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 of to an algebraic closure of K such that the restriction of and 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 which are not equivalent and for which K' would be Henselian; using Exercise 15 (c), show that, if ( being the characteristic exponent of R), but 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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