Bourbaki

Exercise 17, § 8

Commutative Algebra, VI, 468

Bourbaki marks this one as harder than the rest.

Let K be a Henselian field with a valuation v which is not algebraically closed, and L a subfield of K such that [K: L] is finite. Show that under these conditions L is Henselian with the restriction of v. (In the contrary case, there would exist two extensions v1,v2v_1, v_2 of v∣Lv|L to an algebraic closure Ω\Omega of K such that the restriction of v1v_1 and v2v_2 to a finite algebraic extension L' of L would not be equivalent. Deduce that there would exist on a suitable finite quasi-Galois extension K' of L containing K, two valuations v1′,v2′v'_1, v'_2 which are not equivalent and for which K' would be Henselian; using Exercise 15 (c), show that, if E=Lp−∞E = L^{p^{-\infty}} (pp being the characteristic exponent of R), E≠ΩE \neq \Omega but Ω\Omega would be a finite extension of E; complete the argument with the aid of Algebra, Chapter VI, § 2, Exercise 31.)

Solution

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