# Extension of $\mathfrak{p}$-adic valuation

I'm trying to understand the relationship between extensions of valuations and extensions of prime ideals. Let $\mathcal{o}$ a Dedekind domain, $K$ its field of fractions, $L|K$ a finite algebraic separable extension,$\mathfrak{p}\in\mathcal{o}$ a prime ideal and $\mathcal{O}$ the integral closure of $\mathcal{o}$ in $L$. Put on $K$ the $\mathfrak{p}$-adic valuation. Then, if $p\mathcal{O}=\mathfrak{P}_1^{e_1}...\mathfrak{P}_r^{e_r}$ the extensions of the valuation $v_\mathfrak{p}$ to $L$ are $w_i=1/e_iv_\mathfrak{P_i}$. I can't explain why the ramification indeces $e_i$ correspond to the ramification indeces defined for valuation $$e_{w_i}=(w_i(L^*):v_\mathfrak{p}(K^*)).$$The inertia degree is clear. (I'm studying on ANT, Neukirch).

Let $\pi_L\in \mathfrak{P}$ (omitting the subindex) be a uniformizer for $L$, i.e. $v_{\mathfrak{P}}(\pi_L)=1$, or equivalently $w(\pi_L^{e})=1$. Then, if $\pi_K \in \mathfrak{p}$ is a uniformizer for $K$, then $\pi_L^{e}$ and $\pi_K$ as elements in $\mathcal{O}_{\mathfrak{P}}$ differ by a unit in $\mathcal{O}_{\mathfrak{P}}^\times$, i.e. $$\mathfrak{p}\mathcal{O}_{\mathfrak{P}} = \pi_K \mathcal{O}_{\mathfrak{P}}=\pi_L^{e} \mathcal{O}_{\mathfrak{P}}=\mathfrak{P}^{e}\mathcal{O}_{\mathfrak{P}}.$$ Now if you take the factorization of $\mathfrak{p}\mathcal{O}$ into prime ideals and pass to the localization $\mathcal{O}_{\mathfrak{P}}$ you will obtain $\mathfrak{p}\mathcal{O}_{\mathfrak{P}}=\mathfrak{P}^{e'}\mathcal{O}_{\mathfrak{P}}$ (the point is that all the other primes contain a unit of $\mathcal{O}_{\mathfrak{P}}$), where $e'$ denotes the exponent belonging to $\mathfrak{P}$ in the factorization of $\mathfrak{p}\mathcal{O}$. Hence, $e=e'$.

• Thank you, very clear. – Ashley Courtney Oct 23 '17 at 20:29