Bourbaki

Remark

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

Commutative Algebra, chapter VI, § 5. The topology defined by a valuation

For all x∈K^x \in \hat{K} not belonging to A^\hat{A}, there exists x0∈Kx_0 \in K such that v^(x−x0)>0\hat{v}(x - x_0) > 0, v^(x)=v^(x0)=v(x0)<0\hat{v}(x) = \hat{v}(x_0) = v(x_0) < 0; then x0−1x∈A^x_0^{-1} x \in \hat{A} and, as x0−1∈Ax_0^{-1} \in A, it is seen that, if we set S=A−{0}S = A - \{0\}, it is possible to write K^=S−1A^\hat{K} = S^{-1} \hat{A}.

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

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Transcribed from the printed volume. Not translated.