Two definitions of ramification groups, why are they equivalent? Let $L|K$ be a finite galois extension and suppose that $v_k$ is a discrete normalized (non-archimedean) valuation of $K$ with positive residue field characteristic $p$, and that $v_K$ admits a unique extension $w$ to $L.$ Further, let $\mathcal{O}_L$ be the valuation ring of $w.$Denote $v_L = ew$ the associated normalized valuation of $L$ and $\mathcal{B}$ the maximal ideal of $\mathcal{O}_L.$  Then one can define, for any real $i \geq 1$ the $i$th ramification group
$$G_i = \{\sigma \in \operatorname{Gal}(L|K) : v_L(\sigma(x)-x) \geq i+1, \forall x \in \mathcal{O}_L\}.$$
On the other hand, for the field extension $L|K$ one can define the ramification group $R$ as
$$R = \{\sigma \in \operatorname{Gal}(L|K) : \sigma(x)/x-1 \equiv 0 \pmod{\mathcal{B}} , \forall x \in L^*\}.$$
The claim I've seen is then that $G_1= R$ but I can't seem to prove it. This should be very elementary, but I have failed. Any solutions would be more than welcome!
Edit 
I edited a previous error, which gave the wrong definition of $R.$ Instead of being $\equiv 0$ it had $\equiv 1.$
 A: To unlock this, I think three keys will suffice. The first is to observe that both groups are contained in $G_0$, which, from its definition, is the set of all $\sigma\in\mathrm{Gal}^L_K$ that act trivially on the residue field $\kappa=\mathscr O_L/\mathcal B$.
The second key is almost as simple: set up $\varphi\colon G_0\times L^*\to\kappa^*$, by, for $\sigma\in G_0$ and $x\in L^*$, $\varphi_\sigma(x)=\sigma(x)/x\pmod{\mathcal B}$. For fixed $\sigma\in G_0$, this is clearly a homomorphism of $L^*$ into $\kappa^*$, and equally clearly the units of $\mathscr O_L$ are contained in the kernel, by the definition of $G_0$. Thus $\varphi_\sigma$ induces a homomorphism of the cyclic group $L^*/\mathscr O^*$ into $\kappa^*$, which is the trivial map if and only if $\varphi_\sigma(\pi)=1$, where $\pi$ is a prime element of $\mathscr O_L$, i.e. a generator of $\mathcal B$.
So your $R$ is defined to be the set of all $\sigma\in G_0$ such that $\varphi_\sigma(x)=1$ for all $x\in L^*$; according to the above, this is the set of all $\sigma\in G_0$ such that $\varphi_\sigma(\pi)=1$. But this condition is that $v_L(\sigma(\pi)/\pi-1)\ge1$, in other words that $v_L(\sigma(\pi)-\pi)\ge2$.
Now it’s time for the third key: If $F$ is the fixed field of $G_0$, then $F$ is unramified over $K$, and $L$ is totally ramified over $F$, so that in particular every element of $\mathscr O_L$ is congruent modulo $\mathcal B$ to an element of $\mathscr O_F$. Now, for $z\in\mathscr O_L$, let $z=\zeta+\pi\xi$, with $\zeta\in\mathscr O_F$ and
$\xi\in\mathscr O_L$. Then $v_L(\sigma(z)-z)=v_L(\sigma(\pi\xi)-\pi\xi)\ge2$. therefore the condition on $\sigma$ that $v_L(\sigma(\pi)-\pi)\ge2$ is equivalent to the condition that $v_L(\sigma(z)-z)\ge2$ for all $z\in\mathscr O_L$, which is the condition that $\sigma\in G_1$.
