### Henselian implies large

Florian Pop
2010 Annals of Mathematics
In this note we show that the quotient field of a domain which is Henselian with respect to a non-trivial ideal is a large field, and give some applications of this fact, using a specialization theorem for ramified covers of the line over (generalized) Krull fields. technical conditions). The methods of proof in both cases are ingenious and quite technical. These results also seemed to give further new evidence for the fact that the Problems above can be solved in general, as it was generally
more » ... it was generally believed that the above fields K = k((x, y)), and more general K = Quot(R) with R complete Noetherian local and Krull.dim(R) > 1, were not large fields. Note that these fields are definitely not Henselian valued fields! The first point of this short note is to prove that actually K = k((x, y)), and more generally, K = Quot(R) with R a complete Noetherian ring, are large fields, and that the class of large fields is much richer than previously believed. In particular, one can deduce Paran [Pa] from the already known fact that Problem 0 has a positive answer over large base fields K. Second, I give a lower bound for the number of distinct solutions of a non-trivial finite split embedding problem over a Hilbertian large field, a result which represents a wide extension of Harbater-Stevenson [H-S]. Finally, using these results, one can generalize Harbater's result [Ha2], Theorem 4.6, see Theorem 1.3 below, thus giving new evidence for (a stronger form of) Bogomolov's Freeness Conjecture as presented in Positselski [Ps]. In order to announce the results of this note, let first recall that a commutative ring R with identity is said to be Henselian with respect to an ideal a, or that R, a is a Henselian pair, if denoting R := R/a and R[X] → R[X], f (X) → f (X) the reduction map (mod a), for every polynomial f (X) ∈ R[X] the following holds: If a ∈ R is a root of f (X) such that f (a) ∈ R × , then there exists a lifting a ∈ R of a such that f (a) = 0 and f (a) ∈ R × .