Isomorphic Local Fields from Extension of Valuation Suppose a Galois extension of number fields $L/K$ and nonarchimedian valuations $v$ on $K$ is given. (Edit: I initially wrote any extension of global fields, but I restricted the scope here)
Then it seems that, if $w_1, w_2$ are valuations on $L$ extending $v$, then we have isomorphism of completions: $$L_{w_1} \cong L_{w_2} $$
I'd like to know if the following reasoning I have for this is valid, and whether the conclusion $L_{w_1} \cong L_{w_2}$ is true.
Suppose $L=K(\alpha)$ and let $f \in K[x]$ be the minimal polynomial of $\alpha$. Suppose $f = f_1 \cdots f_r \in K_v[x]$ is the decomposition of $f$ into irreducibles in $K_v$. Then for each $j$, $w_j$ corresponds to $f_j$ by $L_{w_j} \cong K_v[x]/(f_j)$, i.e. $L_{w_j}$ is generated by one root from each irreducible component of $f$. Meanwhile, $L \subset L_{w_j}$ and therefore $L_{w_j}$ contains every root of $f$ too. Therefore $L_{w_j}$ is generated by root(s) of $f$ and therefore they are all isomorphic.
In particular, it seems that the field obtained by adjoining distinct primitive $n$th root of unity over $\mathbb Q_p$ are all isomorphic, even though they may have different minimal polynomials. 
 A: I keep your notations $K,L,v,w$ etc. You must distinguish $2$ cases:
1) Suppose $L/K$ is Galois, with group $G$. Then $G$ permutes transitively all the $w$'s above $v$. For $a\in L$ and $\sigma \in G$, by definition $\sigma w (a)=w(\sigma^{-1}a)$. A Cauchy sequence for $w$ in $L$, acted on by $\sigma$, gives a Cauchy sequence for $\sigma w$, and conversely a Cauchy sequence for $\sigma w$, acted on by $\sigma^{-1}$, gives a Cauchy sequence for $w$; so $\sigma$ induces by continuity an isomorphism $L_w \cong L_{\sigma w}$ , and your "conjecture" holds. Note that the decomposition subgroup $G_w$:={$\sigma \in G; \sigma w =w$} is naturally isomorphic to $Gal(L_w /K_v)$. For more details, see Cassels-Fröhlich, "Algebraic Number Theory", chapter 7, §1.
2) The non Galois case can be completely different because the $w$'s above $v$ can behave independently. In general, $K_v \otimes_K L\cong \oplus L_w$ , the direct sum bearing over all the $w$'s above $v$ (op. cit., chap. 2, thm. of §10). Let us construct an example which contradicts your "conjecture". As in @KCd's comment, take $K=\mathbf Q, L=\mathbf Q(\sqrt [3]2)$ and $v=$the $5$-adic valuation. There are exactly $2$ prime ideals of $L$ above $5$, which are $P_1=(5, \alpha +1)$ and $P_2=(5, \alpha^2+3\alpha -1)$, where $\alpha=\sqrt [3]2$ for short. For details, see D. Marcus, "Number Fields", chap. 3, example p. 70, or exercise 12, p.84). For $P_1$(resp. $P_2$), the inertia index is $1$ (resp. $2$), hence the ramification index is $1$ (resp. $1$) because of the usual formula degree $3=e_1f_1 + e_2 f_2$. This means that $L_{w_1}=\mathbf Q_5$ and $L_{w_2}/\mathbf Q_5 $ is quadratic. Note that the difference with 1) is that in the Galois case all the indices $e_i$ (resp. $f_i$) must be the same. Another explanation lies in the proof of the relation $K_v \otimes_K L\cong \oplus L_w$ above (op. cit., chap. 2, §9, lemma) : the minimal polynomial $X^3 -2$ of $\alpha$ over $\mathbf Q$ has no reason to remain irreducible over $\mathbf Q_5$; actually, notice that $3^3=2$ in $\mathbf F_5$ and apply Hensel's lemma.
