Corollary 2

Let K be a field, v a discrete valuation on K with respect to which K is complete and L a finite extension of K of degree n. Then v admits a unique (upto equivalence) extension $v'$ to L, the ring A' of $v'$ is a finitely generated free module over the ring A of v and $e(v'/v) f(v'/v) = n$.

adic topology (where $m = m(A)$); the ring A is complete, for it is closed in K. We conclude that, since $A'/mA'$ is a finite-dimensional vector ($A/m$)-space (no. 4, Proposition 4), A' is a finitely generated A-module (Chapter III, § 2, no. 9, Corollary 3 to Proposition 12). It is therefore free and $e(v'/v) f(v'/v) = n$ by

(no. 2, Proposition 2). *is bijective. For all $x \in L$, the characteristic polynomial $P_{C_{L/K}}(x; X)$ is equal to the product of the characteristic polynomials $P_{C_{\hat{L}_i/\bar{K}}}(x; X)$ ($1 \leq i \leq s$); in particular,

(The last relation in (9) is meaningful, for we may obviously assume that the $v_i'$, which are of height 1 by Corollary 2 to Proposition 1 of no. 1, take, as does v, their values in a subgroup of $\mathbf{R}$.)

As no two of the $v'_i$ are equivalent and they are of height 1, they are independent and Proposition 2 of no. 2 shows therefore that $e(v'_i/v) f(v'_i/v) \leq n_i$ for all $i$ and $\sum_{i=1}^s n_i \leq n$. The first assertion therefore follows from these inequalities and the relation $\sum_{i=1}^s e(v'_i/v) f(v'_i/v) = n$. Under the isomorphism $\phi$ the endomorphism $z \mapsto z(1 \otimes x)$ of $\hat{K} \otimes_K L$ (for $x \in L$) is transformed into the endomorphism of $\prod_{i=1}^s \hat{L}_i$ leaving invariant each of the factors and reducing on each factor to multiplication by $x$ ($L$ being canonically imbedded in its completion $\hat{L}_i$); whence the assertion relating to the characteristic polynomial of $x$ and the first two formulae of (9). Finally, let E be a finite quasi-Galois extension of $\hat{K}$, containing $\hat{L}_i$; as $\hat{K}$ is complete and $\hat{\nu}$ of height 1, there exists only one valuation (upto equivalence) $w$ on E extending $\hat{\nu}$ (no. 2, Corollary 1 to Proposition 2); then, for every $\hat{K}$-automorphism $\sigma$ of E, $w(\sigma(x)) = v'_i(x)$. Therefore

(Algebra, Chapter VIII, § 12, no. 2, formula (15)), which proves the formula of (9).