Lemma 15.123.2. Let $A \subset B$ be an extension of valuation rings with fraction fields $K \subset L$. If the extension $L/K$ is finite, then the residue field extension is finite, the index of $\Gamma _ A$ in $\Gamma _ B$ is finite, and

$[\Gamma _ B : \Gamma _ A] [\kappa _ B : \kappa _ A] \leq [L : K].$

Proof. Let $b_1, \ldots , b_ n \in B$ be units whose images in $\kappa _ B$ are linearly independent over $\kappa _ A$. Let $c_1, \ldots , c_ m \in B$ be nonzero elements whose images in $\Gamma _ B/\Gamma _ A$ are pairwise distinct. We claim that $b_ i c_ j$ are $K$-linearly independent in $L$. Namely, we claim a sum

$\sum a_{ij} b_ i c_ j$

with $a_{ij} \in K$ not all zero cannot be zero. Choose $(i_0, j_0)$ with $v(a_{i_0j_0}b_{i_0}c_{j_0})$ minimal. Replace $a_{ij}$ by $a_{ij}/a_{i_0j_0}$, so that $a_{i_0 j_0} = 1$. Let

$P = \{ (i, j) \mid v(a_{ij}b_ ic_ j) = v(a_{i_0j_0}b_{i_0}c_{j_0}) \}$

By our choice of $c_1, \ldots , c_ m$ we see that $(i, j) \in P$ implies $j = j_0$. Hence if $(i, j) \in P$, then $v(a_{ij}) = v(a_{i_0j_0}) = 0$, i.e., $a_{ij}$ is a unit. By our choice of $b_1, \ldots , b_ n$ we see that

$\sum \nolimits _{(i, j) \in P} a_{ij}b_ i$

is a unit in $B$. Thus the valuation of $\sum \nolimits _{(i, j) \in P} a_{ij}b_ ic_ j$ is $v(c_{j_0}) = v(a_{i_0j_0}b_{i_0}c_{j_0})$. Since the terms with $(i, j) \not\in P$ in the first displayed sum have strictly bigger valuation, we conclude that this sum cannot be zero, thereby proving the lemma. $\square$

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