Lemma 1

Tag 04UQ, permanent. Cite it as [Bourbaki, Tag 04UQ].

Let $x \in K^*, y \in K^*$ and $\alpha \in G$ . If

v(x - y) > \sup(\alpha + 2v(y), v(y)),

then

v(x^{-1} - y^{-1}) > a

$x^{-1} - y^{-1} = x^{-1}(y - x)y^{-1}$ and hence

v(x^{-1} - y^{-1}) = v(x - y) - v(x) - v(y).

v(x - y) > v(y)

, Proposition 1 of § 3, no. 1 implies that

v(x) = v(y)

, since

x = y + (x - y)

. Moreover, if

v(x - y) > a + 2v(y)

, then

v(x^{-1} - y^{-1}) > a + 2v(y) - 2v(y) = a.

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

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