The setup is as follows. Let $\mathfrak{o}$ be a Dedekind domain, $K$ its field of fractions, and $L/K$ a finite separable field extension with $n = [L:K]$. Furthermore, let $\mathfrak{O}$ denote the integral closure of $\mathfrak{o}$ in $L$. Suppose $\mathfrak{p}$ is a non-zero prime ideal in $\mathfrak{o}$. Then, if $k = \mathfrak{o}/\mathfrak{p}$, we have $\dim_k(\mathfrak{O}/\mathfrak{pO}) = n$. This result occurs in the proof of $\sum_{i=1}^r e_i f_i = n$, where $e_i$'s are ramification indices and $f_i$'s are inertial degrees. I have questions about the following proof of $\dim_k (\mathfrak{O}/\mathfrak{pO}) = n$:
Let $\omega_1,\ldots,\omega_m$ be representatives of a basis $\overline{\omega_1},\ldots,\overline{\omega_m}$ of $\mathfrak{O}/p\mathfrak{O}$ over $K$ (we have seen that $\mathfrak{O}$ is a finitely-generated $\mathfrak{o}$-module, so certainly $\dim_k(\mathfrak{O}/\mathfrak{p}\mathfrak{O}) < \infty$.)
- Why is $\dim_k(\mathfrak{O}/\mathfrak{p}\mathfrak{O}) < \infty$?
It is sufficient to show that $\omega_1,\ldots,\omega_m$ is a basis of $L/K$. Assume the $\omega_1,\ldots,\omega_m$ are linearly dependent over $K$, and hence also over $\mathfrak{o}$. Then there are elements $a_1,\ldots,a_m \in \mathfrak{o}$ not all zero such that $$a_1\omega_1 + \ldots + a_m\omega_m = 0$$ Consider the ideal $I = (a_1,\ldots,a_m)$ of $\mathfrak{o}$ and find $a\in I^{-1}$ such that $a\notin I^{-1}p$, hence $aI\not\subset \mathfrak{p}$.
- We can find $a\in I^{-1}$ such that $a\notin I^{-1}\mathfrak{p}$, for if not, $I^{-1} = I^{-1}\mathfrak{p}$ (the other inclusion is obvious) and by the unique factorization of prime ideals (or otherwise) we'd obtain $\mathfrak{p} = \mathfrak{o}$, which is a contradiction since $\mathfrak{p}$ is proper. Why do we have $aI\not\subset \mathfrak{p}$? Note that to define $I^{-1}$, we pick any non-zero $\alpha \in I$ and define $I^{-1} := \{\beta\in o: \beta I\subset (\alpha)\}$.
Then the elements $aa_1,\ldots,aa_m$ lie in $\mathfrak{o}$, but not all belong to $\mathfrak{p}$. The congruence $$aa_1\omega_1 + \ldots + aa_m\omega_m \equiv 0 \bmod \mathfrak{p}$$ thus gives us a linear dependence among the $\overline{\omega_1},\ldots,\overline{\omega_m}$ over $k$, a contradiction. The $\omega_1,\ldots,\omega_m$ are therefore linearly independent over $K$.
- I believe we say that two elements $a,b$ are congruent modulo an ideal $I$ iff $a-b\in I$. Right? I'm trying to understand the congruence above. It seems we pass $\sum_i aa_i\omega_i = 0$ in $\mathfrak{O}$ through the map $\mathfrak{O} \to \mathfrak{O}/p\mathfrak{O}$ to get $\sum_i \overline{aa_i}\overline{\omega_i} = 0$. We have $aa_i \not\in \mathfrak{p}$ for some $i$, but how do we conclude that $\overline{aa_i} \ne 0 \in \mathfrak{O}/\mathfrak{p}\mathfrak{O}$, i.e., $aa_i \not\in \mathfrak{p}\mathfrak{O}$?
I chose to include only the first part of the proof since there are three questions already. The proof goes on to show that $\{\omega_1,\ldots,\omega_m\}$ is a basis for $L/K$, implying $m = n$ as desired. The reference is Neukirch's Algebraic Number Theory. Thank you!