Bourbaki

Exercise 1, § 8

Commutative Algebra, VI, 461

Let KK be a field and A=K[[X,Y,Z]]A = K[[X, Y, Z]] the ring of formal power series in three indeterminates over KK; let v′v' (resp. v′′v'') be the valuation on AA with values in the group Z×Z\mathbf{Z} \times \mathbf{Z} ordered lexicographically, such that v′(X)=(1,0)v'(X) = (1, 0), v′(Y)=(0,1)v'(Y) = (0, 1), v′v' is improper in K[[Z]]K[[Z]] (resp. v′′(X)=(1,0)v''(X) = (1, 0), v′′v'' is improper in K[[Y]]K[[Y]], v′′(Z)=(0,1)v''(Z) = (0, 1)). Let σ\sigma be the automorphism of AA leaving invariant KK and XX and such that σ(Y)=Z,σ(Z)=Y\sigma(Y) = Z, \sigma(Z) = Y, if BB is the subring of AA consisting of the elements invariant under σ\sigma and EE (resp. FF) the field of fractions of AA (resp. BB), the valuations v′v' and v′′v'' (extended canonically to EE) have the same restriction vv to FF, FF is complete with the topology defined by vv and EE is a quadratic extension of FF; the two valuations v′,v′′v', v'' on EE are dependent but not equivalent.

Solution

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Transcribed from the printed volume. Not translated.