Consequences of definitions in Local fields

Question

I am currently taking a course on Local Fields and have some tiny questions I cannot seem to wrap my head around:

Question 1: If $L/K$ is an extension of complete discretely values fields with normalised valuations $v_L, v_K$ and uniformisers $\pi_L, \pi_K$, we define the ramification index $e:=e_{L/K}=v_L(\pi_K)$. This apparently means that $\pi_K O_L = \pi_{L}^{e} O_L$. I am unsure as to why the equality follows?

Question 2: We define the inverse different as $D^{-1}_{L/K}:=\{ y \in L: \operatorname{Tr}_{L/K}(xy) \in O_K, \forall x \in O_L\}$. This is an $O_L$-submodule, but I am not sure why it contains $O_L$?

Further, we define the different $D_{L/K}$ as the inverse of the inverse of $D_{L/K}^{-1}$. I know that $D^{-1}_{L/K}$ is a fractional ideal, but why do we know that an inverse of $D^{-1}_{L/K}$ exists?

Question 3: We define the $s^{th}$ ramification group as $G_s(L/K):=\{ \sigma \in \operatorname{Gal}(L/K) : v_L(\sigma(x)-x) \ge s+1, \forall x \in O_L\}$. Why does this imply that $G_s=\operatorname{ker}(\operatorname{Gal}(L/K) \rightarrow \operatorname{Aut}(O_L/\pi_L^{s+1}O_L))$? In particular, why is $G_0=I_{L/K}$, where $I_{L/K}$ is the inertia subgroup?

Sorry for the many questions, thank you so much!

Answer 1

In order.

1. Since $v_L(\pi_L)=1$ we get $v_L(\pi_L^{e})=e=v_L(\pi_K)$ which implies that $v_L(\pi_L^{e}/\pi_K)=0$, which means that $\pi_L^{e}/\pi_K$ is a unit in $\mathcal{O}_{L}$.
2. If $y$ is an element of $\mathcal{O}_{L}$ then for any $x$ in $\mathcal{O}_{L}$, the product $xy$ lies in $\mathcal{O}_{L}$, hence it is integral over $\mathcal{O}_{K}$ therefore its trace, which is the opposite of the second coefficent of its minimal polynomial over $\mathcal{O}_{K}$, lies in $\mathcal{O}_{K}$. Furthermore, we know that there is an inverse since any fractional ideal over a Dedekind Domain is invertible and any valuation ring is a Dedekind domain.
3. The condition $v_L(\sigma(x)-x) \ge s+1$ for any x in $\mathcal{O}_L$ simply means that for any $x\in \mathcal{O}_L$, the difference $x-\sigma(x)$ lies in $\pi_L^{s+1}O_L$, which is precisely the definition of the kernel. The inertia subgroup is simply the case $n=0$.