Bourbaki

Exercise 20, § 1

Commutative Algebra, V, 359

Bourbaki marks this one as harder than the rest, it is set in small type as supplementary.

Let K0K_0 be a perfect field of characteristic p>0p > 0, KK the field of rational functions K0(Xn)n∈NK_0(X_n)_{n \in \mathbf{N}} and BB the ring of formal power series K[[Z]]K[[Z]], where ZZ is an indeterminate; if mm is the maximal ideal of BB, BB is Hausdorff and complete with the mm-adic topology and is a discrete valuation ring. Let L=K((Z))L = K((Z)) be the field of fractions of BB and EE, the subfield of LL generated by LpL^p and the elements ZZ and XnX_n (for n∈Nn \in \mathbf{N}).

(a) Show that the element c=∑n=0∞XnZnc = \sum_{n=0}^\infty X_n Z^n of LL does not belong to EE.

(b) Let EE be a maximal element of the set of subfields of LL containing EE, and not containing cc; we write A=B∩EA = B \cap E. Show that AA is a discrete valuation ring (and therefore Noetherian) whose field of fractions is EE; if m′m' is the maximal ideal of AA, then m′k=mk∩A{m'}^k = m^k \cap A and AA is dense in BB with the mm-adic topology; L=K(c)L = K(c) is an extension of KK of degree pp and BB is the integral closure of AA in LL but BB is not a finitely generated AA-module (use Chapter 111, § 3, Exercise 9 (b)). \*

Solution

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Transcribed from the printed volume. Not translated.