Bourbaki

Exercise 3, § 8

Commutative Algebra, VI, 461

Let kk be the field of rational functions Fp(Xn)n∈N\mathbf{F}_p(X_n)_{n \in \mathbf{N}} in an infinity of indeterminates over the prime field F\mathbf{F}, and let K=k(U,V)K = k(U, V) be the field of rational functions in two indeterminates over kk.

(a) Show that the element P(U)=∑n=0∞XnpUnpP(U) = \sum_{n=0}^\infty X_n^p U^{np} of the field of formal power series k((U))k((U)) is not algebraic over the field k(U)k(U) of rational functions. The mapping F(U,V)↦F(U,P(U))F(U, V) \mapsto F(U, P(U)) of k[U,V]k[U, V] to k((U))k((U)) can be extended to an isomorphism of KK onto a subfield of k((U))k((U)); the restriction to this subfield of the valuation on k((U))k((U)) equal to the order of the formal power series (§ 3, no. 3, Example 3 ) is a discrete valuation vv on KK, whose residue field is kk.

(b) Let K′K' be the algebraic extension K(V1/p)K(V^{1/p}) of KK so that [K′:K]=p[K': K] = p; if v′v' is the unique valuation on K′K' extending vv, show that e(v′/v)=f(v′/v)=1e(v'/v) = f(v'/v) = 1. The ring of the valuation v′v' is therefore not a finitely generated module over the ring of the valuation vv.

Solution

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