$ and $e = [L:K]$. Here we have $[L : T_p] = e(p/p_T)$ (totally ramified) and also $e(p/p_T) = e(p/p \cap \mathcal O_K)$ (this is another result). If $e(p/p \cap \mathcal O_K)=1$, then $L=T_p$ and $I_p$ is trivial, which is equivalent to saying that $p \cap K$ is not ramified in $L$. – Watson Feb 8 '17 at 9:19
• @ Watson: Thank you! Just one more thing: In your first sentence, shouldn't it be "$p$ of $\mathcal{O}_K$ is totally ramified in $\mathcal{O}_L$" or am I missing something? – SallyOwens Feb 8 '17 at 11:28
• @SallyOwens : I'm not sure what could be wrong in my first sentence. Are you talking about the second one? Anyway, here $p \subset O_L$, and $p_T \subset O_{T_p}$ and the claim is that $p_T$ is totally ramified in $O_L$. But $p \cap O_K \subset O_K$ doesn't have to be totally ramified in $O_L$ ! – Watson Feb 8 '17 at 12:19